General Topology Notes

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GENERAL TOPOLOGY

Contents

1. Relations 31.3. Equivalence relations 31.8. Linear Orders 42. The integers and the real numbers 53. Products and coproducts 64. Finite and infinite sets 75. Countable and uncountable sets 86. Infinite sets and the Axiom of Choice (AC) 11

7. Well-ordered sets 128. The Maximum Principle 149. Topological spaces 1610. Basis for a Topology 1611. Order topologies 1812. The product topology 1812.6. The coproduct topology 1913. The subspace topology 1914. Closed sets and limit points 2214.3. Closure and interior 2214.10. Limit points and isolated points 2314.14. Convergence, the Hausdorff property, and the T 1-axiom 2415. Continuous functions 2415.6. Homeomorphisms and embeddings 2515.10. Maps into product spaces 2615.15. Maps out of coproducts 2716. Quotient maps 2816.1. Open and closed maps 2816.2. Quotient maps 2817. Metric topologies 3417.6. The first countability axiom 3417.12. The uniform metric 3618. Connected spaces 3619. Connected subsets of linearly ordered spaces 3919.5. Path connected spaces 40

19.9. Components and path components 4019.12. Locally connected and locally path connected spaces 4120. Compact spaces 4321. Compact subspaces of linearly ordered spaces 4521.5. Compactness in metric spaces 4722. Limit point compactness and sequential compactness 4823. Locally compact spaces and the Alexandroff compactification 4824. Countability axioms 5225. Separation Axioms 5326. Normal spaces 5527. Second countable regular spaces and the Urysohn metrization theorem 5727.1. An embedding theorem 5727.5. A universal second countable regular space 57

Date : 24th March 2004.

1



2 GENERAL TOPOLOGY

28. Completely regular spaces and the Stone–Cech compactification 5928.6. The Stone–Cech construction 5929. Manifolds 61References 63



GENERAL TOPOLOGY 3

1. Relations

Definition 1.1. A relation R on the set A is a subset R ⊂ A × A.

We shall be interested in these properties of relations

Re: (Reflexivity) aRa for all a ∈ ASy: (Symmetry) aRb ⇒ bRa for all a, b ∈ ATr: (Transitivity) aRb and bRc ⇒ aRc for all a, b, c ∈ ACom: (Comparability) For all a, b ∈ A, we have either a = b, aRb, or bRaNonRe: (Non-reflexivity) aRa for no a ∈ A.

where aRb means (a, b) ∈ R ⊂ A × B.

Example 1.2. We may define a relation D on Z+ by aDb if a divides b. The relation D ⊂ Z+×Z+

is reflexive and transitive. It has none of the other properties.

1.3. Equivalence relations.

Definition 1.4. An equivalence relation on a set A is a relation that is reflexive, symmetric, and transitive.

This means that

Re: a ∼ a for all a ∈ ASy: a ∼ b ⇒ b ∼ a for all a, b ∈ ATr: a ∼ b ∼ c ⇒ a ∼ c for all a, b, c ∈ A

when ∼ is an equivalence relation.The equivalence class containing a ∈ A is the subset

[a] = b ∈ A | a ∼ bof all elements of A that are equivalent to a. The collection of equivalence classes is denoted A/ ∼.The canonical map [ ]: A → A/ ∼ takes the element a ∈ A to the equivalence class [a] ⊂ A.

Example 1.5. (1) Equality is an equivalence relation. The equivalence class [a] =

a

contains

just one element.(2) a mod b mod n is an equivalence relation on Z. The equivalence class [a] = a + nZ consists of all integers congruent to a mod n and the set of equivalence classes is Z/nZ = [0], [1], . . . , [n−1].

(3) x ∼ y def ⇔ |x| = |y| is an equivalence relation in the plane R2. The equivalence class [x] is a

circle centered at the origin and R2/∼ is the collection of all circles centered at the origin.

(4) If f : A → B is any function, a1 ∼ a2def ⇔ f (a1) = f (a2) is an equivalence relation on A. The

equivalence class [a] = f −1(f (a)) ⊂ A is the fibre over f (a) ∈ B. we write A/f for the set of equivalence classes.(5) [Ex 3.2] (Restriction) Let X be a set and A ⊂ X a subset. Declare any two elements of A tobe equivalent and any element outside A to be equivalent only to itself. This is an equivalencerelation. The equivalence classes are A and x for x ∈ X − A. One writes X/A for the set of equivalence classes.

(6) [Ex 3.5] (Equivalence relation generated by a relation) The intersection of any family of equi-valence relations is an equivalence relation. The intersection of all equivalence relations containinga given relation R is called the equivalence relation generated by R.

Lemma 1.6. Let ∼ be an equivalence relation on a set A. Then

(1) a ∈ [a](2) [a] = [b] ⇔ a ∼ b(3) If [a] ∩ [b] = ∅ then [a] = [b]

Proof. (1) is reflexivity, (2) is symmetry, (3) is transitivity: If c ∈ [a] ∩ [b], then a ∼ c ∼ b so a ∼ band [a] = [b] by (2).

This lemma implies that the collection A/∼ is a partition of A, a collection of nonempty, disjointsubsets of A whose union is all of A. Conversely, given any partition of A we define an equivalence

relation by declaring a and b to be equivalent if they lie in the same subset of the partition. Weconclude that an equivalence relation is essentially the same thing as a partition.



4 GENERAL TOPOLOGY

The canonical map [ ]: A → A/ ∼ has the following universal property: Any map f : A → Bthat respects the equivalence relation in the sense that a1 ∼ a2 ⇒ f (a1) = f (a2) factors uniquelythrough A/ ∼. This means that there is a unique map f such that the diagram

Af

[ ]

B

A/∼∃!f

commutes.

Example 1.7. Any map f : A → B is the composite A [ ]

A/f f B . The corestriction

f : A/f → f (A) of f is a bijection between the set of fibres A/f and the image f (A).

1.8. Linear Orders.

Definition 1.9. A linear order on the set A is a relation that satisfies comparability, non-

reflexivity, and transitivity.This means that

Com: If a = b then a > b or b > a for all a, b ∈ ANonRe: a > a for no a ∈ ATr: a > b > c ⇒ a > c for all a, b, c ∈ A

when > is a linear order.

Example 1.10. The usual x > y is a linear order on R.

The open interval from a to b > a is the set

(a, b) = x ∈ A|a < x < b

If (a, b) = ∅ then a is the immediate predecessor of b, and b the immediate successor of a.Definition 1.11. Let (A, <) and (B, <) be linearly ordered sets. An order preserving map is a map f : A → B such that a1 < a2 ⇒ f (a1) < f (a2) for all a1, a2 ∈ A.

An order preserving map is always injective. If there exists an order preserving bijection f : A → B , then we say that (A, <) and (B, <) have the same order type .

Definition 1.12. Let (A, <) and (B, <) be linearly ordered sets. The dictionary order on A × Bis the linear order given by

(a1, b1) < (a2, b2) def ⇔ ( a1 < a2) or ( a1 = a2 and b1 < b2)

What about orders on A ∪ B?Any subset of a linearly ordered set inherits a linear order. The restriction of a dictionary order

to a product subspace is the dictionary order of the restricted linear orders.

Example 1.13. (1) R and (0, 1) have the same order type. [0, 1) amd (0, 1) have distinct ordertypes. −1 ∪ (0, 1) and [0, 1) have the same order type.(2) The dictionary order on Z+ × [0, 1) has the same order type as [1, ∞). In the dictionary orderon [0, 1) × Z+ each element (a, n) has (a, n + 1) as its immediate successor. Thus Z+ × [0, 1) and[0, 1) × Z+ do not have the same order type.

The example shows that, in general, (A, <) × (B, <) = (B, <) × (A, <) as ordered sets.

Definition 1.14. An ordered set (A, <) has the least upper bound property if any subset that is bounded from above has a least upper bound. It is well-ordered if any nonempty subset has a smallest element.

Example 1.15. R has the least upper bound property. Z+ is well-ordered. The order type of thewell-ordered set (Z+, <) is denoted ω. Any well-orderes set has the least upper bound property.



GENERAL TOPOLOGY 5

2. The integers and the real numbers

Theorem 2.1. (Induction Principle) Let A ⊂ Z+. If 1 ∈ A and n ∈ A ⇒ n + 1 ∈ A for every n ∈ Z+, then A = Z+.

Proof. The hypothesis is that A is inductive. By definition, Z+ is the smallest inductive subset of R, so Z+ ⊂ A.

Theorem 2.2. (Well-ordering) Z+ is well-ordered.

Proof. The claim is that any non-empty subset of Z+ has a smallest element.As a first step we show by induction that any nonempty subset of S n has a smallest element.

We claim that the subset

(2.3) n ∈ Z+ | S n is well-orderedin inductive and therefore equal to Z+. Clearly, 1 is in (2.3). Suppose that the integer n belongsto (2.3). We must show that n + 1 is in (2.3). Let therefore B be any nonempty subset of S n+1. If B∩S n+1 = ∅ then the smallest element of B ∩S n+1 is also the smallest element of B . If B ∩S n = ∅,then B =

n

and n is the smallest element of B . This shows that B has a smallest element, i.e.

that n + 1 is in (2.3).Let now A ⊂ Z+ be any nonempty subset. The intersection A ∩ S n is nonempty for some n, so

it has a smallest element. This is also the mallets element of A.

Theorem 2.4. (General Induction Principle) Let A be a subset of Z+. If the implication

S n ⊂ A ⇒ n ∈ A

is true for all n ∈ Z+, then A = Z+.

Proof. Suppose that A is not all of Z+. Let n = min(Z+ − A) be the smallest element outsideA (well-ordering), i.e. x ∈ A ⇒ x ≥ n or, equivalently, S n ⊂ A. Then n ∈ A by hypothesis.Contradiction.

Theorem 2.5. (Archimedean Principle) Z+ has no upper bound in R.

Proof. We assume the opposite and derive a contradiction. Suppose that Z+ is bounded fromabove. Let b = sup Z+ be the smallest upper bound (well-ordering). Since b − 1 is not an upperbound, there is a positive integer n ∈ Z+ such that n > b − 1. Then n + 1 is also an integer (Z+

is inductive) and n + 1 > b contradicting that b is an upper bound for Z+.

The Archimedean Principle says that for any real number x there is a positive integer n greaterthan x.



6 GENERAL TOPOLOGY

3. Products and coproducts

Definition 3.1. An indexed family of sets consists of a collection A of sets, an index set J , and a surjective function f : J → A.

We often denote the set f ( j) by Aj and the whole indexed family by Ajj∈J . Any collectionA can be made into an indexed family by using the identity map A → A as the indexing function.

We define the union and the intersection of the indexed family asj∈j

Aj = a | a ∈ Aj for at least one j ∈ J ,j∈j

Aj = a | a ∈ Aj for all j ∈ J

and we define the Cartesian product to be

(3.2)j∈J

Aj = x : J → Aj | ∀ j ∈ J : x( j) ∈ Aj

Observe that the set of maps into a product

map(B, j∈J

Aj) = j∈J

map(B, Aj)

is a product. In particular there is a map ∆:

j∈J Aj →

j∈J Aj corresponding to the inclusion

maps

j∈J Aj → Aj, j ∈ J . The product comes equipped with projection maps πj :

Aj → Aj , j ∈ J , inducing the above correspondence.

If the index set J = S n+1 = 1, . . . , n then we also write

A1 ∪ · · · ∪ An, A1 ∩ · · · ∩ An, A1 × · · · × An

for

j∈S n+1Aj ,

j∈S n+1

Aj ,

j∈S n+1Aj , respectively. If also and Aj = A for all j ∈ S n+1 we

write An for the product

j∈S n+1A. The elements of An are all n-tuples (a1, . . . , an) of elements

from A.If the index set J = Z+ then we also write

A1 ∪ · · · ∪ An ∪ · · · , A1 ∩ · · · ∩ An ∩ · · · , A1 × · · · × An × · · ·for

j∈Z+Aj ,

j∈Z+

Aj ,

j∈Z+Aj , respectively. If also Aj = A for all j we write Aω for the

product

j∈Z+A, the set of all functions x : Z+ → A, i.e. all sequences (x1, . . . , xn, . . .) of elements

from A. (The reason for this notation is that the order type of ( Z+, <) is often denoted by ω .)

Example 3.3. (1) S 1 ∪ S 2 ∪ · · · ∪ S n · · · = n∈Z+

S n = Z+.

(2) If the collection A = A consists of just one set A then the product

j∈J A = x : J → Ais the set of all functions from J to A.

(3) There is a bijection (which one?) between 0, 1ω = x : Z+ → 0, 1 and P (Z+). Moregenerally, there is a bijection (which one) between the product

j∈J 0, 1 and the power set

P (J ).

We define the copoduct to be the set

(3.4)j∈J

Aj = ( j, a) ∈ J ×

Aj | a ∈ Aj

The coproduct is sometimes also called the disjoint union because

Aj =

Aj is the union of

the disjoint sets Aj = j× Aj ⊂ J ×

j∈J Aj . Observe that the set of maps out of the coproduct

map(j∈J

Aj , B) =j∈J

map(Aj, B)

is a product. In particular there is a map ∇ :

j∈J Aj →

j∈J Aj corresponding to the inclusion

maps Aj

→ j∈J Aj . The coproduct comes equipped with inclusion maps ιj : Aj

→ Aj, j

∈ J ,

inducing the above correspondence.If the index set J = S n+1 = 1, . . . , n then we also write A1 · · · An for the coproduct.



GENERAL TOPOLOGY 7

4. Finite and infinite sets

Definition 4.1. A set A is finite it admits a bijection f : A → S n to some section S n, n ∈ Z+, of Z+. A set is infinite if it is not finite.

Lemma 4.2. Let A be a set, a0 an element of A, and n ∈ Z+ a positive integer. Then There is a bijection A → S n+1 ⇔ There is a bijection A − a0 → S n

Proof. ⇐: Let f 0 : A − a0 → S n be a bijection. Define f : A → S n+1 by

f (a) =

f 0(a) a = a0

n + 1 a = a0

⇒: Let f : A → S n+1 be a bijection. If we are so lucky that f (a0) = n + 1, we just take therestriction of f to A − a0. If we are not lucky, let g = (f (a0), n + 1): S n+1 → S n+1 be thepermutation that interchanges f (a0) and n + 1 and leaves all other elements fixed. The bijectiong f between A and S n+1 takes a0 to n + 1, so we are back in the lucky situation.

Could it happen that there A admits bijections S m+1 A S n+1 to S m+1 and S n+1

where m = n?

Corollary 4.3. Let m ∈ Z+ and n ∈ Z+ be two positive integers. If m = n, then there is nobijection between S m and S n.

Proof. We have just seen that

There is a bijection S m+1 → S n+1 ⇒ There is a bijection S m → S n

or, equivalently,

There is no bijection S m → S n ⇒ There is no bijection S m+1 → S n+1

This is precisely (the nontrivial part of) the statement that

n ∈ Z+ | There is no bijection S m → S n for m = nis inductive.

This means that a finite set is in bijection with S n+1 for precisely one positive integer n ∈ Z+;this integer is the cardinality of A, denoted card A = n.

Corollary 4.4. Let A be a finite set. Any subset B of A is also finite and

B ⊂ A ⇒ card B ≤ card A

B A ⇒ card B < card A

Proof. It suffices to consider subsets of A = S n+1. Using Lemma 4.2 we see that the sets

n ∈ Z+ | Any B ⊂ S n+1 is in bijection with S m+1 for some m ≤ n

n ∈ Z+ | Any B S n+1 is in bijection with S m+1 for some m < nare inductive. Let’s for instance look at the first set. Assume that n belongs to this set. We wantto show that n + 1 belongs to the set. So, let B be a subset of S n+2. If B does not contain n + 1,then B is a subset of S n+1 and then we know from the induction hypothesis that B is in bijectionwith S m+1 for some m ≤ n. If B does contain n + 1, then B − n + 1, a subset of S n+1, is inbijection with S m+1 for some m ≤ n. Lemma 4.2 says that B is in bijection with S m+2 wherem + 1 ≤ n + 1.

In particular, a finite set A can not be in bijection with any of its proper subsets B A (fortheir cardinalities are distinct). We can use this property of finite sets to find an example of aninfinite set.

Corollary 4.5. Z+ is infinite.

Proof. The map x → x + 1, with inverse function x → x − 1, is a bijection between Z+ and itsproper subset Z+ − 1.



8 GENERAL TOPOLOGY

Theorem 4.6 (Characterization of finite sets). Let A be a set. The following statements are equivalent

(1) A is finite (2) There exists a surjection S n+1

→ A for some n

∈ Z+

(3) There exists an injection A → S n+1 for some n ∈ Z+

Proof. (1) ⇒ (2): There even exists a bijection S n+1 → A.(2) ⇒ (3): Let f : S n+1 → A be surjective. If n = 0, A = ∅ and there is nothing to prove. When

n > 0, we construct a right inverse S n+1

f A

g to f . The fibres f −1(a) are nonempty for all

a ∈ A so they have smallest elements since Z+ is well-ordered. Define g : A → S n+1 by

g(a) = min f −1(a)

As a right inverse to f , g is injective.(3) ⇒ (1): If there exists an injection A → S n+1, then there exists a bijection between A and asubset of S n+1. All subsets of the finite set S n+1 are finite (4.4).

Remark 4.7. You may get the idea that any surjective map admits a right inverse: A B

.Can you prove it? Consider for instance the surjective map

(b, B) ∈ A × P (A) | b ∈ B proj−−→ P (A): (b, B) → B

where P (A) = P (A) −∅ is the collection of nonempty subsets of A. The dual statement is that

any injection admits a left inverse: A B

. Can you prove that?

Corollary 4.8. (1) Subsets of finite sets are finite.(2) Images of finite sets are finite.(3) Finite unions of finite sets are finite.(4) Finite Cartesian products of finite sets are finite.

Proof. (2) S n+1 A f (A).(3) We show first that A

∪B is finite if A and B are finite. Suppose f : S n+1

→ A and g : S m+1

→ B

are surjective functions. Then h : S m+n+1 → A ∪ B given by

h(i) =

f (i) i = 1, 2, . . . , n

g(i) i = n + 1, . . . , m + n

is also surjective. Induction now shows that A1 ∪ · · · ∪ An is finite when A1, . . . , An are finitelymany finite sets.(4) Since

A × B =a∈A

a × B

is finite union of finite sets, it is finite. Induction now shows that A1 × ·· · × An is finite whenA1, . . . , An are finitely many finite sets.

5. Countable and uncountable sets

Definition 5.1. A set A is countably infinite if it admits a bijection f : A → Z+. It is countableif it is finite or countably infinite. It is uncountable if it is not countable.

Example 5.2. Z and Z+ × Z+ are countably infinite.

Lemma 5.3. Any infinite subset of Z+ is countably infinite.

Let C ⊂ Z+ be an infinite subset. The claim is that the elements of C can be listed C =c1, c2, . . . as an infinite sequence. The obvious thing to do is let c1 be the smallest element of C ,c2 the next element of C , and so on. However, to do this correctly we must use thePrinciple of recursive definition Given a set A, an element a1 of A, and a formula that for eachi > 1 defines h(i) in terms of h(1), . . . , h(i − 1). Then this formula defines a function h : Z+ → Asuch that h(1) = a1.

This follows from the Induction Principle, but we shall not go into details. Recursive functionscan be computed by a Turing machine.



GENERAL TOPOLOGY 9

Proof of Lemma 5.3. Let C ⊂ Z+ be an infinite set of positive integers. We define a functionh : Z+ → C recursively by

h(1) = min C

h(i) = min(C − h(1), . . . , h(i − 1)), i > 1using that Z+ is well-ordered. Note that C − h(1), . . . , h(i − 1) is nonempty since C is infinite(4.6). We claim that h is bijective.h is injective: The function h is injective (even order preserving) because the descending chain

C C − h(1) · · · C −h(1), . . . , h(i − 1) C − h(1), . . . , h(i − 1), h(i) · · ·implies that h(1) < h(2) · · · < h(i) < h(i + 1) < · · · .h is surjective: Let c ∈ C . We must find a positive integer m such that c = h(m). Our only hopeis

m = minn ∈ Z+ | h(n) ≥ c(Note that this has a meaning since the set n ∈ Z+ | h(n) ≥ c is nonempty as we can notinject the infinite set Z+ into the finite set 1, . . . , c − 1 = S c (4.4). Note also that again we use

well-ordering of Z+.) By definition of m,h(m) ≥ c and h(n) ≥ c ⇒ n ≥ m

The last of these two properties is equivalent to n < m ⇒ h(n) < c, so c ∈ h(1), . . . , h(m − 1),or c ∈ C −h(1), . . . , h(m − 1), and therefore

h(m) = min(C − h(1), . . . , h(m − 1)) ≤ c

by definition of h. Thus h(m) = c.

Theorem 5.4 (Characterization of countable sets). Let A be a set. The following statements are equivalent

(1) A is countable (2) There exists a surjection Z+ → A

(3) There exists an injection A → Z

+

Proof. If A is finite, the theorem is true, so we assume that A is countably infinite.(1) ⇒ (2): Clear.(2) ⇒ (3): If f : Z+ → A is s surjection, then g : A → Z+ given by g(a) = min f −1(a) is a left-inverse to f , hence injective.(3) ⇒ (1): Let g : A → Z+ be injective. The infinite subset g(A) of Z+ is countable by Lemma 5.3and it is in bijection with A since g is injective.

Corollary 5.5. (1) A subset of a countable set is countable (2) The image of a countable set is countable.(3) A countable union of countable sets is countable.(4) A finite product of countable sets is countable.

Proof. (1) B

A

Z+.(2) Z+ A B.(3) Let Ajj∈J be an indexed collection of sets where J is countable and each set Aj is countable.Choose surjective maps f : Z+ → J and gj : Z+ → Aj . Then the map

Z+ × Z+ =

m∈Z+

m × Z+

j∈J

Aj : (m, n) → gf (m)(n)

is surjective. Since Z+ × Z+ is countable (5.6) so is its image (5.5).(4) The product of two countable sets is countable since the product of two surjective maps issurjective. Now use induction.

Example 5.6. The map f : Z+ × Z+ → Z+ given by f (m, n) = 2m3n is injective by uniquenessof prime factorizations. The map g : Z+ × Z+ → Q+ given by g(m, n) = m

n , is surjective. Thus

Z+ × Z+ and Q+ are countable. Q = Q− ∪ 0 ∪ Q+ is countable.You may think that a countable product of countable sets is countable – but that’s false.



10 GENERAL TOPOLOGY

Lemma 5.7. The countable product 0, 1ω =

n∈Z+0, 1 = x : Z+ → 0, 1 of finite sets is

uncountable.

Proof. (Cantor’s diagonal argument) Let g : Z+ → 0, 1ω be any function. We show that g can

not be surjective. Define an element (yn)n∈Z+ ∈ 0, 1ω

by

yn =

0 g(n)n = 1

1 g(n)n = 0

Then (yn) = g(m) for all m ∈ Z+ for ym = g(m)m.

The power set P (Z+) is in bijection with 0, 1ω (3.3.(3)). Thus the countable set Z+ has anuncountable power set P (Z+). It is true in general that card A < card P (A) for any set A.

Theorem 5.8. Let A be any set.

(1) There is no injective map P (A) → A(2) There is no surjective map A → P (A)

Proof. If there exists an injection

P (A) A , then there will also exist a surjection (4.7) in

the other direction: P (A) A

. Thus it suffices to show (2). We assume the converse: That

g : A → P (A) is surjective. The subset

a ∈ A | a ∈ g(a)is of the form g(a0) for some a0 ∈ A. This leads to the contradiction

a0 ∈ g(a0) ⇔ a0 ∈ g(a0)

Conjecture 5.9 (Cantor’s Continuum Hypothesis). Suppose that X ⊂ R is an uncountable set.Then there exists a bijection π : X → R.

Look up the (generalized) continuum hypothesis [Ex 11.8] somewhere [11, 12].



GENERAL TOPOLOGY 11

6. Infinite sets and the Axiom of Choice (AC)

Recall that a set A is infinite if it does not admit a bijection A → S n with some section of thepositive integers.

Theorem 6.1 (Characterization of infinite sets). Let A be a set. The following are equivalent:(1) A is infinite (2) There exists an injective map Z+ → A(3) A is in bijection with a proper subset of itself

The naive student will prove (1) ⇒ (2) like this: Suppose A is infinite. Let a1 be any elementof A. Define h : Z+ → A recursively as follows:

h(1) = a1

h(i) = any element of A − h(1), . . . , h(i − 1)Then h is injective.

The problem here is that there is no formula for h(i) in terms of h( j) for j < i. In fact, for the

proof of this innocent looking theorem we need theAxiom 6.2 (Axiom of Choice (AC)). Given any nonempty collection of nonempty disjoint sets,there exists a set C containing exactly one element from each of the sets in the collection.

Unlike the other axioms os set theory, the AC does not determine C uniquely. In 1938 K. Godelshowed that if the axioms of set theory are consistent (we don’t know if they are), then set theorywith AC is also consistent. The reason could be that AC is a consequence of the other axioms.However, P.J. Cohen showed in 1963 that this is not the case.

Look at (4.7) again.Here is a reformulation of AC.

Theorem 6.3. The following statements are equivalent:

(1) The Axiom of Choice

(2) For any collection A of nonempty sets (not necessarily disjoint) there exists a map (a choice function) c : A →

A∈A A such that c(A) ∈ A for all A ∈ A.(3) The Cartesian product

A∈A A of any nonempty collection A of nonempty sets (not ne-

cessarily disjoint) is nonempty.

Proof. It is clear that (2) and (3) are equivalent by the very definition of Cartesian products.(1) =⇒ (2): Let (A)A∈A be the indexed family of sets

A = (A, a) | a ∈ A ∈ A ⊂ A×

A∈A

A

Then (A)A∈A is a collection of disjoint sets. (Let A1 be two distinct sets in the collection A. Thesets A

1 and A2 are disjoint for the elements, (A1, a), a ∈ A1, of A

1 have first coordinate A1 whilethe elements, (A2, a), a

∈ A2, of A

2 have first coordinate A2

= A1.) By the Axion of Choice there

is set C such that C ∩ A contains a single element for each A. If we write C ∩ A = (A, c(A))then c is a function A →

A∈A A.(2) =⇒ (1): Let c : A →

A∈A A be a choice function for A, a nonempty collection of nonemptydisjoint sets. The set we seek is the image C = c(A) of c: Since c(A) ∈ A and the sets in A aredisjoint, c(A) does not belong to any other of the sets in the collection, so C ∩ A = c(A) for allA ∈ A.

Here is a special, but often used, case. Let A be any set and P (A) = P (A) −∅ the collectionof nonempty subsets of A. Then there exists (6.3.(2)) a choice function c : P (A) → A such thatc(B) ∈ B for any nonempty B ⊂ A. (The choice function picks out an element in each nonemptysubset of A.)

Proof of Theorem 6.1. (1) =⇒ (2): Let P (A) be the collection of nonempty subsets of A andc : P (A) → A a choice function (c(B) ∈ B for each nonempty subset B of A.) Let h(1) be any



12 GENERAL TOPOLOGY

element of A and h : Z+ → A the recursively defined function given by

h(i) = c(A − h(1), . . . , h(i − 1))

Then h is injective (if i < j then h( j)

∈ A

− h(1), . . . , h(i), . . . , h( j

−1)

so h(i)

= h( j)).

(2) =⇒ (3): We view Z+ as a subset of A. Then A = (A−Z+)∪ Z+ is in bijection with the propersubset A − 1 = (A − Z+) ∪ (Z+ − 1).(3) =⇒ (1): We already know that a finite set can not be in bijection with any of its proper subsetsfor cardinality reasons.

7. Well-ordered sets

A very convenient property of (Z+, <) is that any nonempty subset has a smallest element. Wenow focus on this property.

Definition 7.1. A set A with a linear order < is well-ordered if any nonempty subset has a smallest element.

Any well-ordered set has a smallest element. Any element (but the largest) in a well-ordered sethas an immediate successor, the smallest successor [Ex 10.2]. A well-ordered set can not containan infinite descending chain x1 > x2 > · · · , in fact, a linearly ordered set is well-ordered if andonly if it does not contain a copy of the negative integers Z− [Ex 10.4].

Definition 7.2. Let (A, <) be a well-ordered set and α an element of A. The subset

S α(A) = S α = [min A, α) = a ∈ A | a < αis called the section of A by α.

Theorem 7.3 (Transfinite Induction). [Ex 10.7] Let (A, <) be a well-ordered set and J ⊂ A an inductive subset, that is

∀α

∈ A : S α

⊂ J =

⇒ α

∈ J

Then J = A.

Proof. By contradiction. Assume that the inductive set J does not completely fill up A, i.e. thatthe complement A − J is nonempty. The smallest element of A − J belongs to (A − J ) ∩ J : Let αbe the smallest element not in J . Then α ∈ A − J by definition of α, and α ∈ J since S α ⊂ J andJ is inductive. This is absurd.

Let A A × Bπ1 π2 B be the projection maps.

Proposition 7.4. (1) A subset of a well-ordered set is well-ordered.(2) A finite product of well-ordered sets is well-ordered.(3) The union of a well-ordered family of disjoint well-ordered sets is well-ordered [Ex 10.8]

Proof. (1) Clear.(2) Let (A, <) and (B, <) be well-ordered and C a nonempty subset of A ×B. Let a0 = min π1(C )

be the smallest first coordinate of any point in C . Let b0 = min π2(a0×B)∩C ) be the smallestsecond coordinate above a0. Then (a0, b0) = min C is the smallest element in C . This shows thatthe product of two well-ordered sets is

(3) See [Ex 10.8].

Example 7.5. (1) The positive integers (Z+, <) is a well-ordered set.(2) Z and R are not well-ordered sets.(3) Any section S n+1 = 1, 2, . . . , n of Z+ is well-ordered.(4) S n+1 = 1, 2, . . . , n of Z+ is well-ordered. The product S n+1 × Z+ is well-ordered. The finite

products Zn+ = Z+

×Z+

· · · ×Z+ are well-ordered.

(5) The infinite product 0, 1ω is not well-ordered for it contains the infinite descending chain(1, 0, 0, 0, . . .) > (0, 1, 0, 0, . . .) > (0, 0, 1, 0, . . .) > · · · .



GENERAL TOPOLOGY 13

(6) The element ω = 2 × 1 ∈ Z+ × Z+ is the smallest element of 1, 2× Z+ such that the sectionby this element is countably infinite. The well-ordered set S ω = [1 × 1, ω] has ω as its largestelement, the section S ω by ω is countably infinite but every other section of S ω is finite. Anyfinite subset of S ω has an upper bound because the set of not upper bounds is finite (a finite

union of finite sets) while S ω itself is countably infinite.Which of these well-ordereds have the same order type [Ex 10.3]? Draw pictures of examples of

well-ordered sets.

Any finite ordered set is well-ordered. To see this we first prove

Lemma 7.6. Any finite linearly ordered set has a maximal element.

Proof. The proof is by induction over the cardinality of the set. Any ordered set (A, <) of car-dinality 1 has a largest element. (What is it?) Suppose that any finite ordered linearly set of cardinality n has a largest element. Let (A, <) be a finite ordered set of cardinality n + 1. Let abe any element of A. Then maxa, max(A − a) is the largest element of A. (This is a kind of sorting algorithm.)

Theorem 7.7 (Finite order types). [Ex 6.4] Any finite linearly ordered set of cardinality n has the order type of (S n+1, <); in particular, it is well-ordered.

Proof. Again we use induction over the cardinality of the ordered set (A, <). Suppose that weknow that all finite ordered sets of cardinality n satisfy the theorem and let (A, <) be a finiteordered set of cardinality n + 1. Let a be the largest element of A (7.6). By induction hypothesis,A − a has the order type of S n. Then A has the order type of S n+1.

Can you find an explicit order preserving bijection S m+1 × S n+1 → S m+n+1?So there is just one order type of a given finite cardinality n. There are many countably infinite

well-ordered sets (7.5). Is there an uncountable well-ordered set? Our examples, R and 0, 1ω, of uncountable sets are not well-ordered (7.5.(2), 7.5.(5)).

Theorem 7.8 (Well-ordering theorem). (Zermelo 1904) Any set can be well-ordered.

We focus on the minimal criminal, the minimal uncountable well-ordered set .

Lemma 7.9. There exists a well-ordered set S Ω with the following properties:

(1) S Ω has a largest element Ω(2) The section S Ω by Ω is uncountable but every other section is countable.(3) Any countable subset of S Ω has an upper bound (cf 7.5.( 6)).

Proof. (Cf 7.5.(6)) Take any uncountable well-ordered set A. Append a greatest element to A. Callthe result A again. Now A has at least one uncountable section. Let Ω be the smallest element of Asuch that the section by this element is uncountable, that is Ω = minα ∈ A | S α is uncountable.Put S Ω = [a0

×1, Ω] where a0 is the smallest element of A. This well-ordered set satisfies (1) and

(2). Let C be a countable subset of S Ω. We want to show that it has an upper bound. We considerthe set of elements of S Ω that are not upper bounds, i.e

x ∈ S Ω | x is not an upper bound for C = x ∈ S Ω | ∃c ∈ C : x < c =

countable c∈C

S c

uncountable S Ω

This set of not upper bounds is countable for it is a countable union of countable sets. But S Ω isuncountable, so the set of not upper bounds is a proper subset.

See [SupplExI : 8] for an explicit construction of a well-ordered uncountable set. Z+ is thewell-ordered set of all finite (nonzero) order types and S Ω is the well-ordered set of all countable(nonzero) order types. See [Ex 10.6] for further properties of S Ω.

Recall that the ordered set Z+×[0, 1) is a linear continuum of the same order type as [1, ∞) ⊂ R.What happens if we replace Z+ by S Ω [Ex 24.6, 24.12]?



14 GENERAL TOPOLOGY

8. The Maximum Principle

The Axiom of Choice has several reformulations [3, Thm B.18].If we do not insist on comparability in our order relation we obtain:

Definition 8.1. A strict partial order on a set A is a relation ≺ (precedes) on A which is non-reflexive and transitive:

(1) a ≺ a holds for no a ∈ A(2) a ≺ b and b ≺ c implies a ≺ c.

We do not require that any two nonidentical element can be compared. For instance, power setsare strictly partially ordered by proper inclusion .

Theorem 8.2 (Hausdorff’s Maximum Principle). Let A be a set with a strict partial order. Then there exists a maximal linearly ordered subset B of A.

Proof. As a special case, suppose that A is infinitely countable. Then there is some bijectionZ+ → A meaning that we can write A = ai | i ∈ Z+. Define h : Z+ → 0, 1 by recursion as

follows:h(1) = 0

h(i) =

0 The subset aj | j < i,h( j) = 0 ∪ ai is linearly ordered

1 otherwise

Then the set B = ai | h(i) = 0 is linearly ordered and maximal.The general case is formally the same as above. Let I be some set in bijection with A so that

A = ai | i ∈ I . Let < be a well-ordering of I (7.8). Let i0 denote the smallest element of I .Define h : I → 0, 1 by transfinite recursion as follows

h(i0) = 0

h(i) = 0 The subset

aj

| j < i,h( j) = 0

∪ ai

is linearly ordered

1 otherwise

Then the set B = ai | h(i) = 0 is linearly ordered and maximal.

Definition 8.3. Let (A, ≺) be a set with a strict partial order. An element m of A is a maximalif no elements of A are greater than m.

Let B be a subset of A. If b c for all b ∈ B, then c is an upper bound on B.

Maximal elements are often of great interest.

Theorem 8.4 (Zorn’s lemma). A partially strictly ordered set in which every linearly ordered subset has an upper bound contains maximal elements.

Proof. Any upper bound on a maximal linearly ordered subset is maximal: Let B be a maximal

linearly ordered subset. By hypothesis, B has an upper bound m, i.e. b m for all m ∈ B.Suppose that m ≺ d for some d ∈ A. Then b m ≺ d for all elements of B so B ∪ d is linearlyordered, contradicting maximality of B .

We shall later use Zorn’s lemma to prove Tychonoff’s theorem (20.17) that a product of compactspaces is compact. In fact, The Axiom of Choice, Zermelo’s Well-ordering theorem, Hausdorff’smaximum principle, Zorn’s lemma, and Tychonoff’s theorem are equivalent.

Here are two typical applications.A basis for a vector space over a field is a maximal independent subset.

Theorem 8.5. [Ex 11.8] Any vector space has a basis, in fact any linearly independent subset of a vector space is contained in a basis.

Proof. Let M

⊂ V be a linearly independent subset. Apply Zorn’s lemma to the set, strictly

partially ordered by proper inclusion , of independent subsets containing M . A maximal elementis a basis.



GENERAL TOPOLOGY 15

As a corollary we see that R and R2 are isomorphic as vector spaces over Q.A maximal ideal in a ring is a maximal proper ideal.

Theorem 8.6. Any nontrivial ring has a maximal ideal, in fact any proper ideal is contained in

a maximal ideal.Proof. Let I be a proper ideal. Consider the set, strictly partially ordered by proper inclusion, of proper ideals containing the given ideal I .

Other authors prefer to work with partial orders instead of strict partial orders.

Definition 8.7. [Ex 11.2] Let A be a set. A relation on A is said to be a partial order precisely when it is symmetric (that is a a for all a in A), transitive (that is a b and b c implies a c), and anti-symmetric (that is a b and b a implies a = b). anti-symmetric.



16 GENERAL TOPOLOGY

9. Topological spaces

What does it mean that a map f : X → Y between two sets is continuous? To answer thisquestion, and much more, we equip sets with topologies.

Definition 9.1. Let X be a set. A topology on X is a collection T of subsets of X , called opensets, such that

(1) ∅ and X are open (2) The intersection of finitely many (two) open sets is open (3) The union of an arbitrary collection of open sets is open

A topological space is a set X together with a topology T on X .

Example 9.2. (1) The indiscrete topology , or trivial topology , T = ∅, X , has only two opensets.

(2) The discrete topology T = P (X ). All subsets of X are open.(3) The Sierpinski space is the set X = 0, 1 with the topology T = ∅, 0, X . The open sets

are ∅ and the subsets containing the point 0.(4) The standard topology on the real line R is

T =

unions of open intervals

.

(5) More generally, suppose that (X, d) is a metric space. The open r-ball centered at x ∈ X isthe set Bd(x, r) = y ∈ X | d(x, y) < r of points within distance r > 0 from x. The metric topology on X is the collection T d = unions of open balls. The open sets of the topologicalspace (X, T d) and the open sets in the metric space (X, d) are the same. (See §§20–21 for moreon metric topologies.)

(6) The finite complement topology , or cofinite topology , T = U ⊂ X | X − U is finite ∪ ∅.The open sets are ∅ and the cofinite subsets.

The Sierpinski topology and the finite complement topology on an infinite set are not metrictopologies.

Topologies are partially ordered by inclusion. For instance, the finite complement topology(9.2.(6)) on R is contained in the standard topology (9.2.(4), and the indiscrete topology (9.2.(1))

on 0, 1 is contained in the Sierpinski topology (9.2.(3)) is contained in the discrete topology(9.2.(2)).

Definition 9.3 (Comparison of topologies). Let T and T be two topologies on the same set X .

T is finer than T T is coarser than T

def ⇔ T ⊃ T

The finest topology is the topology with the most opens sets, the coarsest topology is the onewith fewest open sets. (Think of sandpaper!) The discrete topology is finer and the indiscretetopology coarser than any other topology: P (X ) ⊃ T ⊃ ∅, X . Of course, two topologies mayalso be incomparable .

10. Basis for a Topology

Any intersection of topologies is a topology [Ex 13.4]. Thus we may (and shall shortly) speakof the coarsest topology containing some given collection of subsets of X . In praxis, we restrictourselves to collections of subsets that cover X .

Definition 10.1. A subbasis is a collection S of subsets of X , called subbasis sets, that cover X ,that is X =

S is the union of the subbasis sets. The topology

T S = unions of finite intersections of subbasis sets is called the topology generated by the subbasis S .

Verify that T S is indeed a topology and that it is the coarsest topology in which the subbasissets are open [Ex 13.5]. (Use the distributive laws [§1] for ∪ and ∩.)

This becomes particularly nice when the subbasis is a basis (as in 9.2.(4) and 9.2.(3)).

Definition 10.2. A collection B of subsets of X , called basis sets, is a basis if (1) X is a union of basis sets.



GENERAL TOPOLOGY 17

(2) The intersection of two basis sets is a union of basis sets.

The topology T B = unions of basis sets

is called the topology generated by the basis B

.

T B is the coarsest topology in which the basis sets are open. When we have a basis rather thena subbasis we do not need first to take finite intersections as these are already contained in thebasis.

A topology is a basis is a subbasis. Conversely, if S is a subbasis, then

(10.3) BS = Finite intersections of subbasis setsis a basis for the topology T S generated by the subbasis S .Lemma 10.4 (Comparison). Let B and B be two bases and S a subbasis.

(1)T B ⊂ T B ⇔ B ⊂ T B ⇔ All B-sets are open in T B

(2)

T B = T B ⇔ All B-sets are open in T B

All B-sets are open in T B

(3)

T B = T S ⇔

All basis sets are open in T S Finite intersections of subbasis sets are open in T B

Proof. (1) is obvious since T B is the coarsest topology containing B. Item (2) is immediate from(1). Item (3) is immediate from (2) with B = BS , the basis generated by the subbasis S .

Two topologies, bases or subbases are said to be equivalent if they generate the same topology.Given a topology T , letting B = T in 10.4.(2) gives a criterion that B is a basis for T (that B andT are equivalent bases); letting B = T in 10.4.(3) gives a criterion that S is a subbasis for T (thatS and T are equivalent subbases).

Example 10.5. (1) In a metric space, the set B = B(x, r) | x ∈ X, r > 0 of open balls is (bydefinition) a basis for the metric topology T d. The collection of open balls of radius 1

n , n ∈ Z+,is an equivalent topology basis for T d.

(2) The collection of rectangular regions (a1, b1) × (a2, b2) in the plane R2 is a topology basisequivalent to the standard basis of open balls B (a, r) = x ∈ R2 | |x − a| < r.

(3) Let f : X → Y be any map. If T is a topology on Y with basis B or subbasis S , then thepull-back f −1(T ) is a topology, the initial topology for f , on X with basis f −1(B) and subbasisf −1(S ).

(4) More generally, let f j : X → Y j , j ∈ J , be a set of maps form a set X into spaces Y j . Let T jbe a topology on Y j with basis Bj or subbasis S j , j ∈ J . Then

f −1j (T j),

f −1j (Bj),

f −1j (S j)

are equivalent subbases on X . The topology they generate is called the initial topology for themaps f j , j ∈ J .

Example 10.6. [Topologies on R] We consider three topologies on R:R: The standard topology with basis the open intervals (a, b).R: The lower limit or right half-open interval topology with basis the right half-open inter-

vals [a, b).RK : The K -topology with basis (a, b) ∪ (a, b) − K where K = 1, 1

2 , 13 , 1

4 , . . ..

The lower limit topology is strictly finer than the standard topology because (10.4) (a, b) =a<α[α, b), and [0, 1) is open in R but not in R (an open interval containing 0 is not a subset

of [0, 1).) The K -topology is strictly finer than the standard topology because its basis containsthe standard basis and R − K is open in RK but not in R (an open interval containing 0 is not asubset of R − K ). The topologies R and RK are not comparable [Ex 13.6].

Example 10.7. The collection of of open intervals is a basis (10.6) and the collection of open rays

S = (−∞, b) ∪ (a, +∞)is a subbasis for the standard topology on R by 10.4.(3).



18 GENERAL TOPOLOGY

11. Order topologies

We associate a topological space to any linearly ordered set. You may view the topological spaceas a means to study the ordered set, to find invariants, or you may view this construction as a

provider of interesting examples of topological spaces.Let (X, <) be a linearly ordered set containing at least two points. The open rays in X are thesubsets

(−∞, b) = x ∈ X | x < b, (a, +∞) = x ∈ X | a < xof X . The collection of all open rays is clearly a subbasis (just as in 10.7).

Definition 11.1. The order topology T < on the linearly ordered set X is the topology with all open rays as subbasis. A linearly ordered space is a linearly ordered set with the order topology.

The open intervals in X are the subsets of the form

(a, b) = (−∞, b) ∩ (a, +∞) = x ∈ X | a < x < b, a, b ∈ X, a < b.

Lemma 11.2. The collection of all open rays together with all open intervals is a basis for the

order topology T <.Proof. We get a basis from a subbasis by taking all finite intersections of subbasis sets (10.3).

If X has a smallest element a0 then the open ray (−∞, b) = [a0, b). If X has no smallest element,the the open ray (−∞, b) =

a<b(a, b) is a union of open intervals and we do not need this open

ray in the basis. Similar remarks apply to the greatest element when it exists. In particular, if X has neither a smallest nor a greatest element then the collection of open intervals is a basis.

Lemma 11.3. If X has no smallest and no largest element, then the set (a, b) of open intervals is a basis for the order topology.

Example 11.4. (1) The order topology on the ordered set (R, <) is the standard topology (10.6).(2) The order topology on the ordered set R2 has as basis the collection of all open intervals

(a1 × a2, b1 × b2). An equivalent basis (10.4) consists of the open intervals (a × b1, a × b2).(3) The order topology on Z+ is the discrete topology because (n − 1, n + 1) = n is open.(4) The order topology on Z+ × Z+ is not discrete. Any open set that contains the element 2 × 1

also contains elements from 1 × Z+. Thus the set 1 × 2 is not open.(5) The order topology on Z × Z is discrete.(6) Is the order topology on S Ω discrete?(7) I 2 = [0, 1]2 with the order topology is denoted I 2o and called the ordered square . The open sets

containing the point x ×y ∈ I 2o look quite different depending on whether y ∈ 0, 1 or 0 < y < 1.

12. The product topology

Let (X j)j∈J be an indexed family of topological spaces. Let πk :

j∈J X j → X k be the projec-tion map. An open cylinder is a subset of the product space of the form

π−1j (U j), U j ⊂ X j open, j ∈ J.

The collection of all open cylinders is clearly a subbasis. The set π−1k (U k) consists of the points

(xj) ∈X j with kth coordinate in U k.

Definition 12.1. The product topology on

j∈J X j is the topology which has the collection of all open cylinders as a subbasis.

The product topology is the coarsest topology making all the projection maps πj :

X j → X j , j ∈ J , continuous.

Lemma 12.2. The collection of all products

j∈J V j where V j ⊂ X j is open in X j and V j = X j for all but finitely many j ∈ J is a basis for the product topology.

Proof. We get a basis from a subbasis by taking all finite intersections of subbasis sets (10.3).

This becomes particularly simple when we consider finite products.



GENERAL TOPOLOGY 19

Lemma 12.3. Let X = X 1 × X 2 × · · · X k be a finite Cartesian product. The collection

B = U 1 × U 2 × · · · × U k | U 1 open in X 1, U 2 open in X 2, . . . , U k open in X kof all products of open sets is a basis for the product topology.

Corollary 12.4. Suppose that Bj is a basis for the topology on X j, j = 1, . . . , k. Then B1×·· ·×Bk

is a basis for the product topology on X 1 × · · · × X k.

Proof. Note that B1 × · · · × Bk is indeed a basis. Compare it to the basis of 12.3 using 10.4.

When (X, <) and (Y, <) are linearly ordered sets, we now have two topologies on the Cartesianproduct X × Y : The product topology of the order topologies, (X, T <) × (Y, T <), and the ordertopology of the product dictionary order, (X × Y, T <). These two topologies are in general not identical or even comparable. It is difficult to imagine any general relation between them since X <×Y < is essentially symmetric in X and Y whereas the dictionary order has no such symmetry. Buteven when X = Y there does not seem to be a general pattern: The order topology (Z+ × Z+, T <)is coarser than the product topology (Z+, T <)× (Z+, T <) (which is discrete) (11.4.(4)–(5)). On theother hand, the order topology (R × R, T <) is finer than the product topology (R, T <) × (R, T <)

which is the standard topology, see 11.4.(1)–(2) and [Ex 16.9].Corollary 12.5 (Cf [Ex 16.9]). Let (X, <) and (Y, <) be two linearly ordered sets. Suppose that Y does not have a largest or a smallest element. Then the order topology (X × Y )< = X d × Y <where X d is X with the discrete topology. This topology is finer than X < × Y <.

Proof. The equations

(−∞, a × b) =x<a

x × Y ∪ a × (−∞, b), (a × b, +∞) = a × (b, +∞) ∪x>a

x × Y

show that the order topology is coarser than the product of the discrete and the order topology. Onthe other hand, the sets x×(c, d) is a basis for = X d×Y < (11.3, 12.4) and x×(c, d) = (x×c, x×d)is open in the order topology since it is an open interval.

The corollary shows that the order topology on X ×

Y does not yield anything new in case thesecond factor is unbounded in both directions. As an example where this is not the case we havealready seen Z+ × Z+ and I 2o (11.4.(4), 11.4.(7)).

12.6. The coproduct topology. The coproduct topology on the coproduct

X j (3.4) is thefinest topology making all the inclusion maps ιj : X j →

X j , j ∈ J , continuous.

This means that

U ⊂j∈J

X j is open ⇐⇒ ι−1j (U ) is open for all j ∈ J

Alternatively, the open sets of the coproduct

X j are the coproducts

U j of open set U j ⊂ X j .

13. The subspace topology

Let (X,T

) be a topological space and Y ⊂

X a subset. We make Y into a topological space.

Definition 13.1. The subspace topology on Y is the topology T Y = Y ∩ U | U ∈ T . A subset V ⊂ Y is open relative to Y if it open in the subspace topology.

It is immediate that T Y is indeed a topology. The subset V ⊂ Y is open relative to Y if andonly if V = Y ∩ U for some open set U ⊂ X .

Lemma 13.2. If B is a basis for T , then Y ∩ B = Y ∩ U | U ∈ B is a basis for the subspace topology T Y . If S is a subbasis for T , then Y ∩ S = Y ∩ U | U ∈ S is a subbasis for the subspace topology T Y .

Proof. This is 10.5.(3) applied to the inclusion map A → X .

If V ⊂ Y is open, then V is also relatively open. The converse holds if Y is open.

Lemma 13.3. Assume that A ⊂ Y ⊂ X . Then (1) A is open in Y ⇔ A = Y ∩ U for some open set U in X



20 GENERAL TOPOLOGY

(2) If Y is open then: A is open in Y ⇔ A is open in X

Proof. As the first point is clear by definition of the subspace topology, we only consider the secondpoint.

⇐: True for any Y .⇒: We have A = Y ∩ U for some open set U ⊂ X . Since both Y and U are open, A is open.

The lemma says that an open subset of an open subset is open.Subspace topology and product topology commute.

Theorem 13.4. Let Y j ⊂ X j, j ∈ J . The subspace topology that

Y j inherits from

X j is the product topology of the subspace topologies on Y j.

Proof. The collection of setsY j

∩ π−1

k (U k), U k ⊂ X k open, (πk :

X j → X k)

is a subbasis for the subspace topology (13.2, 12.1). The collection of sets

π−1k (Y k

∩U k), U k

⊂ X k open, (πk : Y j

→ Y k)

is a subbasis for the product topology on the product

Y j of subspaces (12.1, 13.1). These two

collections are the same because π−1k (Y k ∩ U k)

[Ex 2.2]= π−1

k (Y k) ∩ π−1k (U k) = (

Y j) ∩ π−1

k (U k).

When (X, <) is a linearly ordered set and Y ⊂ X a subset we now have two topologies on Y .We can view Y as a subspace of the topological space X with the order topology or we can viewY as a sub-ordered set of X and give Y the order topology. In general, these two topologies arenot the same (13.6.(2)).

Lemma 13.5 (Y < ⊂ Y ∩X <). Let (X, <) be a linearly ordered set and Y ⊂ X a subset. The order topology (Y, T <) is coarser than the subspace topology Y ∩ T <. These two topologies are identical when Y is convex, but in general, they are distinct.

Proof. When a and B are points in Y , the equalities

(−∞, b)Y = Y ∩ (−∞, b)X , (a, +∞)Y = Y ∩ (a, +∞)X

show that the subbasis for the order topology (11.1) on Y is contained in subbasis for the subspacetopology on Y (13.2). These two topologies are not the same when X = R (with order topology)and Y = [0, 1) ∪ 2 ⊂ R. (Any basis set (11.2) for the order topology on Y containing the point2 also contains elements of [0, 1) so 2 is not open.)

If Y is convex and a a point outside Y , then a is either a lower or an upper bound for Y as wecan not have y1 < a < y2 for two points y1, y2 of Y . Therefore

Y ∩ (a, +∞)X =

(a, +∞)Y a ∈ Y

Y a ∈ Y, a a lower bound for Y

∅ a ∈ Y, a an upper bound for Y

This and the similar statement for Y ∩ (−∞, b)X show that the subbasis for the subset topologyon Y is contained in the subbasis for the order topology on Y .

Example 13.6. (1) S 2×1 = 1 × Z+ ∪ 2 × 1 = [1 × 1, 2 × 1] is a convex subset of Z+ × Z+. Thesubspace topology (11.4.(4)) is the same as the order topology.

(2) The subset Z+ × Z+ is not convex in Z × Z so we expect the subspace topology to be strictlyfiner than the order topology. Indeed, the subspace topology that Z+ × Z+ inherits from thediscrete space Z × Z is discrete but the order topology is not discrete (11.4.(4)–(5)).

(3) Consider X = R2 and Y = [0, 1]2 with order topologies. Y is not convex so we expect thesubspace topology to be strictly finer than the order topology. Indeed, the subspace topology on[0, 1]2, which is [0, 1]d × [0, 1], is strictly finer than I 2o (11.4.(7)): The set

[0, 1

2] × [0, 1] = ([0, 1] × [0, 1]) ∩ (−∞,

1

2 × 2)

is open in the subspace topology on Y but it is not open in the order topology on Y as any basisopen set (11.2) containing 1

2 × 1 also contains points with first coordinate < 1

2.



GENERAL TOPOLOGY 21

(4) Consider X = R2 and Y = (0, 1)2 with order topologies. Y is not convex so we expect thesubspace topology to be strictly finer than the order topology. But it isn’t! The reason is that(0, 1) does not have a greatest nor a smallest element (12.5).

(5) The subset Q of the linearly ordered set R is not convex but nevertheless the subspace topology

inherited from R is the order topology. Again, the reason seems to be that Q does not have agreatest nor a smallest element.



22 GENERAL TOPOLOGY

14. Closed sets and limit points

Let (X, T ) be a topological space and A a subset of X .

Definition 14.1. We say that the set A is closed if its complement X

−A is open.

Note that X and ∅ are closed (and open), a finite union of closed sets is closed, an arbitraryintersection of closed sets is closed. The finer the topology, the more closed sets.

Example 14.2. (1) [a, b] is closed in R, [a, b) is neither closed nor open, R is both closed andopen.

(2) Let X = [0, 1] ∪ (2, 3). The subsets [0, 1] and (2, 3) of X are both closed and open (they areclopen ) as subsets of X . The interval [0, 1] is closed in R while (2, 3) is open in R.

(3) K = 1, 12

, 13

, . . . is not closed in R but it is closed in RK .(4) Recall that R is the set of real numbers equipped with the topology generated by the basis

sets [a, b) of right half-open intervals. All sets that are (open) closed in the standard topologyR are also (open) closed in the finer topology R. Sets of the form (−∞, a), [a, b), and [a, ∞) =

a<x[a, x) are both open and closed. Sets of the form (−∞, b] or [a, b] are closed and not open

since they are not unions of basis sets. Sets of the form (a, ∞) are open and not closed. Sets of the form (a, b] are neither open nor closed.(5) Let X be a well-ordered set. In the order topology, sets of the form (a, ∞) = [a+, ∞),

(−∞, b] = (∞, b+) and (a, b] = (a, ∞) ∩ (−∞, b] are closed and open. (Here, b+ denotes b if b isthe largest element and the immediate successor of b if b is not the largest element.)

14.3. Closure and interior. We consider the largest open set contained in A and the smallestclosed set containing A.

Definition 14.4. The interior of A is the union of all open sets contained in A:

Int A =

U ⊂ A | U open = A

The closure of A is the intersection of all closed sets containing A:

Cl A = C ⊃ A | C closed = ANote or prove [6, Ex 17.6, 17.7, 17.8, 17.9] [3, p 9] [5, 7.1.7–9] that

• A ⊂ A ⊂ A• A is open, it is the largest open set contained in A, and A is open iff A = A• A is closed, it is the smallest closed set containing A, and A is closed iff A = A• (A ∩ B) = A ∩ B, A ∪ B = A ∪ B, X − A = X − A, (X − A) = X − A• For any A ⊂ X and B ⊂ Y , we have (A × B) = A × B and A × B = A × B in the

Cartesian product X × Y (15.12) [4, p 88].

• A ⊂ B ⇒

Int A ⊂ Int BCl A ⊂ Cl B

Definition 14.5. A neighborhood of the point x ∈ X is an open set containing x.

Proposition 14.6. Let A ⊂ X and let x be a point in X . Then

x ∈ A ⇔ There exists a neighborhood of x that is contained in A

x ∈ A ⇔ All neighborhoods of x intersect A nontrivially

and X − A = (X − A), X − A = X − A.

Proof. The first assertion is clear as A is the union of open sets contained in A. We now compute

X − A = X −

C ⊃ A | C closed =

X − C ⊂ X − A | X − C open=

U ⊂ X − A | U open = (X − A)

and, similarly, X − A = X − A. The second assertion is just a reformulation,

x ∈ A ⇔ x ∈ (X − A)

⇔ x has a neighborhood that does not intersect A,of the above formula.



GENERAL TOPOLOGY 23

Proposition 14.7 (Closure with respect to subspace). Let A ⊂ Y ⊂ X . Then

(1) A is closed in Y ⇐⇒ A = Y ∩ C for some closed set C in X (2) If Y is closed then: A is closed in Y ⇐⇒ A is closed in X (3) ClY (A) = Y

∩A.

Proof. (1) We have A = Y ∩ C ⇔ Y − A = Y ∩ (X − C ) for any subset C of X (make a drawing).The claim therefore follows from (is equivalent to) the corresponding claim for relatively open sets.(2) is now obvious.(3) The set ClY (A) is the intersection of all relatively closed sets containing A. The relativelyclosed sets are the sets of the form Y ∩ C where C is closed in X . This means that

ClY (A) =

Y ∩ C | C ⊃ A, C closed = Y

C | C ⊃ A, C closed = Y ∩ A

by a direct computation.

Definition 14.8. A subset A ⊂ X is said to be dense if A = X , or, equivalently, if every open subset of X contains a point of A.

Proposition 14.9. Let A be a dense an U an open subset of X . Then A∩

U = U .

Proof. The inclusion A ∩ U ⊂ U is general. For the other inclusion, consider a point x ∈ U . LetV be any neighborhood of x. Then V ∩ (A ∩ U ) = (V ∩ U ) ∩ A is not empty since V ∩ U is aneighborhood of x and A is dense. But this says that x is in the closure of A ∩ U .

14.10. Limit points and isolated points. Let X be a topological space and A a subset of X .

Definition 14.11. A point x ∈ X is a limit point of A if every neighborhood of x contains a point from A − x. The set of limit points 1 of A is denoted A. A point a ∈ A is an isolated point if ahas a neighborhood that intersects A only in a.

Equivalently, x ∈ X is a limit point of A iff x ∈ A − x, and a ∈ A is an isolated point of A iff a is open in A. These two concepts are almost each others opposite:

x is not a limit point of A ⇐⇒ x has a neighborhood U such that U ∩ A ⊂ x⇐⇒ x has a neighborhood U such that U ∩ (A ∪ x) = x⇐⇒ x is an isolated point of A ∪ x

Proposition 14.12. Let A be a subset of X and A the set of limit points of A. Then A ∪ A = Aand A ∩ A = A − a ∈ A | a is isolated so that

A ⊃ A ⇐⇒ A is closed

A ⊂ A ⇐⇒ A has no isolated points

A ∩ A = ∅ ⇐⇒ A is discrete

A = ∅ ⇐⇒ A is closed and discrete

If B ⊂

A then B

⊂ A.

Proof. Since A − A ⊂ A ⊂ A we get that A ∪ A = A. The intersection A ∩ A is precisely the setof nonisolated points of A. If A ∩ A = ∅ then all all points of A are isolated so that the subspaceA has the discrete topology. We have A = ∅ ⇐⇒ A ⊃ A, A ∩ A = ∅.

Example 14.13. (1) Q = R = R − Q (R−Q contains (Q−0)√

2). Cl Int Q = ∅, Int Cl Q = R.(2) Let C r ⊂ R2 be the circle with center (0, r) and radius r > 0. Then

n∈Z+

C 1/n =

n∈Z+

C 1/n,

n∈Z+

C n = (R × 0) ∪

n∈Z+

C n

The first set of decreasing circles (known as the Hawaiian Earring [6, Exmp 1 p 436]) is closedbecause it is also the intersection of a collection (which collection?) of closed sets.

(3) The set of limit points of K =

1n

| n

∈ Z+

is

0

in R and R and

∅ in RK .

1The set of limit points of A is sometimes denotes Ad and called the derived set of A



24 GENERAL TOPOLOGY

14.14. Convergence, the Hausdorff property, and the T 1-axiom. Let xn, n ∈ Z+, be asequence of points in X , that is a map Z+ → X .

Definition 14.15. The sequence xn converges to the point x ∈ X if for any neighborhood U of x

there exists some N such that n > N ⇒ xn ∈ U .If a sequence converges in some topology on X it also converges in any coarser topology but not

necessarily in a finer topology.

Example 14.16. In the Sierpinski space X = 0, 1, the sequence 0, 0, · · · converges to 0 and to1. In R with the cofinite topology, the sequence 1, 2, 3, · · · converges to any point. In RK , thesequence 1

n does not converge; in R and R it converges to 0 (and to no other point). The sequence

− 1n converges to 0 in R and RK , but does not converge in R. In Z+ × Z+, the sequence 1 × n

converges to 2 × 1 (and to no other point).

Definition 14.17. A topological space X is Hausdorff (or T 2) if there are enough open sets toseparate points: For any two distinct points x1 = x2 in X there exist disjoint open sets, U 1 and U 2, such that x1 ∈ U 1 and x2 ∈ U 2.

All linearly ordered spaces are Hausdorff [Ex 17.10]. R and RK are Hausdorff because thetopologies are finer than the standard Hausdorff topology R.

Theorem 14.18. A sequence in a Hausdorff space can not converge to two distinct points.

A property of a topological space is said to be (weakly) hereditary if any (closed) subspace of a space with the property also has the property. Hausdorffness is hereditary and also passes toproduct spaces.

Theorem 14.19. [Ex 17.11, 17.12] Any subset of a Hausdorff space is Hausdorff. Any product of Hausdorff spaces is Hausdorff.

Definition 14.20. A topological space is T 1-space if points are closed.

Equivalently, X is T 1 iff all finite subsets are closed. Cofinite topologies are T 1 by construction.

All Hausdorff spaces are T 1. Not all T 1-spaces are Hausdorff as R with the cofinite topology shows.The Sierpinski space is not T 1.

Theorem 14.21. Suppose that X is T 1. Let A be a subset of and x a point in X . Then

x is a limit point ⇔ All neighborhoods of x intersect A in infinitely many points

Proof. ⇐: If all neighborhoods of x intersect A in infinitely many points, then, clearly, they alsointersect A in a point that is not x.⇒: Assume that x is a limit point. Let U be a neighborhood of x. Then U contains a pointof A different from x. Remove this point from U . Since points are closed, the result is a newneighborhood of x. This new neighborhood of x contains a point of A different from x. In thisway we recursively find a whole sequence of distinct points in U ∩ A.

15. Continuous functionsLet f : X → Y be a map between two topological spaces.

Definition 15.1. The map f : X → Y is continuous if all open subsets of Y have open preimages in X : V open in X ⇒ f −1(V ) open in Y

If T X is the topology on X and T Y is the topology on Y then

(15.2) f is continuous ⇔ f −1(T Y ) ⊂ T X ⇔ f −1(BY ) ⊂ T X ⇔ f −1(S Y ) ⊂ T Xwhere BY is a basis and S Y a subbasis for T Y . The finer the topology in Y and the coarser thetopology in X are, the more difficult is it for f : X → Y to be continuous.

Theorem 15.3. Let f : X → Y be a map between two topological spaces. The following are equi-valent:

(1) f is continuous (2) The preimage of any open set in Y is open in X : V open in Y ⇒ f −1(V ) open in X



GENERAL TOPOLOGY 25

(3) The preimage of any closed set in Y is closed in X : V closed in Y ⇒ f −1(V ) closed in X (4) f −1(B) ⊂ (f −1(B)) for any B ⊂ Y

(5) f −1(B) ⊂ f −1(B) for any B ⊂ Y

(6) f (A)

⊂ f (A) for any A

⊂ X

(7) For any point x ∈ X and any neighborhood V ⊂ Y of f (x) there is a neighborhood U ⊂ X of x such that f (U ) ⊂ V .

Proof. It is easy to see that the first three conditions are equivalent.

(2) ⇒ (4): f −1(B) ⊂ f −1(B)

f −1(B) open

⇒ f −1(B) ⊂ (f −1(B)).

(4) ⇒ (2): Let B be any open set in Y . Then f −1(B) B=B

= f −1(B)(4)⊂ (f −1(B)) ⊂ f −1(B) so

f −1(B) is open since it equals its own interior.(3) ⇔ (5): Similar to (2) ⇔ (4).

(3) ⇒ (6): A ⊂ f −1(f (A))

f −1(f (A)) is closed

⇒ A ⊂ f −1(f (A)).

(6) ⇒ (5): f (f −1

(B))

(6)

⊂ f (f −1

(B))

f (f −1(B))⊂B

⊂ B(2) ⇔ (7): This is just a reformulation.

Example 15.4. The function f (x) = −x is continuous R → R, but not continuous RK → RK

(K is closed in RK but f −1(K ) is not closed (14.16)) and not continuous R → R ([0, ∞) is openin R but f −1([0, ∞)) = (−∞, 0] is not open (14.2.(4))). Since (a, b] is closed and open in R themap 1[a,b) : R → 0, 1, with value 1 on [a, b) and value 0 outside [a, b), is continuous.

Theorem 15.5. (1) The identity function is continuous.(2) The composition of two continuous functions is continuous.(3) Let B be a subspace of Y , A a subspace of X , and f : X → Y a map taking values in B.

Then

f : X → Y is continuous =⇒ f |A : A → Y is continuous (restriction )

f : X → Y is continuous ⇐⇒ B|f : X → B is continuous (corestriction )

(4) Let f j : X j → Y j, j ∈ J , be an indexed set of maps. Then f j :

X j →

Y j is continuous ⇐⇒ f j : X j → Y j is continuous for all j ∈ J

(5) Suppose that X =

U α is a union of open sets. Then

f : X → Y is continuous ⇐⇒ f |U α is continuous for all α

for any map f : X → Y .(6) Suppose that X = C 1 ∪ · · · ∪ C n is a finite union of closed sets. Then

f : X → Y is continuous ⇐⇒ f |C i is continuous for all i = 1, . . . , n

for any map f : X → Y .

Proof. This easily checked. The =⇒-part of (4) uses 15.11 below.

15.6. Homeomorphisms and embeddings. One of the central problems in topology is to decideif two given spaces are homeomorphic.

Definition 15.7. A homeomorphism is a bijective continuous map f : X → Y with a continu-ous inverse X ← Y : f −1. An embedding is an injective continuous map f : X → Y such that f : X → f (X ) (where f (X ) has the topology as a subspace of Y ) is a homeomorphism.

A bijection f : X → Y is a homeomorphism iff

U is open in X ⇐⇒ f (U ) is open in Y

holds for all subsets U of X iff

f −1

(V ) is open in X ⇐⇒ V is open in Y holds for all subsets V of Y .



26 GENERAL TOPOLOGY

An injective map f : X → Y is an embedding if

U is open in X ⇐⇒ f (U ) is open in f (X )

holds for all subsets U of X . An embedding is a homeomorphism followed by an inclusion.

Example 15.8. (1) The map f (x) = 3x + 1 is a homeomorphism R → R.(2) The identity map R → R is bijective and continuous but not a homeomorphism.(3) The map [0, 1) → S 1 : t → (cos(2πt), sin(2πt)) is continuous and bijective but not a home-

morphism. The image of the open set [0, 12

) is not open in S 1.(4) The obvious bijection [0, 1)∪2 → [0, 1] is continuous but not a homeomorphism. There does

not exist any continuous surjection in the other direction.

(5) The spaces [1 × 1, 2 × 1] ⊂ Z+ × Z+ and 1n | n ∈ Z+ ⊂ R are homeomorphic.

(6) The map R → R × R : t → (t, t) is an embedding. For any continuous map f : X → Y , themap X → X × Y : x → (x, f (x)) is an embedding; see (27.4) for a generalization.

(7) Rn embeds into S n via stereographic projection.(8) R2 embeds in R3. Does R3 imbed in R2?(9) Are the spaces C n and C 1/n of 14.13.(2) homeomorphic?

(10) A knot is in embedding of S 1 in R3 (or S 3). Two knots, K 0 : S 1 → R3 and K 1 : S 1 → R3,are equivalent if there exists a homeomorphism h of R3 such that h(K 0) = K 1. The fundamentalproblem of knot theory [1] is to classify knots up to equivalence.

Lemma 15.9. If f : X → Y is a homeomorphism (embedding) then the corestriction of the restric-tion f (A)|f |A : A → f (A) ( B|f |A : A → B) is a homeomorphism (embedding) for any subset A of X (and any subset B of Y such that f (A) ⊂ B ). If the maps f j : X j → Y j are homeomorphisms (embeddings) then the product map

f j :

X j → Y j is a homeomorphism (embedding).

Proof. In case of homeomorphisms there is a continuous inverse in both cases. In case of embed-dings, use that an embedding is a homeomorphism followed by an inclusion map.

15.10. Maps into product spaces. There is an easy test for when a map into a product spaceis continuous. Recall that the product topology has as basis the collection of all product sets U j

where U j is open in X j for all j and U j = X j for all except finitely many j . The collection of allsets of the from π−1

j (U j) (ie all factors but one is X j) is a subbasis.

Theorem 15.11. Let f : X → j∈J Y j be a map into a product space. Then

f : X →j∈J

Y j is continuous ⇐⇒ πj f : X → Y j is continuous for all j ∈ J

where πj :

j∈J Y j → Y j is the projection map.

Proof. Let T X be the topology on X and T j the topology on Y j . Then

j∈J π−1j (T j) is a subbasis

for the topology on

j∈J Y j. Therefore

f : X →j∈J

Y j is continuous ⇐⇒ f −1(

j∈J

π−1j (T j)) ⊂ T

⇐⇒ j∈J

f −1(π−1j (T j)) ⊂ T

⇐⇒ ∀ j ∈ J : (πj f )−1(T j) ⊂ T ⇐⇒ ∀ j ∈ J : πj f is continuous

by definition of continuity (15.2).

Similarly there was (15.5.(3)) an easy test for when a map f : X → B into a subspace B ⊂ Y is continuous: X → B is continuous ⇐⇒ X → B → Y is continuous. The secret is that thesubspace is equipped with the initial topology for the inclusion map and that the product space isequipped with the initial topology for the projection maps.

Let (X j)j∈J be an indexed family of topological spaces with subspaces Aj ⊂ X j . Then Aj is

a subspace of X j .Theorem 15.12.

Aj =

Aj.



GENERAL TOPOLOGY 27

Proof. Let (xj) be a point of

X j . Then

(xj) ∈

Aj ⇐⇒

U j ∩

Aj = ∅ for any basis neighborhood

U j of (xj)

⇐⇒ U j ∩

Aj

=

∅ for any basis neighborhood U j of (xj)

⇐⇒ ∀ j : U j ∩ Aj = ∅ for any neighborhood U j of xj

⇐⇒ ∀ j : xj ∈ Aj

⇐⇒ (xj) ∈

Aj

It follows that a product of closed sets is closed. (Whereas a product of open sets need not beopen in the product topology.)

Definition 15.13. The box toplogy on the set

X j is the toplogy with basis consisting of all products

U j of open set U j ⊂ X j.

The box topology is finer than the product topology (which has as basis the collection of allproducts U j of open set U j ⊂ X j where U j = X j except for finitely many j ∈ J ) and equal tothe product topology for finite products. The projection maps πj are continuous and open also inthe box topology and the above theorem also holds for the box topology (with the same proof).However, since there are more open sets in the box topology it is more difficult for a map into theproduct to be continuous. A concrete example is this.

Example 15.14. Let ∆: R →Z+

R be the diagonal map of R into the product of countably

many copies of R. ∆ is continuous if we use the product topology (15.11) but not continu-ous if we use the box topology; for instance

n∈Z+

(−1/n, 1/n) is open in the box topology but

∆−1

n∈Z+(−1/n, 1/n)

=

n∈Z+(−1/n, 1/n) = 0 is not open.

Thus the easy characterization (15.11) of continuous maps into the product topology does not

work for the box topology.15.15. Maps out of coproducts.

Theorem 15.16. Let f :

j∈J X j → Y be a map out of a coproduct space. Then

f :j∈J

X j → Y is continuous ⇐⇒ f ιj : X j → Y is continuous for all j ∈ J

where ιj : X j →

j∈J X j is the inclusion map.

Let f j : X j → Y j , j ∈ J , be an indexed set of maps. Thenf j :

X j →

Y j is continuous ⇐⇒ f j : X j → Y j is continuous for all j ∈ J



28 GENERAL TOPOLOGY

16. Quotient maps

In this section we will look at the quotient space construction. But first we consider open andclosed maps.

16.1. Open and closed maps. [2, I.§5] Let X and Y be topological spaces and f : X → Y amap.

Definition 16.1. The map f : X → Y is

open

closed

if for all U ⊂ X we have:

U

open

closed

in X ⇒ f (U )

open

closed

in Y

The restriction (16.10) of an open (or closed) map to an arbitrary subspace need not be open(closed). However,

Proposition 16.2. [6, Ex 22.5] The restriction of an open (or closed) map f : X → Y to an open (closed) subspace A

⊂ X is an open (closed) map f

|A : A

→ f (A) or f

|A : A

→ Y .

Proof. Let U ⊂ A be any subset. The implications

U is open in A (A is open in X )

⇒ U open in X (f is open)

⇒ f (U ) open in Y

⇒ f (U ) open in f (A)

show that f |A : A → f (A) is open.

A bijective continuous and open or closed map is a homeomorphism.

Example 16.3. (1) The projection map π1 : R × R → R is continuous and open. It is not aclosed map for H =

(x, y)

∈ R

×R

| xy = 1

is closed but π1(H ) = R

− 0

is not closed.

(2) The map f : [−1, 2] → [0, 1] given by

f (x) =

0 −1 ≤ x ≤ 0

x 0 ≤ x ≤ 1

1 1 ≤ x ≤ 2

is continuous and closed (20.9.(1)). It is not open for f ([−1, −1/2)) = 0 is not open.(3) The map R → 0, 1 with value 1 on [0, 1) and value 0 elsewhere is continuous because

(14.2.(4)) the half-open interval [0, 1) is both closed and open. This map is closed and openbecause all subsets of the discrete space 0, 1 are closed and open. See 16.13 for another exampleof a closed and open map.

16.2. Quotient maps. Quotient maps are continuous surjective maps that generalize both con-

tinuous, open surjective maps and and continuous, closed surjective maps.Let X and Y be topological spaces and p : X → Y a continuous surjective map.

Definition 16.4. The continuous surjective map p : X → Y is a quotient map if for all V ⊂ Y

p−1(V ) is open in X ⇒ V is open in Y

Sets of the form p−1(V ) =

y∈V p−1(y), consisting of unions of fibres, are called saturated sets.

The saturation of A ⊂ X is the union f −1f (A) of all fibres that meet A.

Proposition 16.5. For a surjective map p : X → Y the following are equivalent:

(1) p : X → Y is a quotient map

(2) For all V ⊂ Y we have: p−1(V ) is

open

closed

in X ⇔ V is

open

closed

in Y .

(3) p : X → Y is continuous and maps saturated open closed

sets to open closed

open sets



GENERAL TOPOLOGY 29

Proof. Condition (1) and condition (2) with the word “open” are clearly equivalent. Suppose nowthat: p−1(V ) is open ⇔ V is open. Then we get

p−1(C ) is closed ⇔ X − p−1(C ) is open ⇔ p−1(Y − C ) is open

⇔ Y − C is open ⇔ C is closed

for all C ⊂ Y . This shows that the two conditions of (2) are equivalent. The content of (3) is justa reformulation of (2).

Thus a quotient map p : X → Y induces a bijective correspondenceSaturated open (closed) sets

in X

U → p(U )

p−1(V )←V

Open (closed) setsin Y

Just as for open (closed) continuous bijective maps we evidently have:

Corollary 16.6. A bijective continuous map is a quotient map if and only if it is a homeomorph-ism.

Quotient maps behave nicely with respect to composition:

Corollary 16.7. Let X f

Y g

Z be continuous maps. Then

f and g are quotient ⇒ g f is quotient ⇒ g is quotient

Proof. The first assertion is a tautology: Assume that f and g are quotient. Then

(g f )−1(V ) open in X ⇔ f −1g−1(V ) open in X f is quotient⇒ g−1(V ) open in Y

g is quotient⇒ V open in Z

for any set V ⊂ Y . Next, suppose that X f −→ Y

g−→ Z is a quotient map. Then the last map g issurjective and

g−1(V ) is open in Y f continuous⇒ (g f )−1(V ) is open in X

g f quotient⇒ V is open in Z

for any set V ⊂ Z .

If p : X → Y is quotient there is a simple criterion for continuity of maps out of Y .

Lemma 16.8. Let p : X → Y be a quotient map and let f : Y → Z be some map of Y into the topological space Z . Then

Y f −→ Z is continuous ⇔ X

p−→ Y f −→ Z is continuous

Proof. This is because for all U ⊂ Z ,

f −1(U ) is open in Y

⇔ (f

p)−1(U ) = p−1f −1(U ) is open in Y

since p is a quotient map.

Corollary 16.9. Any

open

closed

continuous surjective map is a quotient map. A quotient map

f : X → Y is

open

closed

if and only if all

open

closed

sets A ⊂ X have

open

closed

saturations f −1f (A).

Proof. This is immediate from the defintion of the quotient topology:

f is open ⇐⇒ f (A) is open in Y for all open sets A ⊂ X

⇐⇒ f −1f (A) is open in X for all open sets A ⊂ X

There are quotient maps that are neither open nor closed [6, Ex 22.3] [2, Ex 10 p 135] and thereare (non-identity) quotient maps that are both open and closed (16.13) [2, Ex 3 p 128].



30 GENERAL TOPOLOGY

Example 16.10. (1) The projection map π1 : R × R → R is (16.3.(1)) open, continuous, andsurjective so it is a quotient map. The restriction π1|H ∪ (0, 0) is continuous and surjective,even bijective, but it is not a quotient map (16.5) for it is not a homeomorphism: (0, 0) is openand saturated in H

∪ (0, 0)

but π1(

(0, 0)

=

0

is not open.

Thus the restriction of a quotient map need not be a quotient map in general. On the positiveside we have

Proposition 16.11. The restriction of a quotient map p : X → Y to an open (or closed) saturated subspace A ⊂ X is a quotient map p|A : A → p(A).

Proof. (Similar to the proof of 16.2.) Let p : X → Y be a quotient map and B ⊂ Y an open set.(The case where B is closed is similar.) The claim is that p| p−1(B) : p−1(B) → B is quotient. Forany U ⊂ B the implications

p−1(U ) open in p−1(B) ( p−1(B) is open)

⇒ p−1(U ) open in X ( p is quotient)

⇒ U is open in Y ⇒ U is open in B

show that p| p−1(B) : p−1(B) → B is quotient.

We can always make a surjective map defined on a topological space into a quotient map.

Definition 16.12. Let X be a topological space, Y a set, and p : X → Y a surjective map. The quotient topology on Y is the collection

U ⊂ Y | p−1(U ) is open in X of subsets of Y .

The quotient topology2 is the finest topology on Y such that p : X →

Y continuous.A typical situation is when R is an equivalence relation on the space X and X → X/R is the

map that takes a point to its equivalence class. We call X/R with the quotient topology for thequotient space of the equivalence relation R. A set of equivalence classes is an open subset of X/R if and only if the union of equivalence classes is an open subset of X : V ⊂ X/R is open ⇔

[x]∈V [x] ⊂ X is open. We shall often say that X/R is the space obtained by identifying equivalent

points of X .

Example 16.13. (1) The real projective n-space RP n is the quotient space of S n by the equi-valence relation with equivalence classes ±x, x ∈ S n. The quotient map p : S n → RP n is bothopen and closed since (16.9) the saturation ±U of an open (closed) set U ⊂ S n is open (closed)because x → −x is a homeomorphism. Elements of RP n can be thought of as lines through theorigin of Rn+1. A set of lines is open if the set of intersection points with the unit sphere is open.

Does RP 2

embed in R3

?(2) More generally, let G × X → X be the action of a discrete group G on a space X . Give theorbit space G\X the qutient topology and let pG : X → G\X be the quotient map. The saturationof any set A ⊂ X is the union GA =

ginG gA of orbits intersecting A. If A is open, its saturation

GA is also open. Thus pG is an open map (16.9). The saturated open sets are precisely the sets of the form GA where A is open in X . The quotient map is closed if G is finite.

A continuous map f : X → Y respects the equivalence relation R if equivalent points haveidentical images, that is if x1Rx2 ⇒ f (x1) = f (x2). The quotient map X → X/R respectsthe equivalence relation R and it is the universal example of such a map.

Theorem 16.14. Let f : X → Y be a continuous map. Then

(1) The map p : X → X/R respects the equivalence relation R.

2The quotient topology is sometimes called the final topology [2, I.§2.4] with respect to the map f



GENERAL TOPOLOGY 31

(2) If the continuous map f : X → Y respects R then there exists a unique continuous mapf : X → Y such that

X f

p

Y

X/R

f

commutes. (We say that f factors uniquely through X/R.) Conversely, if f factors through X/R then f respects R.

(3) If f exists then: f is quotient ⇔ f is quotient

Proof. If f exists then clearly f respects R. Conversely, if f respects R then we can definef [x] = f (x) and this map is continuous by 16.8 and it is the only possibility. The rest follows from16.7.

The theorem says that there is a bijective correspondence

Continuous maps X →

Y that respect R

f →f

g p←g Continuous mapsX/R → Y

taking quotient maps to quotient maps.

Example 16.15. (1) Let f : X → Y be any surjective continuous map. Consider the equivalencerelation corresponding to the partition X =

y∈Y f −1(y) of X into fibres fibres f −1(y), y ∈ Y .

Let X/f denote the quotient space. Thus X/f is the set of fibres equipped with the quotienttopology. By constuction, the map f respects this equivalence relation so there is a uniquecontinuous map f such that the diagram

X

f Y

X/f

f

commutes. Note that f is bijective. The bijective continuous map f : X/f → Y is a homeomorph-ism if and only if f is quotient (16.6, 16.14). Thus all quotient maps have up to homeomorphismthe form X → X/R for some equivalence relation R on X .

(2) Let f : X → Y be any surjective continuous map. The induces map f : X/f → Y is a continu-ous bijection but in general not a homeomorphism. Instead, the quotient space X/f is homeo-morphic to Y with the quotient topology which is finer than the given topology. For instance,X/f is Hausdorff if Y is Hausdorff.

(3) Let f : X → Y be any continuous map. Then f has a canonical decomposition

X p−→ X/f

f −→ f (X ) ι−→ Y

where p is a quotient map, f is a continuous bijection, and ι is an inclusion map.(4) Let X be a topological space and A1, A2, . . . , Ak a finite collection of closed subsets. Considerthe equivalence relation where the equivalence classes are the sets A1, A2, . . . , Ak together withthe sets x for x ∈ A1 ∪ A2 ∪ · · · ∪ Ak. The quotient space X/(A1, . . . , Ak) is obtained from X by identifying each of the sets Ai to the point p(Ai). The quotient map p : X → X/(A1, . . . , Ak)is closed because (16.9) closed sets A ⊂ X have closed saturations A ∪Ai∩A=∅ Ai. A continuous

map f : X → Y factors through the quotient space X/(A1, . . . , Ak) if and only if it sends each of the sets Ai ⊂ X to a point in Y (16.14). The restriction p|X −(A1∪·· ·∪Ak) : X −(A1∪·· ·∪Ak) →X/(A1, . . . , Ak) − p(A1), . . . , p(Ak) to the complement of A1 ∪ · · · ∪ Ak is a homeomorphism(16.2). In case of just one closed subspace A ⊂ X , the quotient space is denoted X/A.

(5) The standard map f : [0, 1] → S 1 that takes t ∈ [0, 1] to (cos(2πt), sin(2πt)) is quotient be-cause it is continuous and closed. (If you can’t see this now, we will prove it later (20.8.(1)).)The induced map [0, 1]/

0, 1

→ S 1 is a homeomorphism. More generally, the standard map

Dn/S n−1 → S n is a homeomorphism where Dn ⊂ Rn, the unit disc, is the set of vectors of length ≤ 1.



32 GENERAL TOPOLOGY

(6) Let R be the equivalence relation “zero or not zero” on R. The quotient space R/R ishomeomorphic to Sierpinski space 0, 1.

(7) There is an obvious continuous surjective map f : Z+ × S 1 →

C n (14.13.(2)) that takes

Z+

× 1

to the point common to all the circles. This map is continuous because its restriction

to each of the open sets n × S 1 is continuous (15.5.(5). However, f is not a quotient map(16.5) for the image of the closed saturated set consisting of the points n × (cos( π

2 ), sin(π2 ))

is not closed as it does not contain all its limit points. The induced bijective continuous mapf : Z+ ×S 1/Z+ ×1 →

C n is therefore not a homeomorphism. There is an obvious continuoussurjective map g : Z+ × S 1 →

C 1/n (14.13.(2)) that takes Z+ × 1 to the point common to

all the circles. This map is continuous because its restriction to each of the open sets n × S 1

is continuous (15.5.(5)). However, g is not a quotient map for the image of the closed saturatedset Z+ × −1 is not closed as it does not contain all its limit points. The induced bijectivecontinuous map g : Z+ × S 1/Z+ × 1 →

C 1/n is therefore not a homeomorphism either.

(Actually (17.10.(5)), the quotient space Z+ × S 1/Z+ × 1, known as the countable wedge of circles

n∈Z+

S 1 [6, Lemma 71.4], is not homeomorphic to any subspace of the plane.)

(8) [6, p 451] Let P 4g be a regular 4g-gon and with edges labeled a1, b1, a1, b1, . . . , ag , bg , ag, bg in

counter-clockwise direction. The closed orientable surface M g of genus g ≥ 1 is (homeomorphicto) the quotient space P 4g/R where R is the equivalence relation that makes the identifications

a1b1a−11 b−1

1 · · · agbga−1g b−1

g on the perimeter and no identifications in the interior of the polygon.See [6, p 452] for the case g = 2. Are any of these surfaces homeomorphic to each other? [6, Thm77.5]

(9) [6, p 452] Let P 2g be a regular 2g-gon with edges labeled a1, a1, . . . , ag, ag in counter-clockwisedirection. The closed non-orientable surface N g of genus g ≥ 1 is the quotient space P 2g/R whereR is the equivalence relation that makes the identifications a2

1 · · · a2g on the perimeter and no

identifications in the interior of the polygon. For g = 1 we get the projective plane RP 2 and forg = 2 we get the Klein Bottle [6, Ex 74.3].

Example 16.16. (The adjunction space) [6, Ex 35.8] [4, p 93] [7, Chp 1, Exercise B p 56] Consider

the set-up X Y

i

f Y consisting of a space X and a continuous map f : A → Y definedon the closed subspace A ⊂ X . Let R be the smallest equivalence relation on X Y such that

aRf (a) for all a ∈ A; the equivalence classes of R are a f −1(a) for a ∈ A, x for x ∈ X − A,anf f −1(y) y for y ∈ Y . The adjunction space is the quotient space

X ∪f Y = X Y /R

for the equivalence relation R. Let f : X → X ∪f Y be the map X → X Y → X ∪f Y and leti : Y → X ∪f Y be the map Y → X Y → X ∪f Y . These two continuous maps agree on A in the

sense that f i = i f and the adjunction space is the universal space with this property. For anyother space Z receiving maps X → Z ← Y that agree on A there exists a unique continuous mapX ∪f Y such that the diagram

A

i

f Y

i

X f

X ∪f Y

∃!

Z

commutes. The map i : Y → X ∪f Y is closed for closed sets B ⊂ Y ⊂ X Y have closedsaturations f −1(B) B . Since i is injective it is an embedding; its image is a closed subspaceof X ∪f Y homeomorphic to Y . The map f |X − A : X − A → X ∪f Y is open for open sets

U ⊂ X −A ⊂ X Y have open saturations U ∅. Since f |X −A is injective it is an embedding; itsimage is an open subspace of X ∪f Y homeomorphic to X −A. The quotient map X Y → X ∪f Y is closed if the map f is closed for then also closed subsets B

⊂ X

⊂ X

Y have closed saturations

B ∪ f −1f (B ∩ A) f (A ∩ B). We shall later see [6, Ex 35.8] that X ∪f Y is normal when X andY are normal.



GENERAL TOPOLOGY 33

Only few topological properties are preserved by quotient maps. The reason is that surjectiveopen maps and surjective closed maps are quotient maps so that any property invariant underquotient maps must also be invariant under both open and closed maps.

As we saw in 16.15.(6) the quotient space of a Hausdorff space need not be Hausdorff, not even

T 1. In general, the quotient space X/R is T 1 if and only if all equivalence classes are closed sets.(For instance, if X and Y are T 1 then also the adjunction space X ∪f Y is T 1.) The quotient spaceX/R is Hausdorff if and only if any two distinct equivalence classes are contained in disjoint opensaturated sets. We record an easy criterion for Hausdorfness even though you may not yet knowthe meaning of all the terms.

Proposition 16.17. If X is regular and A ⊂ X is closed then the quotient space X/A is Hausdorff.

The product of two quotient maps need not be a quotient map [2, I.§5.3]. See [6, Ex 29.11] foran important example where the product of two quotient maps is quotient, though.



34 GENERAL TOPOLOGY

17. Metric topologies

If X is a set with a metric d : X × X → [0, ∞) the collection Bd(x, ε) | x ∈ X,ε > 0 of ballsBd(x, ε) = y ∈ X | d(x, y) < ε is a basis for the metric topology T d induced by d.

Definition 17.1. A metric space is the topological space associated to a metric set. A topological space is metrizable if the topology is induced by some metric on X .

Hausdorff dimension, fractals, or chaos are examples of metric, not topological, concepts.Which topological spaces are metrizable? To address this question we need to build up an

arsenal of metrizable and non-metrizable spaces and to identify properties that are common to allmetrizable spaces. Here are the first such properties: All metric spaces are Hausdorff and firstcountable.

Theorem 17.2 (Continuity in the metric world). Let f : X → Y be a map between metric spaces with metrics dX and dY , respectively. The following conditions are equivalent:

(1) f is continuous (2) ∀x ∈ X ∀ε > 0∃δ > 0∀y ∈ X : dX(x, y) < δ ⇒ dY (f (x), f (y)) < ε

(3) ∀x ∈ X ∀ε > 0∃δ > 0 : f (BX(x, δ )) ⊂ BY (f (x), ε)Proof. Essentially [6, Ex 18.1].

Proposition 17.3 (Comparison of metric topologies). Let d and d be metrics on X and T d, T dthe associated metric topologies. Then

T d ⊂ T d ⇔ ∀x ∈ X ∀ε > 0∃δ > 0 : Bd(x, δ ) ⊂ Bd(x, ε)

Proof. T d ⊂ T d if and only if the identity map (X, T d) → (X, T d) is continuous [6, Ex 18.3] .

Lemma 17.4 (Standard bounded metric). Let d be a metric on X . Then d(x, y) = mind(x, y), 1is a bounded metric on X (called the standard bounded metric corresponding to d) that induce the same topology on X as d.

Proof. Either use the above proposition or use that for any metric the collection of balls of radius

< 1 is a basis for the metric topology. These bases are the same for the two metrics.

Theorem 17.5. Any subspace of a metrizable space is a metrizable. Any countable product of metrizable spaces is metrizable.

Proof. See [6, Ex 21.1] for the first assertion. To prove the second assertion, let X n, n ∈ Z+, be acountable collection of metric spaces. We may assume that each X n has diameter at most 1 (17.4).Put

d((xn), (yn)) = sup 1

ndn(xn, yn) | n ∈ Z+

for points (xn) and (yn) of

X n and convince yourself that d is a metric.Since d(x, y) < ε/n ⇒ dn(x, y) < ε, the projection maps πn :

X n → X n are continuous (17.2)

for all n and therefore the metric topology is finer than the product topology.Conversely, let B ((xn), ε)

⊂X n be an open ball in the metric topology. Then

B1(x1, ε) × · · · BN (xN , ε) × X N +1 × X N +2 × · · · ⊂ B((xn), ε)

when N is so big that 1/N < ε. As the left hand side is open in the product topology, this showsthat any set that is open in the metric topology is also open in the product topology.

An uncountable product of metric spaces need not be metrizable (17.10.(4)).

17.6. The first countability axiom. Let X be a topological space.

Definition 17.7. The space X has a countable basis at the point x ∈ X if there is a countable collection U nn∈Z+ of neighborhoods of x such that any neighborhood U of x contains at least one of the neighborhoods U n. The space X is first countable if all points of X have a countable basis.

Any metrizable space is first countable.

Proposition 17.8. [6, Thm 30.2] Any subspace of a first countable space is first countable. Any countable product of first countable spaces is first countable.



GENERAL TOPOLOGY 35

Proof. The first assertion is immediate. Let

X n be a countable product of first countable spaces.Let (xn) be a point of

X n. Let Bn be countable basis at xn ∈ X n. The collection of all products

U n where U n ∈ Bn for finitely many n and U n = X n for all other n is then a countable (5.5)

basis at (xn).

An uncountable product of first countable (even metric) spaces and the quotient of a firstcountable space may fail to be first countable (17.10.(4), 17.10.(5)). Some linearly ordered spacesare first countable, some are not (17.10.(3)).

In a first countable (eg metric) space X , the points of the closure of any A ⊂ X can beapproached by sequences from A in the sense that they are precisely the limit points of convergentsequences in A. This is not true in general (17.10.(3)).

Lemma 17.9. Let X be a topological space.

(1) (The sequence lemma) Let A ⊂ X be a subspace and x a point of X . Then

x is the limit of a sequence of points from A ⇒ x ∈ A

The converse holds if X is first countable.

(2) (Continuous map preserve convergent sequences) Let f : X → Y be a map of X into a space Y . Then

f is continuous ⇒ f (xn) → f (x) whenever xn → x for any sequence xn in X

The converse holds if X is first countable.

Proof. (1) The direction ⇒ is clear. Conversely, suppose that x ∈ A. Let U n be a countable basisat x. For each n choose a point xn ∈ A ∩ U 1 ∩ . . . ∩ U n. Then xm ∈ U n for all m ≥ n. We claimthat the sequence (xn) converges to x. Let U be any neighborhood of x. Then U n ⊂ U for somen so that xm ∈ U n ⊂ U for m ≥ n.(2) The direction ⇒ is clear. Conversely, suppose that f (xn) → f (x) whenever xn → x. Theidea is to approximate points in the closure of A ⊂ X by sequences, which we can do in the firstcountable space X , to prove that f (A)

⊂ f (A) (15.3.(6)) for any A

⊂ X . Let x

∈ A. Since X is

first countable, there is by (1) a sequence of points xn ∈ A converging to x. By hypothesis, thesequence f (xn) ∈ f (A) converges to f (x). Thus f (x) ∈ A by (1) again.

We say that X satisfies the sequence lemma1 if for any A ⊂ X and for any x ∈ A there is asequence of points in A converging to x.

To summarize:

X is metrizable ⇒ X is first countable ⇒ X satisfies the sequence lemma

Examples show that neither of these arrows reverse.

Example 17.10. (1) R is first countable as is any metric space.(2) R is first countable. The collection of half-open intervals [a, b) where b > a is rational is a

countable basis of neighborhoods at the point a. Is R metrizable? (24.3) [6, Ex 30.6]

(3) S Ω does not satisfy the sequence lemma (so is not first countable and not metrizable). Thelargest element Ω lies in the closure of S Ω but it can not be the limit of any sequence of pointsfrom S Ω for any such such sequence has an upper bound in S Ω. The section S Ω is first countable:1 = [1, 2) is open so S Ω has a finite basis at the first element 1. For any other element, α > 1, wecan use the collection of neighborhoods of the form (β, α] for β < α. Is S Ω metrizable? (22.3)[6,Exmp 3 p 181, Ex 30.7].

(4) RJ is not first countable when J is uncountable: Let U nn∈Z+ be any countable collection

of neighborhoods of, say, the point (0)j∈J of RJ . The idea is to construct a neighborhood thatdoes not contain any of the U n. I claim that there is an index j0 ∈ J such that πj0(U n) = R forall n. Indeed, the set of j s for which this is not true

j ∈ J | ∃n ∈ Z+ : πj(U n) = R =

n∈Z+ j ∈ J | πj(U n) = R

1Aka a Frechet space



36 GENERAL TOPOLOGY

is a countable union of finite sets, hence countable (5.5.(3)). Then the neighborhood π−1j0

(R−1)

of (0) does not contain any of the neighborhoods U n in the countable collection. (RJ does noteven satisfy the sequence lemma [6, Exmp 2 p 133].)

(5) The quotient map p : n∈Z+S 1

→ n∈Z+S 1 is closed (16.15.(4)). The domain n∈Z+

S 1 =

Z+ × S 1 ⊂ R × R2 is first countable (17.8) but the image n∈Z+S 1 (16.15.(7)) is not: Let

U nn∈Z+ be any collection of saturated neighborhoods of Z+×1. The saturated neighborhoodU which at level n equals U n with one point deleted (cf Cantor’s diagonal argument (5.7)) doesnot contain any of the U n. It follows (17.8) that

n∈Z+

S 1 does not embed in R2 nor in any

other first countable space. For instance, the universal property of quotient spaces (16.14) givesa factorization

(17.11)

n∈Z+S 1

f

p

n∈Z+

S 1

n∈Z+

S 1f

of the continuous map f such that πm(f |n × S 1) is the identity function when m = n and theconstant function when m = n. The induced map f is an injective continuous map. It can not bean embedding for the countable product

S 1 of circles is first countable (17.8). The topology

on

S 1 is finer than the subspace topology inherited from

S 1.(6) The closed (16.15.(4)) quotient map R → R/Z takes the first countable space R to a space

that is not first countable [4, 1.4.17] (at the point corresponding to Z). This can be seen by anargument similar to the one from item (5): Let U nn∈Z be any countable collection of openneighborhoods of Z ⊂ R. Let U be the open neighborhood of Z such that U ∩ (n, n + 1), n ∈ Z,equals U n ∩ (n, n + 1) with one point deleted. Then U does not contain any of the U n.

17.12. The uniform metric. Let Y be a metric space with a bounded metric d and let J be aset. We shall discuss uniform convergence which is a metric, not a topological, concept.

Definition 17.13. [6, pp 124,266] The uniform metric on Y J

= j∈J Y is the metric given by d((xj), (yj)) = sup j ∈ J | d(xj, yj).

It is easy to see that this is indeed a metric.

Theorem 17.14. On RJ we have: product topology ⊂ uniform topology ⊂ box topology

Proof. Omitted.

The elements of Y J are functions f : J → Y . Note that a sequence of functions f n : J → Y converges in the uniform metric to the function f : J → Y if and only if

∀ε > 0∃N > 0∀ j ∈ J ∀n ∈ Z+ : n > N ⇒ d(f n( j), f ( j)) < ε

We say that the sequence of functions f n : J → Y converges uniformly to the function f : J → Y .

Theorem 17.15 (Uniform limit theorem). (Cf [6, Thm 43.6]) Suppose that J is a topological space and that Y is a metric space. The uniform limit of any sequence (f n) of continuous functions f n : J → Y is continuous.

Proof. Well-known.

18. Connected spaces

Definition 18.1. The topological space X is connected if it cannot be represented as the union X = X 1 ∪ X 2 of two disjoint open non-empty subsets X 1 and X 2.

Two subsets A and B are separated if A ∩ B = ∅ = A ∩ B . This means that the two setsare disjoint and neither contains a limit point of the other. Two disjoint open (closed) sets areseparated. If C ⊂ A and D ⊂ B and A and B are separated, then C and D are separated. A

separation of X consists of two separated non-empty subsets A and B with union X = A ∪ B.Theorem 18.2. The following are equivalent:



GENERAL TOPOLOGY 37

(1) X is connected (2) The only closed and open subsets of X are ∅ and X (3) If X = A ∪ B where A and B are separated then A = ∅ or B = ∅.(4) Every continuous map X

→ 0, 1

to the discrete space

0, 1

is constant.

Proof. (1) ⇒ (2): If C is a closed and open subset of X then also the complement X − C is closedand open. Thus X = C ∪ (X − C ) is the union of two disjoint open subsets and therefore one of these sets must be empty.(2) ⇒ (3): Suppose that X = A ∪ B where A and B are separated. Then A ⊂ A for A doesnot meet B. Thus A is closed. Similarly, B is closed. Therefore A is both closed and open. Byhypothesis, A is either the empty set or all of X .(3) ⇒ (4): We prove the contrapositive. Suppose that there exists a continuous map f : X → 0, 1that is not constant. Then X = f −1(0) ∪ f −1(1) is the union of two non-empty separated subsets.(4) ⇒ (1): We prove the contrapositive. Suppose that X is non-connected. Then X = U 1 ∪ U 2where U 1 and U 1 are disjoint, open, and non-empty. The map f : X → 0, 1 given by f (U 1) = 0and f (U 1) = 1 is a non-constant continuous map.

Theorem 18.3. The image under a continuous map of a connected space is connected.Proof. We use the equivalence of (1) and (4) in Theorem 18.2. Let f : X → Y be a continuousmap. If f (X ) is not connected, there is a non-constant continuous map f (X ) → 0, 1 and hencea non-constant continuous map X → 0, 1. So X is not connected.

Example 18.4. (1) R −0 = R− ∪ R+ is not connected for it is the union of two disjoint opennon-empty subsets.

(2) We shall later prove that R is connected (19.3) and that the connected subsets of R areprecisely the intervals, rays, R, and ∅.

(3) R is not connected for [a, b) is a closed and open subset whenever a < b. In fact, any subsetY of R containing at least two points a < b is disconnected as Y ∩ [a, b) is closed and open butnot equal to ∅ or Y . (R is totally disconnected .)

(4) Q is totally disconnected (and not discrete): Let Y be any subspace of Q containing at leasttwo points a < b. Choose an irrational number t between a and b. Then Y ∩ (t, ∞) = Y ∩ [t, ∞)closed and open but not equal to ∅ or Y .

(5) RK is connected [6, Ex 27.3].(6) The Sierpinski space 0, 1 is connected.

A subspace of X is said to be connected if it is connected in the subspace topology. A subspaceof a connected space need obviously not be connected.

Lemma 18.5. A subspace of X is connected if and only if it is not the union of two non-empty separated subsets of X .

Proof. Let Y ⊂ X and suppose that Y = Y 1 ∪ Y 2. Observe that

Y 1 and Y 2 are separated in Y

⇔ Y 1 and Y 2 are separated in X

because ClY (Y 1) ∩ Y 2 = Y 1 ∩ Y ∩ Y 2 = Y 1 ∩ Y 2 by 14.7.(3). The lemma now follows immediatelyfrom 18.2, the equivalence of (1) and (3).

Corollary 18.6. Suppose that Y ⊂ X is a connected subspace. For every pair A and B of separated subsets of X such that Y ⊂ A ∪ B we have either Y ⊂ A or Y ⊂ B.

Proof. The subsets Y ∩ A and Y ∩ B are separated (since the bigger sets A and B are separated)with union Y . By 18.5, one of them must be empty, Y ∩ A = ∅, say, so that Y ⊂ B .

Theorem 18.7. Let (Y j)j∈J be an indexed family of connected subspaces of the space X . Suppose that there is an index j0 ∈ J such that Y j and Y j0 are not separated for any j ∈ J . Then the union

j∈J Y j is connected.

Proof. We have to prove that the subspace Y = j∈J Y j is connected. We use the criterion of

Lemma 18.5. Suppose that Y = A ∪ B is the union of two separated subsets A and B of X . Eachof the sets Y j is by Corollary 18.6 contained in either A or B . Let’s say that Y j0 is contained in A.



38 GENERAL TOPOLOGY

Then Y j must be contained in A for all j ∈ J for if Y j ⊂ B for some j then Y j and Y j0 would beseparated. Therefore Y = A.

Corollary 18.8. The union of a collection of connected subspaces with non-empty intersection is

connected.Proof. No two subspaces in the collection are separated.

Corollary 18.9. Let C be a connected subspace of X . If C ⊂ Y ⊂ C then Y is connected.

Proof. Apply Theorem 18.7 to the collection consisting of C and y for y ∈ Y − C . Then C andy are not separated since y ∈ C .

Corollary 18.10. Suppose that for any two points in X there is a connected subspace containing both of them. Then X is connected.

Proof. Let x0 be some fixed point of X . For each point x ∈ X , let C x be a connected subspacecontaining x0 and x. Then X =

C x is connected since

C x = ∅ (18.8).

Theorem 18.11. Products of connected spaces are connected.

Proof. We prove first that the product X × Y of two connected spaces X and Y is connected. Infact, for any two points (x1, y1) and (x2, y2) the subspace X × y1 ∪ x2 × Y contains the twopoints and this subspace is connected since (18.8) it is the union of two connected subspaces witha point in common. Thus X × Y is connected by Corollary 18.10.

Next, induction shows that the product of finitely many connected spaces is connected.Finally [6, Ex 23.10], consider an arbitrary product

X j of connected spaces X j, j ∈ J . Choose

a point xj in each of the spaces X j (assuming that all the spaces of the product are non-empty).For every finite subset F of J let C F ⊂

X j be the product of the subspaces X j if j ∈ F and

xj if j ∈ F . Since C F is connected for each finite subset F and these subsets have the point(xj)j∈J in common, the union

F ∈F C F , where F is the collection of all finite subsets of J , is

connected. This union is not all of

X j but its closure is, so that Corollary 18.9 tells us that

F ∈F C F = j∈J X j is connected.



GENERAL TOPOLOGY 39

19. Connected subsets of linearly ordered spaces

Recall that a subset C of a linearly ordered set is convex if a, b ∈ C ⇒ [a, b] ⊂ C .

Lemma 19.2. Let X be a linearly ordered space and C a subspace of X . Then

C is connected ⇒ C is convex

Proof. Let C ⊂ X a subspace that is not convex. Then there exist points a < x < b in X suchthat a and b are in C and x is outside C . Since C ⊂ (−∞, x) ∪ (x, +∞) is contained in the unionof two separated subsets and meet both of them, C is not connected (18.6).

The converse to 19.2 is of course not true as for instance Z+ is convex but not connected (thereare gaps).

Recall that a linear continuum is a linearly ordered set with the least upper bound propertywhere (x, y) = ∅ for any two points x < y (no gaps). We shall now identify the linearly orderedspaces that are connected. (You may read the following theorem and its proof as if we discussedX = R only.)

Theorem 19.3. [2, Ex 7 p 382] Let X be a linearly ordered space. Then

X is connected in the order topology ⇔ X is a linear continuum

The connected subsets of a linear continuum are intervals, rays, the whole space and the empty set.

Proof. ⇒: This is an exercise [6, Ex 24.4] [4, Problem 6.3.2] [2, Ex 7 p 382].⇐: Assume X is a linear continuum. We will show that for C ⊂ X we have

C is connected ⇔ C is convex

Thanks to 19.2 we only need to show that convex subsets are connected. It is enough (18.8) toshow that all closed intervals [a, b] in X are connected (any convex set C =

b∈C [a0, b] is the

union of the closed intervals in C emanating from some fixed point a0 ∈ C ). Suppose that [a, b] isnot connected. Then [a, b] = A

∪B is the union of two separated sets A

=

∅ and B

=

∅ (18.5).

We may assume that b ∈ B. The sets A and B are closed and open in [a, b] (since they form aseparation of [a, b]) and closed in X (since (14.7) [a, b] is closed (11.1) in X ). Let c be the leastupper bound of the bounded set A. Then c ∈ A since A is closed in X . (The least upper bound of any bounded set always belongs to the closure of the set since otherwise it wouldn’t be the least upper bound.) In particular, c < b so A ⊂ [a, b). But an element c of an open subset A of [a, b)(with the subspace topology or, what is the same (13.5), the order topology) can not be an upperbound for A because when c ∈ A also [c, e) ⊂ A for some e > c and [c, e) is not empty when X isa linear continuum. (In short, c ∈ A since A is closed in X and c ∈ A since A is open in [a, b).)

We now know that the connected and the convex subsets of X are the same. In particular, thelinear continuum X itself, which certainly is convex, is connected in the order topology.

We finally show [2, Prop 1 p 336] that the non-empty convex subsets of X are the (bounded orunbounded, closed, open, or half-open) intervals. Let C be a convex subset of X . If C is neither

bounded from above nor below it must be all of X , for if x is any point of X there exist a, b ∈ C so that a < x < b and then also x ∈ C . If C is bounded from above but not from below, let c beits least upper bound. Then for any x < c, x is neither a lower nor an upper bound for C so thereexist a, b ∈ C such that a < x < b ≤ c so that x ∈ C . Thus C is either (−∞, c) or (−∞, c]. Theargument is similar for the other cases. Note that X also has the greatest lower bound property[6, Ex 3.13].

The real line R, the ordered square I 2o , the (ordinary) (half) line [1, ∞) = Z+ × [0, 1), and thelong (half) line S Ω × [0, 1) [6, Ex 24.6, 24.12] are examples of linear continua.

Theorem 19.4. [Intermediate Value Theorem] Let f : X → Y be a continuous map of a connected topological space X to a linearly ordered space Y . Then f (X ) is connected, hence convex. If Y is a linear continuum, f (X ) is an interval (bounded or unbounded, closed, half-open, or open)

Proof. f (X ) is connected (18.3) and therefore convex (19.2). For subsets of a linear continuum weknow (19.3) that connected = convex = interval.



40 GENERAL TOPOLOGY

Any linearly ordered space containing two consecutive points, two points a and b with (a, b) = ∅is not connected as X = (−∞, b) ∪ (a, +∞) is the union of two disjoint open sets.

Any well-ordered set X containing at least two points is totally disconnected in the ordertopology. For if C

⊂ X contains a < b then a

∈ C

∩(a, b]

b is closed and open in C since (a, b]

is closed and open in X .

19.5. Path connected spaces. Path connectedness is a stronger property than connectedness.

Definition 19.6. The topological space X is path connected if for any two points x0 and x1 in X there is a continuous map (a path) f : [0, 1] → X with f (0) = x0 and f (1) = x1.

The image under a continuous map of a path connected space is path connected, cf 18.3. Anyproduct of path connected spaces is path connected [6, Ex 24.8], cf 18.7.

Example 19.7. The punctured euclidian plane Rn −0 is path connected when n ≥ 2 since anytwo points can be joined by a path of broken lines. Thus also the (n − 1)-sphere S n−1, which isthe image of Rn − 0 under the continuous map x → x/|x|, is path connected for (n − 1) ≥ 1.

Since the unit interval I is connected (19.3), all paths f (I ) are also connected (18.3) so all path

connected spaces are, as unions of paths emanating from one fixed point, connected (18.10). Theconverse is not true, not all connected spaces are path connected.

Example 19.8. (1) (Topologist’s sine curve) Let S be the graph of the function sin(1/x), 0 <x ≤ 1, considered as a subspace of the plane R2. The (closed) topologist’s sine curve is thesubspace S = S ∪ (0 × [−1, 1]) of the plane. It follows from 18.3 and 18.9 that S is connected.However, S is not path connected. For suppose that there existed a continuous map f : [0, 1] → S from f (0) = (0, 0) to some point in S , say f (1) = ( 1

2π , 0) ∈ S . We may arrange it so thatf (t) ∈ S for t > 0. For each positive integer n ∈ Z+ we have π2f ([0, 1/n]) = [−1, 1]. Thereason is that the image π1f ([0, 1

n ]) on the first axis is an interval (19.4) so that f maps [0, 1/n]onto S ∩ (π1f ([0, 1/n]) × R) and then π2f ([0, 1/n]) = π2(S ∩ (R × π1f ([0, 1/n])) = [−1, 1].This means that for each n ∈ Z+ there is a point tn ∈ [0, 1/n] such that the second coordinateπ2f (tn) = (

−1)n. The sequence (tn) converges to 0 but the image sequence π2f (tn) = (

−1)n

does not converge, contradicting continuity of f (17.9.(2)).(2) The ordered square I o

2 (11.4.(7)) is connected since it is a linear continuum [6, Ex 3.15] butit is not path connected. For suppose that f : [0, 1] → I 2o is a path from the smallest element(the lower left corner) f (0) = 0 × 0 to the largest element (the upper right corner) f (1) = 1 × 1.Then f is surjective for the image contains (19.4) the interval [0 × 0, 1 × 1] = I 2o . But this isimpossible since the ordered square I o2 contains uncountably many open disjoint subsets (egx × (0, 1) = (x × 0, x × 1), 0 ≤ x ≤ 1) but [0, 1] does not contain uncountably many open disjointsubsets (choose a rational number in each of them).

(3) RK is connected and not path-connected space [6, Ex 27.3].

19.9. Components and path components. The relations on X defined by

x ∼ y ⇔ There is a connected subset C ⊂ X such that x ∈ C and y ∈ C

x p

∼ y ⇔ There is a path in X between x and yare equivalence relations. Check transitivity!

Definition 19.10. The equivalence classes of the first (second) equivalence relation are called components (path components) of X .

The component containing the point x ∈ X is, by definition, the union of all connected subspacescontaining x. The path component of the point x ∈ X consists, by definition, of all points with apath to x.

Theorem 19.11. (1) The components are connected closed disjoint subsets with union X . Any connected subset of X is a subset of precisely one component.

(2) If X =

C α where the C α are connected subspaces such that for α = β there is a separation

of A

∪B of X with C α

⊂ A and C β

⊂ B, then the C α are the components of X .

(3) The path components of X are path connected disjoint subsets with union X . Any path con-nected subset is contained in precisely one of the path components.



GENERAL TOPOLOGY 41

(4) The path components are connected. The components are unions of path components.

Proof. (1) The components form a partition of X since they are equivalence classes. The compon-ents are closed since the closure of any connected subset is connected. All the points of a connected

subset are equivalent so they are contained in the same equivalence class.(2) All points in C α are equivalent. No point in C α can be a equivalent to any point in C β forα = β : Assume that there is a connected set C ⊂ X that intersects both C α and C β . Choose aseparation X = A ∪ B such that C α ⊂ A and C β ⊂ B . Then C α ∪ C ∪ C β ⊂ X = A ∪ B and theconnected (18.8) set C α ∪ C ∪ C β intersects both A and B, which is impossible (18.6).

If X has only finitely many components the components are closed and open. If X = C 1∪·· ·∪C nwhere the finitely many subspaces C i are connected and separated then the C i are the componentsof X .

The path components can be closed, open, closed and open, or neither closed nor open. However,for locally connected spaces (19.13) components and path components coincide (19.16).

19.12. Locally connected and locally path connected spaces.

Definition 19.13. The space X is locally (path) connected at the point x if every neighborhood of x contains a (path) connected neighborhood of x. The space X is locally (path) connected if it is locally (path) connected at each of its points.

Thus a space is locally (path) connected iff it has a basis of (path) connected subsets.Consider this table

R [−1, 0) ∪ (0, 1] S Q

Connected YES NO YES NOLocally connected YES YES NO NO

before you draw any conclusions about the relationship between connected and locally connectedspaces. Note in particular that a space can be connected (even path connected 19.17) and notlocally connected.

On the positive side, note that open subsets of locally (path) connected spaces are locally (path)

connected and that quotient spaces of locally connected spaces are locally connected [6, Ex 25.8][2, I.§11.6].

Proposition 19.14. The space X is locally (path) connected if and only if open subsets have open (path) components (open in X or, what is the same ( 13.3), open in the open set).

Proof. Assume that X is locally connected and let U be an open subset. Consider a componentC of U . The claim is that C is open. Let x be a point in C . Choose a connected neighborhoodV of x such that V ⊂ U . Since V is connected in X , V is connected in U so V is contained in acomponent of U (19.11) and this component must be C ; V ⊂ C . This shows that C is open.

Conversely, assume that open subsets have open components. Let x be a point of X and U aneighborhood of x. Let C be the component of U containing x. Then C is an open connectedneighborhood of x contained in U .

It follows for instance that the open subsets of R are unions of at most countably many openintervals. For any open set is the union of its components which are open, since R is locallyconnected, and connected, so they are open intervals. Since each of the open intervals contains arational number, there are at most countably many components. Any closed subset of R is thecomplement to a union of at most countably many open intervals.

Proposition 19.15. [6, Ex 25.8] [2, I.§11.6] Locally connected spaces have locally connected quo-tient spaces.

Is it true that locally path connected spaces have locally path connected quotient spaces?

Theorem 19.16. The components and the path components are the same in a locally path con-nected space.

Proof. The path components P α of X are connected and open, since X is locally path connected(19.14). Therefore (19.11.(2)) they are the components.



42 GENERAL TOPOLOGY

For instance, the components and the path components of a locally Euclidean space, such as amanifold, are the same.

Example 19.17. (1) The comb space X = ([0, 1]×0)∪((0∪1/n | n ∈ Z+)× [0, 1]) ⊂ R2 is

clearly path connected but it is not locally connected for (19.16) the open set U = X − [0, 1]×0has a component (use (19.11.(2)) to identify the components) that is not open. (The same is truefor any (small) neighborhood of (0, 1/2).) See [6, Ex 25.5] for a similar example.

(2) The closed topologist’s sine curve S has two path path-components, S = S ∩ (R+ × R), whichis open and not closed for its closure is S , and 0 × [0, 1], which is closed and not open for S is not closed. Since there are two path components and one component, S is not locally pathconnected (19.16). S is not even locally connected: Let U be a small neighborhood (well, not toobig) around (0, 1/2), then U has component that is not open. For instance, U = S − ([0, 1] × 0has a component (19.11.(2)), 0 × (0, 1] that is not open (no neighborhood of (0, ±1

2) is disjoint

from S ). (The Warsaw circle is an example of a path connected not locally connected space, sois the next example.)

(3) The space X =

C n (14.13.(2)) is even path connected but still not locally connected for the

component R+

× 0

of the open set X

− (0, 0)

(use (19.11.(2)) to identify the components)

is not open (any neighborhood of (1, 0) contains points from the circles). Alternatively, X is notlocally connected since any small neighborhood of (1, 0) has a separation.

(4) Any linear continuum is locally connected since the basis for the topology consists of connectedsets (19.3). It need not be locally path connected [6, Ex 25.3].



GENERAL TOPOLOGY 43

20. Compact spaces

Definition 20.2. A topological space X is compact if it satisfies one of the following two equivalent conditions:

(1) Every collection of open subsets whose union is X contains a finite subcollection whose union is X .(2) Every collection of closed subsets whose intersection is ∅ contains a finite subcollection

whose intersection is ∅.

A collection of open subsets whose union is the whole space is called an open covering . Thus aspace is compact if any open covering contains a finite open covering.

Any finite space is compact as there are only finitely many open subsets. Any compact discretespace is finite. The real line R is not compact as the open covering (−n, n) | n ∈ Z+ has nofinite subcovering.

Theorem 20.3. The image under a continuous map of a compact space is compact.

Proof. Let f : X

→ Y be continuous. Let

A be an open covering of f (X ). Then f −1(

A) =

f −1(A) | A ∈ A is an open covering of X . Assuming that X is compact, there is a finite subsetB of A such that f −1(B) is a covering of X . Then B is a finite covering of f (X ).

Theorem 20.4. Every closed subset of a compact space is compact.

Proof. The theorem follows immediately from the second definition of compactness (20.2.(2)) whenwe observe that closed subsets of a closed subset are closed.

Proposition 20.5. Let Y ⊂ X . Then

Y is compact ⇐⇒Any collection of open sets (in X ) that cover Y contains a subcollection that covers Y

Proof. ⇒: Let A = Aj | j ∈ J be a collection of open sets Aj ⊂ X such that Y ⊂ Aj . Then

Y ∩ Aj | j ∈ J is an open covering of Y . Since Y is compact, there is a finite subcovering of Y .The corresponding finitely many sets of the collection A then cover Y .⇐: Let B = Bj | j ∈ J be an open covering of Y . Then Bj = Y ∩ Aj for some open set Aj ⊂ X .The collection Bj | j ∈ J covers Y and so it contains a finite subcollection that covers Y . Thecorresponding subcollection of B then also covers Y .

Lemma 20.6. Let X be a Hausdorff space and let C ⊂ X , x0 ∈ X be a compact subset and a point such that x0 ∈ C . Then there exist open sets U , V such that U ∩ V = ∅ and C ⊂ U and x0 ∈ V .

Proof. For each point a ∈ C , the Hausdorff property implies that there are open sets U a, V a suchthat U ∩ V = ∅ and a ∈ U a and x0 ∈ V a. Since C is compact (20.5), C ⊂ U a1 ∪ . . . ∪ U ak for finitelymany points a1, . . . , ak ∈ C . Set U = U a1 ∪ . . . ∪ U ak and V = V a1 ∩ . . . ∩ V ak .

Theorem 20.7. Every compact subset of a Hausdorff space is closed.

Proof. This is an immediate consequence of 20.6.

Corollary 20.8. Let X be a compact Hausdorff space.

(1) Let C ⊂ X a subset. Then C is compact ⇐⇒ C is closed.(2) If A and B are disoint closed sets in X then there exist disjoint open sets U , V such that

A ⊂ U and B ⊂ V .

Proof. (1) ⇒ is 20.7 and ⇐ is 20.4.(2) Let A and B be disjoint closed subsets of X . Then A and B are compact as just shown. Nowapply 20.6 and compactness to find disjoint open sets U ⊃ A, V ⊃ B .

Corollary 20.9. (1) Every continuous map of a compact space into a Hausdorff space is closed

(and a quotient map onto its image).(2) Every injective continuous map of a compact space into a Hausdorff space is an embedding.



44 GENERAL TOPOLOGY

(3) Every bijective continuous map of a compact space onto a Hausdorff space is a homeo-morphism.

Proof. Let f : X → Y be a continuous map of a compact space X into a Hausdorff space Y .

(1) We haveC is closed in X

20.4⇒ C is compact 20.3⇒ f (C ) is compact

20.7⇒ f (C ) is closed in Y

which shows that f is a closed map.(2 Every injective continuous and closed map is an embedding.(3) Every bijective continuous and closed map is a homeomorphism.

Example 20.10. The (unreduced) suspension of the space X is the quotient space

SX = [0, 1] × X/(0 × X, 1 × X )

What is the suspension of the n-sphere S n? Define f : [0, 1] × S n → S n+1 to be the (continuous)map that takes (t, x), t ∈ [0, 1], to the geodesic path from the north pole (0, 1) ∈ S n+1, throughthe equatorial point (x, 0)

∈ S n

⊂ S n+1, to the south pole (0,

−1)

∈ S n+1. (Coordinates in

Rn+2 = Rn+1 × R.) By the universal property of quotient spaces there is an induced continuousand bijective map f : S S n → S n+1. Since SS n is compact (as a quotient of a product of twocompact spaces) and S n+1 is Hausdorff (as a subspace of a Hausdorff space), f is a homeomorphism(20.9). We often write SS n = S n+1, n ≥ 0, where the equality sign stands for ‘is homeomorphicto’.

Corollary 20.11. The Hausdorff image of a locally connected compact space is locally connected compact.

Proof. We now know that f is a quotient map. The quotient image of a locally connected space islocally connected (19.15).

Lemma 20.12 (The tube lemma). Let X and Y be topological spaces where Y is compact. Let x0

be a point of X . For any neighborhood N ⊂ X × Y of the slice x0 × Y there is a neighborhood U of x0 such that x0 × Y ⊂ U × Y ⊂ N .

Proof. For each point y ∈ Y there is a product neighborhood such that x0 × y ⊂ U y × V y ⊂ N .Since Y is compact, there are finitely many points y1, . . . , yk ∈ Y such that Y = V 1 ∪ . . .∪V k whereV i = V yi . Put U = U 1 ∩ . . . U k.

Theorem 20.13. (Cf 20.17) The product of finitely many compact spaces is compact.

Proof. We show that the product X ×Y of two compact spaces X and Y is compact. The theoremfollows from this by induction. Suppose that A is an open covering of X × Y . We show that thereis a finite subcovering. For each point x ∈ X , the slice x × Y , homeomorphic to Y , is compactand therefore x × Y ⊂ A | A ∈ A(x) for some finite set A(x) ⊂ A. Thanks to the tubelemma (20.12) we know that the open set A | A ∈ A(x) actually contains whole tube U (x)×Y

for some neighborhood U (x) of x. By compactness, X = U (x1) ∪ . . . ∪ U (xk) can be covered byfinitely many of the neighborhoods U (x). Now

X × Y =k

i=1

U (xi) × Y =k

i=1

A | A ∈ A(xi) =

A | A ∈

ki=1

A(xi)

where k

i=1 A(xi) ⊂ A is finite (4.8).

Here are two small lemmas that are used quite often.

Lemma 20.14. A compact space can not contain an infinite closed discrete subspace.

Proof. Any closed discrete subspace of a compact space is compact (20.4) and discrete, hencefinite.

Lemma 20.15. [6, Ex 28.5] [4, 3.10.2] Let C 1 ⊃ C 2 ⊃ ·· · ⊃ C n ⊃ · · · be a descending sequence of closed nonempty subsets of a compact space. Then

C n = ∅.



GENERAL TOPOLOGY 45

Proof. If

C n = ∅ then C n = ∅ for some n ∈ Z+ by 20.2.(2).

Theorem 20.16. [Cf [6, Ex 27.5] ] Any nonempty compact Hausdorff space without isolated points ( 14.11) is uncountable.

Proof. Let xn be a sequence of points in X . It suffices (5.4) to show that xn | n ∈ Z+ = X .Since X is Hausdorff and has no isolated points there is a descending sequence

V 1 ⊃ V 2 ⊃ ·· · ⊃ V n−1 ⊃ V n ⊃ · · ·of nonempty open sets such that xn ∈ V n for all n. These are constructed recursively. Put V 0 = X .Suppose that V n−1 has been constructed for some n ∈ Z+. Since xn is not isolated, xn = V n−1.Choose a point yn ∈ V n−1 distinct from xn and choose disjoint (separated) open sets U n ⊂ X ,V n ⊂ V n−1 such that xn ∈ U n and yn ∈ V n. Then U n ∩ V n = ∅ so xn ∈ V n.

By construction, the intersection

V n contains none of the points xn of the sequence and bycompactness (20.15),

V n = ∅. Thus X contains a point that is not in the sequence.

Is it true that a connected Hausdorff space is uncountable?

Theorem 20.17 (Tychonoff theorem). The product

j∈J X j of any collection (X j)j∈J of compact spaces is compact.

Proof. Put X =

j∈J X j. Let us say that a collection of subsets of X is an FIP-collection (finite

intersection property) if any finite subcollection has nonempty intersection. We must (20.2.(2))show that

A = ∅ for any FIP-collection A of closed subsets. We show that in fact A | A ∈

A = ∅ for any FIP-collection A (of not necessarily closed sets). This means that we must find apoint x in

A∈A A.

Step 1. The FIP-collection A is contained in a maximal FIP-collection.The set (supercollection?) A of FIP-collections containing A is a strictly partially ordered by strictinclusion. Any linearly ordered subset (subsupercollection?) B ⊂ A has an upper bound, namelythe FIP-collection B = B∈B

B containing

A. Now Zorn’s lemma (8.4) says that A has maximal

elements.We can therefore assume that A is a maximal FIP-collection.

Step 2. Maximality of A implies

(20.18) A1, . . . , Ak ∈ A =⇒ A1 ∩ · · · ∩ Ak ∈ Aand for any subset A0 ⊂ X we have

(20.19) ∀A ∈ A : A0 ∩ A = ∅ =⇒ A0 ∈ ATo prove (20.18), note that the collection A∪A1 ∩· · ·∩Ak is FIP, hence equals A by maximality.To prove (20.19), note that the collection A ∪ A0 is FIP: Let A1, . . . , Ak ∈ A. By (20.18),A1 ∩ · · · ∩ Ak ∈ A so A1 ∩ · · · ∩ Ak ∩ A = ∅ by assumption.Step 3. πj(A) | A ∈ A is FIP for all j ∈ J .

πj(A1) ∩ · · · ∩ πj(Ak) ⊃ πj(A1 ∩ · · · ∩ Aj) = ∅ for any finite collection of sets A1, . . . , Ak ∈ A.Since X j is compact and πj(A) | A ∈ A is an FIP collection of closed subsets of X j , the

intersection

A∈A πj(A) ⊂ X j is nonempty (20.2.(2)). Choose a point xj ∈

πj(A) for each j ∈ J and put x = (xj) ∈ X .

Step 3. x ∈A∈A A.

Note first that π−1j (U j) ∈ A for any neighborhood U j of xj . This follows from (20.19) since

π−1j (U j) ∩ A = ∅, or equivalently, U j ∩ πj(A) = ∅, for any A ∈ A as xj ∈ πj(A). Since A has

the FIP it follows that π−1j1

(U j1) ∩ · · · ∩ π−1jk

(U jk) ∩ A = ∅ whenever U ji are neighborhoods of the

finitely many points xji ∈ X ji and A ∈ A. Thus x ∈A∈A A.

21. Compact subspaces of linearly ordered spaces

Lemma 21.1. Any compact subset C = ∅ of a linearly ordered space Y has a smallest and a largest element.



46 GENERAL TOPOLOGY

Proof. The proof is by contradiction. Assume that C has no largest element. This means thatfor any a ∈ C there is a c ∈ C such that a < c or, in other words, that C ⊂c∈C (−∞, c). Bycompactness

C

⊂ (

−∞, c1)

∪ · · · ∪(

−∞, ck)

⊂ (

−∞, c)

where c = maxc1, . . . , ck is the greatest (7.6) of the elements c1, . . . , ck ∈ C . But this is acontradiction as c ∈ C and c ∈ (∞, c).

We shall now identify the compact intervals of the real line R. The result says that all closedand bounded intervals [a, b] ⊂ R are compact. The argument works just as well in any linearlyordered space X with the least upper bound property (1.14) so for the sake of generality we shallwork in this context.

Theorem 21.2. Let X be a linearly ordered space. Then

X has the least upper bound property ⇐⇒ Every closed and bounded interval in X is compact

Proof. ⇐: [6, Ex 27.1].⇒: Let [a, b] ⊂ X be a closed interval and A and open covering of [a, b] (with the subspace topology

which is the order topology). We must show that [a, b] is covered by finitely many of the sets fromthe collection A. The set

C = y ∈ [a, b] | [a, y] can be covered by finitely many members of Ais nonempty (a ∈ C ) and bounded from above (by b). Let c = sup C be the least upper bound of C . Then a ≤ c ≤ b. We would like to show that c = b.Step 1 If x ∈ C and x < b then y ∈ C for some y ∈ (x, b]Proof of Step 1. Suppose first that x has an immediate successor y > x. We can not have y > b > xfor then y would not be an immediate successor. So x < y ≤ b. Clearly [a, y] = [a, x] ∪ y can becovered by finitely many members of A. Suppose next that x has no immediate successor. Choosean open set A ∈ A containing x. Since A is open in [a, b] and contains x < b, A contains aninterval of the form [x, d) for some d ≤ b. Since d is not an immediate successor of x there is apoint y

∈ (x, d). Now [a, y] = [a, x]

∪[x, y]

⊂ [a, x]

∪[x, d)

⊂ [a, x]

∪A can be covered by finitely

members of A.Step 2 c ∈ C .Proof of Step 2 . The claim is that [a, c] can be covered by finitely many members of A. From Step1 we have that C contains elements > a so the upper bound c is also > a. Choose A ∈ A suchthat c ∈ A. Since A is open in [a, b] and a < c, A contains an interval of the form (d, c] for some dwhere a ≤ d < c. Since c is the least upper bound for C , there must be points from C in (d, c]. Letz be such a point. Now [a, c] = [a, z] ∪ (d, c] where [a, z] can be covered by finitely many membersof A and (d, c] is contained in A. Thus [a, c] can be covered by finitely many members of A.Step 3 c = b.Proof of Step 3 . Since c ∈ C by Step 2, Step 1 says that if c < b, then c can not be an upperbound.

For instance, the ordered square I 2o = [0

×0, 1

×1] and S Ω = [1, Ω] are compact linearly ordered

spaces [6, Ex 10.1]. More importantly, since the linearly ordered space R has the least upper boundproperty, we get

Corollary 21.3. The compact and connected subsets of a linear continuum are precisely the closed and bounded intervals.

Proof. Let X be a linear comtinuum. All closed and bounded intervals in X are compact andconnected (21.2, 19.3). Conversely, if C ⊂ X is compact then C ⊂ [a, b] for some a, b in C (21.1)and if C is also connected then C is convex (19.3) so [a, b] ⊂ C .

In particular, the compact and connected subsets of R are precisely the closed and boundedintervals.

Theorem 21.4 (Extreme Value theorem). Let f : X

→ Y be a continuous map of a topological

space X into a linearly ordered space Y .(1) If X is compact then there exist m, M ∈ X such that f (m) ≤ f (x) ≤ f (M ) for all x ∈ X .



GENERAL TOPOLOGY 47

(2) If X is compact and connected then there exist m, M ∈ X such that f (X ) = [f (m), f (M )].

Proof. (1) Since the image f (X ) is compact (20.3) it has (21.1) a smallest and a greatest element,f (m) ∈ f (X ) and f (M ) ∈ f (X ), so that f (X ) ⊂ [f (m), f (M )].

(2) f (X )

21.1

⊂ [f (m), f (M )]

19.2

⊂ f (X ) since f (X ) is compact and connected (20.3, 18.3).

21.5. Compactness in metric spaces.

Lemma 21.6 (Lebesgue lemma). Let X be a compact metric space. For any open covering A of X there exists an ε > 0 (the Lebesgue number of the open covering A) such that any subset of X with diameter < ε is contained in a member of A.

Proof. For each point x ∈ X , choose εx > 0 such that the ball B (x, 2εx) is contained in a memberof A. The collection B(x, εx)x∈X covers X so by compactness

X = B(x1, εx1) ∪ B(x2, εx2) ∪ · · · ∪ B(xk, εxk)

for finitely many points x1, x2, . . . , xk ∈ X . Let ε be the smallest of the numbers ε1, ε2, . . . , εk.The triangle inequality implies that this ε works. (Suppose that A ⊂ X with diam A < ε. Choose

any point a in A. There is an i such that d(xi, a) < ε. Let b be any point in A. Since d(b, xi) ≤d(b, a) + d(a, xi) < ε + εi ≤ 2εi, the point b is contained in B(xi, 2εi) which is contained in amember of A.)

Definition 21.7. A map f : X → Y between metric spaces is uniformly continuous if for all ε > 0there is a δ > 0 such that dY (f (x1), f (x2)) < ε whenever dX(x1, x2) < δ :

∀ε > 0∃δ > 0∀x1, x2 ∈ X : dX(x1, x2) < δ ⇒ dY (f (x1), f (x2)) < ε

Theorem 21.8 (Uniform continuity theorem). Let f : X → Y be a continuous map between metric spaces. If X is compact, then f is uniformly continuous.

Proof. Given ε > 0. Let δ be the Lebesgue number of the open covering f −1B(y, ε) | y ∈ Y .Then we have for all x1, x2 ∈ X

dX(x1, x2) < δ ⇒

diam

x1, x2

< δ

⇒ ∃y ∈ Y : x1, x2 ⊂ f −1B(y, ε)

⇒ ∃y ∈ Y : f (x1), f (x2) ⊂ B(y, ε)

⇒ dY (f (x1), f (x2)) < 2ε

The following is a theorem that applies not to metric spaces in general but rather to the specialmetric space Rn.

Theorem 21.9 (Heine–Borel). Consider the metric space Rn with its standard metric. A subset C ⊂ Rn is compact if and only if it is closed and bounded.

Proof. Let C ⊂ Rn be compact. Since C is compact and (−R, R)n | R > 0 an open covering of

Rn

, there is an R > 0 such that C ⊂ [−R, R]n

. Thus C is bounded. As a compact subset of aHausdorff space, C is closed (20.7).

Conversely, if C is closed and bounded then C is a closed subset of [−R, R]n for some R > 0.But [−R, R]n is compact (21.3, 20.13) and closed subsets of compact spaces are compact (20.4);in particular, C is compact.

The theorem is not true if we replace the standard metric with the bounded metric associatedto the standard metric.

Can you now answer question 15.8.(9)?



48 GENERAL TOPOLOGY

22. Limit point compactness and sequential compactness

Definition 22.1. A space X is

(1) limit point compact if any infinite subset of X has a limit point

(2) sequentially compact if any sequence in X has a convergent subsequence.

In other words, a space is limit point compact if it contains no infinite closed discrete subspaces(14.12).

We shall not say much about these other forms of compactness (see [4, §3.10] for a more thoroughdiscussion).

Theorem 22.2. (Cf [6, Ex 28.4, Ex 35.3]) For any topological space X we have

X is compact =⇒ X is limit point compact X 1st countable

=⇒ X is sequentially compact

All three forms of compactness are equivalent for a metrizable space.

Proof. X is compact =⇒ X is limit point compact: A subset with no limit points is closed and

discrete (14.12), hence finite (20.14).X is limit point compact and 1st countable =⇒ X is sequentially compact: Let (xn) be a sequencein X . Consider the set A = xn | n ∈ Z+. If A is finite there is a constant subsequence (4.8.(3)).If A is infinite, A has a limit point x by hypothesis. If X is 1st countable, then x is (17.9) the limitof a sequence of points from A. By rearranging, if necessary, we get (!) a subsequence convergingto x.X is sequentially compact and metrizable =⇒ X compact: This is more complicated (but shouldalso be well-known from your experience with metric spaces) so we skip the proof.

Example 22.3. (1) The ordered space S Ω of all countable ordinals is limit point compact andsequentially compact but it is not compact. S Ω is not compact (21.1) for it is a linearly orderedspace with no greatest element [6, Ex 10.6]. On the other hand, any countably infinite subsetof S Ω is contained in a compact subset (7.9.(3), 21.2). Therefore (22.2) any countably infinitesubset, indeed any infinite subset (14.12), has a limit point and any sequence has a convergentsubsequence. (Alternatively, use that S Ω is first countable (17.10.(3)).) It follows (22.2) that S Ωis not metrizable.

(2) The Stone–Cech compactification β Z+ of the positive integers is compact but it is not limitpoint compact [6, Ex 38.9] since no point of the remainder β Z+ − Z+ is the limit of a sequencein Z+ (so β Z+ is not first countable and not metrizable).

23. Locally compact spaces and the Alexandroff compactification

Definition 23.1. A space X is locally compact at the point x ∈ X if there exist an open set U ⊂ X and a compact set C ⊂ X such that x ∈ U ⊂ C . X is locally compact if X is locally compact at each of its points.

Compact spaces are locally compact (U = X = C ). The real line R (more generally, Rn) islocally compact but not compact. All linearly ordered spaces with the least upper bound propertyare locally compact. The space Q ⊂ R of rational numbers is not locally compact [6, Ex 29.1].

Let X be a locally compact Hausdorff space. Let ω X = X ∪ ω denote the union of X with aset consisting of a single point ω . The collection

T = U | U ⊂ X open ∪ ωX − C | C ⊂ X compactis a topology on ωX : It is easy to see that finite intersections and arbitrary unions of open sets of the first (second) kind are again open of the first (second) kind. It follows that T is closed underarbitrary unions. It is also closed under finite intersections since U ∩ (ωX − C ) = U − C is openin X . And ω X = ωX − ∅ and ∅ are open.

Theorem 23.2 (Alexandroff). Let X be a locally compact Hausdorff space. Then ωX is a compact Hausdorff space and the inclusion map X → ωX is an embedding.



GENERAL TOPOLOGY 49

If c : X → cX is another embedding of X into a compact Hausdorff space cX whose image is cX but one point then there is a unique homeomorphism cX → ωX such that the diagram

X

c

cX ωX

commutes.

Proof. We must verify the following points:The subspace topology on X is the topology on X : The subspace topology X ∩ T is clearly theoriginal topology on X .ωX is Hausdorff : Let x1 and x2 be two distinct points in ω X . If both points are in X then thereare disjoint open sets U 1, U 2 ⊂ X ⊂ ωX containing x1 and x2, respectively. If x1 ∈ X and x2 = ω,choose an open set U and a compact set C in X such that x ∈ U ⊂ C . Then U x and ωX −C ωare disjoint open sets in ω X .ωX is compact : Let

U j

j∈J be any open covering of ωX . At least one of these open sets contains

ω. If ω ∈ U k, k ∈ J , then U k = (X − C ) ∪ ω for some compact set C ⊂ X . There is a finite setK ⊂ J such that U jj∈K covers C . Then U jj∈K ∪k is a finite open covering of ωX = X ∪ω.

Let now c : X → cX be another embedding of X into a compact Hausdorff space cX such thatcX − X = ω consists of just one point ω. Define f : cX → ωX by f (x) = x for all x ∈ X and f (ω) = ω . For any open set U ⊂ X , f −1(U ) is open in X and hence in cX since X is openin cX (13.3.(2)). For any compact set C ⊂ X , f −1(ωX − C ) = cX − C is open in cX since thecompact set C is closed in the Hausdorff space cX (20.8.(1)). This shows that f is continuous. Byconstruction, f is bijective and hence (20.9) a homeomorphism.

The space ωX is called the Alexandroff or the one-point compactification of X .

Example 23.3. (1) If X is compact Hausdorff then ω X is X together with an isolated point ω .(ω = ωX − X is a neighborhood of ω .)

(2) The n-sphere S n

= ωR

n

is the Alexandroff compactification of the locally compact Hausdorff space Rn for there is (15.8.(7)) a homeomorphism of Rn onto the complement of a point in S n.(3) Let X be a linearly ordered space with the least upper bound property and let [a, b) be

a half-open interval in the locally compact space X . Then ω[a, b) = [a, b]. For instance theAlexandroff compactifications of the half-open intervals [0, 1) ⊂ R, Z+ = [1, ω) ⊂ Z+ × Z+, andS Ω = [1, Ω) ⊂ S Ω are [0, 1], [1, ω], [1, Ω] = S Ω, respectively.

(4) ω(Z+ × R) =

n∈Z+C 1/n, the Hawaiian Earring (14.13.(2)).

Corollary 23.4. Let X be a topological space. The following conditions are equivalent:

(1) X is locally compact Hausdorff (2) X is homeomorphic to an open subset of a compact Hausdorff space (3) X is Hausdorff and for any point x ∈ X and any neighborhood U x there is a neighbor-

hood V

x such that V

⊂ U and V is compact.

Proof. (1) =⇒ (2): Use 23.2.(2) =⇒ (3): Suppose that Y is a compact Hausdorff space and that X ⊂ Y is an open subset. X is Hausdorff since subspaces of Hausdorff spaces are Hausdorff (14.19). Let x be a point of X andU ⊂ X a neighborhood of x. Since Y − U is compact (20.8) and does not contain x, there are(20.6) disjoint open set V x and W ⊃ Y − U . Then x ∈ V ⊂ Y − W ⊂ U . Here V ⊂ Y − W ⊂ U since Y − W is closed and V is compact since it is closed (20.8).(3) =⇒ (1): Take x ∈ U = X and find V ⊂ X such that C = V is compact.

Corollary 23.5. Any closed subset of a locally compact space is locally compact. Any open or closed subset of a locally compact Hausdorff space is locally compact Hausdorff.

Proof. Let X be a locally compact space and A ⊂ X a closed subspace. We use the definitiondirectly to show that A is locally compact. Let a

∈ A. There are subsets a

∈ U

⊂ C

⊂ X such

that U is open and C is compact. Then A ∩ U ⊂ A ∩ C where A ∩ U is a neighborhood in A andC ∩ A is compact as a closed subset of the compact set C (20.4).



50 GENERAL TOPOLOGY

If X is locally compact Hausdorff and A ⊂ X open, it is immediate from 23.4 that A is locallycompact Hausdorff.

An arbitrary subspace of a locally compact Hausdorff space need not be locally compact

(23.6.(1)). The product of finitely many locally compact spaces is locally compact but an ar-bitrary product of locally compact spaces need not be locally compact [6, Ex 29.2]; for instanceZω

+ is not locally compact. The image of a locally compact space under an open continuous [6,29.3] or a perfect map [6, Ex 31.7] [4, 3.7.21] is locally compact but the image under a generalcontinuous map of a locally compact space need not be locally compact; indeed, the quotient of alocally compact space need not be locally compact (23.6.(2)) [4, 3.3.16].

Example 23.6. (1)

n∈Z+C n ⊂ R2 (14.13.(2)) is not locally compact at the origin: Any neigh-

borhood of 0 × 0 contains a countably infinite closed discrete subspace so it can not be containedin any compact subspace (20.14).

(2) The quotient space R/Z+ (17.10.(6)) is not locally compact at the point corresponding toZ+: Any neighborhood of this point contains an infinite closed discrete subspace so it can not becontained in any compact subspace (20.14).

(3) In diagram (17.11), the space S 1 is locally compact, S 1 is compact, and S 1 is notlocally compact (at the one point common to all the circles).

(4) (Wedge sums and smash products) A pointed space is a topological space together with oneof its points, called the base point. The wedge sum of two pointed disjoint spaces (X, x0) and(Y, y0) is the quotient space

X ∨ Y = (X Y )/(x0 ∪ y0)

obtained from the disjoint union of X and Y (12.6) by identifying the two base points. Letf : X ∪ Y → X × Y be the continuous map given by f (x) = (x, y0), x ∈ X , and f (y) = (x0, y),y ∈ Y . The image f (X ∪ Y ) = (X ×y0) ∪ (x0 × Y ) = π−1

2 (y0) ∪ π−11 (x0) is closed when

X and Y are T 1–spaces. In fact, f is a closed map for if C ⊂ X is closed then f (C ) = (x, y0) |x ∈ C = π−1

1 (C ) ∩ π−12 (y0) is also closed. Thus f is a quotient map onto its image and its

factorization f as in the commutative diagram (16.14.(2))

X ∪ Y

f X × Y

X ∨ Y

f

is an embedding. We can therefore identify X ∨ Y and X ×y0∪ x0× Y . The smash productis defined to be

X ∧ Y = (X × Y )/(X ∨ Y )

the quotient of the product space X × Y by the closed subspace X ∨ Y (× − ∨ = ∧).If A ⊂ X and B ⊂ Y are closed subspaces, the universal property of quotient maps (16.14.(2))

produces a continuous bijection g such that the diagram

X × Y

gX×gY X/A × Y /BgX/A×Y/B X/A ∧ Y /B

(X × Y )/(X × B ∪ A × Y )

g

commutes. However, g may not be a homeomorphism since the product gX ×gY of the two closedquotient maps gX : X → X/A and gY : Y → Y /B may not be a quotient map. If X/A and Y arelocally compact Hausdorff spaces ([Ex 31.7] may be useful here) then gX × gY is quotient [Ex29.11] so that g is a homeomorphism (16.7, 16.14.(3), 16.6) and

(23.7) (X × Y )/(X × B ∪ A × Y ) = X/A ∧ Y /B

in this case. Here is an application. The n–sphere S n is (homeomorphic to) the quotient spaceI n/∂I n. Using the formula [Exam June 2003] for the boundary of the product of two sets we find



GENERAL TOPOLOGY 51

that the smash product of an m–sphere with an n–sphere

S m ∧ S n = I m/∂I m ∧ I n/∂I n (23.7)

= I m × I n/(I m × ∂I n ∪ ∂I m × I n) = I m+n/∂I m+n = S m+n

is an (m + n)–sphere.

(5) (The Alexandroff compactification of a product space) If X and Y are locally compact Haus-dorff spaces the map X × Y → ωX × ωY → ωX ∧ ωY is an embedding. This follows from (16.11,16.6) when we note that the first map embeds X × Y into an open saturated subset of ωX × ωY (15.9). The universal property of the Alexandroff compactification (23.2) implies that

(23.8) ω(X × Y ) = ωX ∧ ωY

for any two locally compact Hausdorff spaces X and Y . Once again we get

S m ∧ S n = ωRm ∧ ωRn (23.8)= ω(Rm × Rn) = ωRm+n = S m+n

when we view the spheres as Alexandroff compactifications of euclidian spaces.

The locally compact Hausdorff space X is open in ωX . This is actually true whenever X isembedded as a dense subspace of a Hausdorff space.

Proposition 23.9. Let X be a locally compact and dense subspace of the Hausdorff space Y . Then X is open.

Proof. Let x ∈ X . Since X is locally compact Hausdorff the point x has a neighborhood X ∩ U ,where U is open in X , such that its relative closure

ClX(X ∩ U ) 14.7

= X ∩ X ∩ U 14.9

= X ∩ U

is compact and hence closed in the Hausdorff space Y (20.7). But no part of U can stick outsideX ∩ U for since U − (X ∩ U ) is open

U − (X ∩ U ) = ∅ X dense=⇒ (X ∩ U ) − (X ∩ U ) = ∅

which is absurd. Thus we must have U ⊂ X ∩ U and of course X ∩ U ⊂ X so U ⊂ X . ThereforeX is open.

Some mathematicians prefer to include Hausdorffness in the definition of (local) compactness.What we call a (locally) compact Hausdorff space they simply call a (locally) compact space; whatwe call a (locally) compact space they call a (locally) quasicompact space.



52 GENERAL TOPOLOGY

24. Countability axioms

Let us recall the basic question: Which topological spaces are metrizable? In order to furtheranalyze this question we shall look at a few more properties of topological spaces.

We have already encountered the first countability axiom. There are three more axioms usingthe term ’countable’. Here are the four countability axioms.

Definition 24.1. A topological space

• that has a countable neighborhood basis at each of its points is called a first countable space • that has a countable basis is called a second countable space • that contains a countable dense subset is said to contain a countable dense subset 1

• in which every open covering has a countable subcovering is called a Lindelof space

A subset A ⊂ X is dense if A = X , that is if every nonempty open subset contains a point fromA.

Any second countable space is first countable. Any subspace of a (first) second countable spaceis (first) second countable. A countable product of (first) second countable spaces is (first) secondcountable, see 17.8.

The real line R is second countable for the open intervals (a, b) with rational end-points forma countable basis for the topology. The real line R with the right half-open interval topology isfirst but not second countable (25.4).

All metric spaces are first countable but not all metric spaces are second countable [6, Example2 p 190].

Any compact space is Lindelof. Any closed subset of a Lindelof space is Lindelof [6, Ex 30.9](with a proof that is similar to that of 20.4). A product of two Lindelof spaces need not be Lindelof (25.5).

The next result implies that R, in fact any subset of the 2nd countable space R, is Lindelof.

Theorem 24.2. Let X be a topological space. Then

X has a countable dense subset X is 2nd countable

X is Lindel¨ of

X is 1st countable

If X is metrizable, the three conditions of the top line equivalent.

Proof. Suppose first that X is second countable and let B be a countable basis for the topology.Pick a point bB ∈ B in each basis set. The image of the map B → X : B → bB is dense. Let now U be an open covering of X . For each basis set B that is contained in some member of U pick aU B ∈ U such that B ⊂ U B. Then the image of the map (from a subset of) B → U : B → U B is anopen covering: Let x be any point in X . Since x is contained in a member of U and every openset is a union of basis sets we have x ∈ B ⊂ U for some basis set B and some U ∈ U . But thenalso x ∈ B ⊂ U B.

To prove the arrows in the other direction, let X be a metric space with a countable dense

subset A ⊂ X . Then the collection B(a, r) | a ∈ A, r ∈ Q+ of balls centered at points in A andwith a rational radius is a countable (5.5.(4)) basis for the topology: It suffices to show that forany y ∈ B(x, ε) there are a ∈ A and r ∈ Q+ such that y ∈ B(a, r) ⊂ B(x, ε). Let r be a positiverational number such that 2r < ε − d(x, y) and let a ∈ A ∩ B(y, r). Then y ∈ B (a, r), of course,and B(a, r) ⊂ B(x, ε) for if d(a, z) < r then d(x, z) ≤ d(x, y) + d(y, z) ≤ d(x, y) + d(y, a) + d(a, z) <d(x, y) + 2r < ε.

Finally, let X be a metric Lindelof space. For each positive rational number r, let Ar be acountable subset of X such that X =

a∈Ar

B(a, r). Then A =

r∈Q+Ar is a dense countable

(5.5.(3)) subset: Any open ball B(x, ε) contains a point of Ar when 0 < r < ε, r ∈ Q.

Example 24.3. R has a dense countable subset and is not second countable (25.4). The orderedsquare I 2o is Lindelof and is not second countable since it contains uncountably many disjoint opensets (x

×0, x

×1), x

∈ I . Thus R and I 2

o

are not metrizable [6, Ex 30.6].

1Or to be separable



GENERAL TOPOLOGY 53

25. Separation Axioms

Definition 25.1. A space X is called a

• T 1-space if points x are closed in X

• T 2-space or a Hausdorff space if for any pair of distinct points x, y ∈ X there exist disjoint open sets U and V such that x ∈ U and y ∈ V • T 3-space or a regular space if points are closed and for every closed set B ⊂ X such that

x ∈ B there exist disjoint open sets U and V such that x ∈ U and B ⊂ V • T 4-space or a normal space if points are closed and for every par of disjoint closed sets

A, B ⊂ X there exist disjoint open sets U and V such that A ⊂ U and B ⊂ V .

We have the following sequence of implications

X is normal =⇒ X is regular =⇒ X is Hausdorff =⇒ X is T 1

where none of the arrows reverse (25.5, 25.6, [6, Ex 22.6]).

Lemma 25.2. Let X be a T 1-space. Then

(1) X is regular

⇐⇒ For every point x

∈ X and every neighborhood U of x there exists an

open set V such that x ∈ V ⊂ V ⊂ U .(2) X is normal ⇐⇒ For every closed set A and every neighborhood U of A there is an open

set V such that A ⊂ V ⊂ V ⊂ U .

Proof. Let B = X − U .

Theorem 25.3. (Cf 14.19) Any subspace of a regular space is regular. Any product of regular spaces is regular.

Proof. Let X be a regular space and Y ⊂ X a subset. Then Y is Hausdorff (14.19)). Considera point y ∈ Y and a relatively closed set B ⊂ Y such that y ∈ B. Then B = ClY B = B ∩ Y so y ∈ B. By definition of regularity, there exist disjoint open sets U and V such that y ∈ U and B ⊂ V . The relatively open sets U ∩ Y and V ∩ Y are disjoint and they contain y and B,respectively.

Let X =

X j be the Cartesian product of regular space X j , j ∈ J . Then X is Hausdorff (14.19)). We use 25.2.(1) to show that X is regular. Let x = (xj) be a point in X and U =

U j

a basis neighborhood of x. Put V j = X j whenever U j = X j . Otherwise, choose V j such that

xj ∈ V j ⊂ V j ⊂ U j. Then V =

V j is a neighborhood of x in the product topology and

V =

V j ⊂

U j = U . Thus X is regular.

A subspace of a normal space need not be normal (28.5) and the product of two normal spacesneed not be normal ((25.5),[6, Example 2 p 203], [4, 2.3.36]). Any closed subspace of a normalspace is normal [6, Ex 32.1].

Example 25.4 (Sorgenfrey’s half-open interval topology). The half-open intervals [a, b) form abasis for the space R (10.6, 14.2.(4), 17.10.(2), 18.4.(3)).

R is 1st countable: At the point x the collection of open sets of the form [x, a) wherea > x is rational, is a countable local basis at x.

R is not 2nd countable: Let B be any basis for the topology. For each point x choose amember Bx of B such that x ∈ Bx ⊂ [x, x + 1). The map R → B : x → Bx is injective forx = inf Bx.

R has a countable dense subset: Q is dense since any (basis) open set in R containsrational points.

R is Lindelof: It suffices (!) to show that any open covering by basis open sets containsa countable subcovering, ie that if R =

j∈J [aj , bj) is covered by a collection of right

half-open intervals [aj , bj) then R is actually already covered by countably many of theseintervals. Write

R =

j∈J

(aj , bj) ∪

R −

j∈J

(aj , bj)

as the (disjoint) union of the corresponding open intervals and the complement of thisunion. The first set can be covered by countably many of the intervals (aj , bj) for any



54 GENERAL TOPOLOGY

subset of R (with the standard topology) is Lindelof (24.2). Also the second set can becovered by countably many of the intervals (aj , bj) simply because it is countable: Thesecond set

R

− j∈J

(aj , bj) = j∈J

[aj , bj)−

j∈J

(aj , bj) =

x ∈

k∈J ak

| ∀ j ∈

J : ak

∈ (aj , bj)

consists of some of the left end-points of the intervals. For any element x of this set, choosean interval I (x) = [ak, bk) from the collection such that x = ak is the left end-point of I (x). The corresponding open intervals Int I (x) = (ak, bk) are disjoint for the left end-point of Int I (x) is not in Int I (y) when x = y so there can be no overlap. Therefore themap I is injective. But there is only room for countably many open disjoint intervals inR (choose a rational point in each of them) so the second set contains at most countablymany elements.

R is normal: Let A, B ⊂ R be disjoint closed sets. For each point a ∈ A ⊂ R − B thereis an xa ∈ R such that [a, xa) ⊂ R − B. Let U A =

a∈A[a, xa). Define U B =

b∈B[b, xb)

similarly. Then U A and U B are open sets containing A and B , respectively. If U A ∩U B = ∅then [a, xa)

∩[b, xb)

=

∅ for some a

∈ A, b

∈ B . Say a < b; then b

∈ [a, xa)

⊂ R

−B which

is a contradiction So U A and U B are disjoint. (R is even completely normal [6, Ex 32.6]by this argument.)

Example 25.5 (Sorgenfrey’s half-open square topology). The half-open rectangles [a, b) × [c, d)form a basis (12.4) for the product topology R × R. The anti-diagonal L = (x, −x) | x ∈ R isa closed discrete (L ∩ [x, ∞) × [−x, ∞) = (x, −x)) subspace of the same cardinality as R. Q × Q

is a countable dense subspace.

R × R is not Lindelof: A Lindelof space can not contain an uncountable closed discretesubspace (cf 20.14, [6, Ex 30.9]).

R × R is not normal: A normal space with a countable dense subset can not containa closed discrete subspace of the same cardinality as R [4, 2.1.10]. (Let X be a spacewith a countable dense subset. Since any continuous map of X into a Hausdorff space isdetermined by its values on a dense subspace [6, Ex 18.13], the set of continuous mapsX → R has at most the cardinality of RZ. Let X be any normal space and L a closeddiscrete subspace. The Tietze extension theorem (26.4) says that any map L → R extendsto continuous map X → R. Thus the set of continuous maps X → R has at least thecardinality of RR which is greater (5.8) than the cardinality of RZ.) See [6, Ex 31.9] for aconcrete example of two disjoint closed subspaces that can not be separated by open sets.

R × R is completely regular: since it is the product (28.3) of two compeletely regular(even normal (25.4)) spaces.

Example 25.4 shows that the arrows of Theorem 24.2 do not reverse. Example 25.5 shows thatthe product of two normal spaces need not be normal, that the product of two Lindel of spacesneed not be Lindelof, and provides an example of a (completely) regular space that is not normal.

Example 25.6 (A Hausdorff space that is not regular). The open intervals plus the sets (a, b)

−K

where K = 1n | n ∈ Z+ form a basis for the topology RK (10.6, 19.8.(3)). This is a Hausdorff topology on R since it is finer than the standard topology. But RK is not regular. The set K is closed and 0 ∈ K . Suppose that U 0 and V ⊃ K are disjoint open sets. We may chooseU to be a basis open set. Then U must be of the form U = (a, b) − K for the other basis setscontaining 0 intersect K . Let k be a point in (a, b) ∩ K (which is nonempty). Since k is in theopen set V , there is a basis open set containing k and contained in V ; it must be of the form (c, d).But U ∩ V ⊃ ((a, b) − K ) ∩ (c, d) and ((a, b) − K ) ∩ (c, d) = ∅ for cardinality reasons. This is acontradiction



GENERAL TOPOLOGY 55

26. Normal spaces

The two most important theorems about normal spaces are the Urysohn lemma and the Tietzeextension theorem.

Theorem 26.1 (Urysohn lemma). Let X be a normal space and let A and B be disjoint closed subsets of X . Then there exists a continuous function (a Urysohn function) f : X → [0, 1] such that f (a) = 0 for a ∈ A and f (b) = 1 for b ∈ B.

Proof. We shall recursively define open sets U r ⊂ X for all r ∈ Q ∩ [0, 1] such that (make adrawing!)

(26.2) r < s =⇒ A ⊂ U 0 ⊂ U r ⊂ U r ⊂ U s ⊂ U s ⊂ U 1 ⊂ X − B

To begin, let U 1 be any open set such that A ⊂ U 1 ⊂ X − B , for instance U 1 = X − B . Bynormality (25.2.(2)) there is an open set U 0 such that A ⊂ U 0 ⊂ U 0 ⊂ U 1. To proceed, arrangethe elements of Q ∩ [0, 1] into a sequence

Q ∩ [0, 1] = r0 = 0, r1 = 1, r2, r3, r4, . . .such that the first two elements are r0 = 0 and r1 = 1. Assume that the open sets U ri satisfyingcondition (26.2) have been defined for i ≤ n, where n ≥ 1. We shall now define U rn+1. Supposethat if we put the numbers r0, r1, . . . , rn, rn+1 in order

0 = r0 < · · · < r < rn+1 < rm · · · < r1 = 1

the immediate predecessor of rn+1 is r and rm is the immediate successor. Then U r ⊂ U rm by(26.2). By normality (25.2.(2)) there is an open set U rn+1 such that U r ⊂ U rn+1 ⊂ U rn+1 ⊂ U rm .The sets U ri , i ≤ n + 1, still satisfy (26.2).

Consider the function f : X → [0, 1] given by

f (x) =

inf r ∈ Q ∩ [0, 1] | x ∈ U r x ∈ U 1

1 x ∈ X − U 1

Then f (B) = 1 by definition and f (A) = 0 since A ⊂ U 0. But why is f continuous? It suffices(15.2) to show that the subbasis intervals (11.1) [0, a), a > 0, and (b, 1], b < 1, have open preimages.Since

f (x) < a ⇐⇒ ∃r < a : x ∈ U r ⇐⇒ x ∈r<a

U r

f (x) > b ⇐⇒ ∃r > b : x ∈ U r ⇐⇒ ∃r > b : x ∈ U r ⇐⇒ x ∈r>b

(X − U r)

the sets

f −1([0, a)) =r<a

U r and f −1((b, 1]) =r>b

(X − U r)

are open.

Example 26.3. Let X = R and let A = (−∞, −1] and B = [2, ∞). If we let U r = (−∞, r) forr ∈ Q ∩ [0, 1] then

f (x) =

0 x ≤ 0

x 0 < x < 1

1 1 ≤ x

is the Urysohn function with f (A) = 0 and f (B) = 1.

Theorem 26.4 (Tietze extension theorem). Every continuous map from a closed subspace A of a normal space X into (0, 1), [0, 1) or [0, 1] can be extended to X .

Lemma 26.5. Let X be a normal space and A ⊂ X a closed subspace. For r > 0 and any continuous map f 0 : A → [−r, r] there exists a continuous map g : X → R such that

(26.6) ∀x ∈ X : |g(x)| ≤ 13

r and ∀a ∈ A : |f 0(a) − g(a)| ≤ 23

r



56 GENERAL TOPOLOGY

Proof. Since the sets f −10 ([−r,−1

3 r]) ⊂ A and f −10 ([ 1

3 r, r]) ⊂ A are closed and disjoint in A they

are also closed and disjoint in X . Choose (26.1) a Urysohn function g : X → [−13

r, 13

r] such that

g(x) = −13

r on f −10 ([−r,−1

3r]) and g(x) = 1

3r on f −1

0 ([ 13

r, r]). From

f −10 ([−r, r]) = f −10 ([−r, −1

3 r]) g=− 1

3r

∪ f −10 ([−1

3 r, 1

3 r]) |g|≤ 1

3r

∪ f −10 ([1

3 r, r]) g= 1

3r

we see that the second inequality of (26.6) is satisfied.

Proof of 26.4. We shall first prove the theorem for functions from A to [0, 1] or rather to thehomeomorphic interval [−1, 1] which is more convenient for reasons of notation.

Given the continuous function f : A → [−1, 1] we shall now recursively define a sequence g1, g2, · · ·of continuous functions X → R such that

(26.7) ∀x ∈ X : |gn(x)| ≤ 1

3

2

3

n−1

and ∀a ∈ A : |f (a) −n

i=1

gi(a)| ≤

2

3

n

To obtain g1 we apply 26.5 with f 0 = f and r = 1. Suppose that g1, . . . , gn−1 have been defined. If

we apply 26.5 with f 0 = f −n−1i=1 gi

|A and r = 23n−1 then we obtain a function gn satisfying

(26.7).By the first inequality in (26.7), the series

∞n=1 gn(x) converges uniformly. The sum function

F = ∞

n=1 gn is continuous by the uniform limit theorem (17.15). By the second inequality in(26.7), F (a) = f (a) for all a ∈ A.

Assume now that f : A → (−1, 1) maps A into the open unit interval. We know that we canextend f to a continuous function F 1 : X → [−1, 1] into the closed interval. We want to modify F 1so that it does not take the values −1 and 1 and stays the same on A. The closed sets F −1

1 (±1)and A are disjoint. There exists (26.1) a Urysohn function U : X → [0, 1] such that U = 0 onF −1

1 (±1) and U = 1 on A. Then F = U · F 1 is an extension of f that maps X into (−1, 1).A similar procedure applies to functions f : A → [−1, 1) into the half-open interval.

The properties mentioned in 26.1 and 26.4 characterize normal spaces among T 1

-spaces.

Corollary 26.8. Let X be a T 1-space. Then the following conditions are equivalent:

(1) X is normal (2) For any pair A, B of disjoint closed subspaces of X there exists a continuous function

f : X → [0, 1] such that f (A) = 0 and f (B) = 1(3) For any closed subset A of X and any continuous function f : A → [0, 1] there exists a

continuous function F : X → [0, 1] such that F |A = f .

Proof. The Urysohn lemma (26.1) says that (1) =⇒ (2) and the converse is clear. The Tietzeextension theorem (26.4) says that (1) =⇒ (3). Only (3) =⇒ (1) remains. Assume that X is a T 1-space with property (3). Let A, B be two disjoint closed subsets. The function f : A ∪ B → [0, 1]given by f (A) = 0 and f (B) = 1 is continuous (15.5.(6)). Let F : X → [0, 1] be a continuousextension of f . Then A

⊂ f −1[0, 1

4) and B

⊂ f −1( 3

4, 1] where f −1[0, 1

4) and f −1( 3

4, 1] are open and

disjoint. Hence X is normal.

Many familiar classes of topological spaces are normal.

Theorem 26.9. Compact Hausdorff spaces are normal.

Proof. See 20.8.(2).

In fact, every regular Lindelof space is normal [6, Ex 32.4] [4, 3.8.2].

Theorem 26.10. Metrizable spaces are normal.

Proof. Let X be a metric space with metric d and let A, B be disjoint closed sets. The continuousfunction f : X → [0, 1] given by

f (x) = d(x, A)

d(x, A) + d(x, B)is a Urysohn function with f (A) = 0 and f (B) = 1. This shows that X is normal (26.8).



GENERAL TOPOLOGY 57

Theorem 26.11. [4, Problem 1.7.4]. Linearly ordered spaces are normal.

Proof. We shall only prove the special case that every well-ordered space is normal. The half-openintervals (a, b], a < b, are (closed and) open (14.2.(5)). Let A and B be two disjoint closed subsets

and let a0 denote the smallest element of X . Suppose that neither A nor B contain a0. For anypoint a ∈ A there exists a point xa < a such that (xa, a] is disjoint from B . Similarly, for any pointb ∈ B there exists a point xb < b such that (xb, b] is disjoint from A. The proof now proceeds asthe proof (25.4) for normality of R. Suppose next that a0 ∈ A ∪ B, say a0 ∈ A. The one-point seta0 = [a0, a+

0 ) is open and closed (as X is Hausdorff). By the above, we can find disjoint opensets U , V such that A − a0 ⊂ U and B ⊂ V . Then A ⊂ U ∪ a0 and B ⊂ V − a0 where theopen sets U ∪ a0 and V − a0 are disjoint.

In particular, the well-ordered spaces S Ω and S Ω are normal (the latter space is even compactHausdorff).

27. Second countable regular spaces and the Urysohn metrization theorem

We investigate closer the class of regular spaces. We start with an embedding theorem that is

used in other contexts as well.

27.1. An embedding theorem. We discuss embeddings into product spaces.

Definition 27.2. A set f j : X → Y j | ∈ J of continuous functions is said to separate points and closed sets if for any point x ∈ X and any closed subset C ⊂ X we have

x ∈ C =⇒ ∃ j ∈ J : f j(x) ∈ f (C )

Lemma 27.3. Let f : X → Y be a map that separates points and closed sets. Assume that X is T 1 or that f is injective. Then f is an embedding.

Proof. The map separates points since points are closed in the T 1-space X . In particular, f isinjective. The assumption is that

f (x)

∈ f (C ) =

⇒ x

∈ C

when x ∈ X and C ⊂ X is closed. This implies that f (X ) ∩ f (C ) ⊂ f (C ). Then actually

f (X ) ∩ f (C ) = f (C ) for the other inclusion is obvious. But this equality says that f (C ) is closedin f (X ). Hence the continuous map f : X → f (X ) is a homeomorphism.

Theorem 27.4 (Evaluation embedding theorem). Let f j : X → Y j | ∈ J be a family of con-tinuous functions that separates points and closed sets. Assume that X is T 1 or that at least one of the functions f j is injective. Then the evaluation map f = (f j) : X →

j∈J Y j is an embedding.

Proof. Also the map f separates points and closed sets because

f (x) ∈ f (C ) =⇒ ∀ j ∈ J : πjf (x) ∈ πj(f (C )) 15.3.(6)

=⇒ ∀ j ∈ J : πjf (x) ∈ πj(f (C ))

⇐⇒ ∀ j

∈ J : f j(x)

∈ f j(C )

(f j) separates=

⇒ x

∈ C

for all points x ∈ X and all closed subsets C ⊂ X . The theorem now follows from 27.3.

In particular, if one of the functions f j is injective and separates points and closed sets thenf = (f j) is an embedding. For instance, the graph X → X × Y : x → (x, g(x)) is an embeddingfor any continuous map g : X → Y .

27.5. A universal second countable regular space. We say that a space X is universal forsome property if X has this property and any space that has this property embeds into X .

Theorem 27.6 (Urysohn metrization theorem). The following conditions are equivalent for a second countable space X :

(1) X is regular (2) X is normal

(3) X is metrizable The Hilbert cube [0, 1]ω is a universal second countable metrizable (or normal or regular) space.



58 GENERAL TOPOLOGY

Proof. (1) =⇒ (2): Let X be a regular space with a countable basis B. We claim that X is normal.Let A and B be disjoint closed sets in X . By regularity (25.2.(1)), each point a ∈ A ⊂ X − B hasa basis neighborhood U a ∈ B such that U a is disjoint from B. Let U 1, U 2, . . . be the elements inthe image of the map A

→ B: a

→ U a. Then

A ⊂∞

n=1

U n and U n ∩ B = ∅ for n = 1, 2 . . .

Similarly, there is a sequence V 1, V 2, . . . of basis open sets such that

B ⊂∞

n=1

V n and V n ∩ A = ∅ for n = 1, 2 . . .

The open sets ∞

n=1 U n and∞

n=1 V n may not be disjoint. Consider instead the open sets

U 1 = U 1 − V 1 V 1 = V 1 − U 1

U 2 = (V 1 ∪ V 2) V 2 = V 2 − (U 1 ∪ U 2)

..

.

..

.U n = U n − (V 1 ∪ · · · ∪ V n) V n = V n − (U 1 ∪ · · · ∪ U n)

Since U n covers as much of A as does U n and V n covers as much of B as does V n we have

A ⊂∞

n=1

U n and B ⊂∞

n=1

V n

where∞

n=1

U n ∩∞

n=1

V n =

m≤n

U m ∩ V n

∪

m>n

U m ∩ V n

= ∅

because U m ⊂ U m does not intersect V n ⊂ X − U m if m ≤ n and U m ⊂ X − V n does not intersectV n ⊂ V n if m > n.

(2) =⇒ (3): Let X be a normal space with a countable basis B. We show that there is a countableset f UV of continuous functions X → [0, 1] that separate points and closed sets.For any closed subset C and any point x ∈ C , or x ∈ X − C , there are (25.2) basis open sets

U , V such that x ∈ U ⊂ U ⊂ V ⊂ X − C (choose V first). Let f UV : X → [0, 1] be a Urysohnfunction (26.1) such that f UV (U ) = 0 and f UV (X − V ) = 1. Then f UV (x) = 0 and f UV (C ) = 1 so

that f UV (x) ∈ f UV (C ). Since there are only countably many basis sets there are only countablemany of these Urysohn functions. According to the Diagonal embedding theorem (27.4) there isan embedding X → [0, 1]ω with the f UV as coordinate functions. Since [0, 1]ω is metrizable, thesubspace X is metrizable (17.5).(3) =⇒ (1): Any metrizable space is regular, even normal (26.10).

The point is that in a second countable regular hence normal space there is a countable set of (Urysohn) functions that separate points and closed sets. Therefore such a space embeds in the

Hilbert cube.We have (27.6, 24.2) the following identities

2nd countableregular

=

2nd countable

normal

=

2nd countable

metrizable

=

Countable dense subset

metrizable

=

Lindelof metrizable

=

Subspace of Hilbert cube

between classes of topological spaces.



GENERAL TOPOLOGY 59

28. Completely regular spaces and the Stone–Cech compactification

There is no version of the Tietze extension theorem (26.4) for regular spaces. Instead we havea new class of spaces.

Definition 28.1. A space X is completely regular if points are closed in X and for any closed subset C of X and any point x ∈ C there exists a continuous function f : X → [0, 1] such that f (x) = 0 and f (C ) = 1.

Clearlynormal =⇒ completely regular =⇒ regular =⇒ Hausdorff =⇒ T 1

and none of these arrows reverse (25.5, [6, Ex 33.11]).In a completely regular space there are enough continuous functions X → [0, 1] to separate

points and closed sets.

Theorem 28.2. The following conditions are equivalent for a topological space X :

(1) X is completely regular (2) X is homeomorphic to a subspace of [0, 1]J for some set J (3) X is homeomorphic to a subspace of a compact Hausdorff space

Proof. (1) =⇒ (2): If X is completely regular then the set C(X ) of continuous maps X → [0, 1]separates points and closed sets. The evaluation map

∆ : X → j∈C(X)[0, 1], πj(∆(x)) = j(x), j ∈ C(X ), x ∈ X,

is therefore an embedding (27.4).(2) =⇒ (3): [0, 1]J is compact Hausdorff by the Tychonoff theorem (20.17).(3) =⇒ (1): A compact Hausdorff space is normal (26.9) and hence completely regular. It is easyto see that any subspace of a completely regular space is completely regular.

Corollary 28.3. (Cf 14.19, 25.3) Any subspace of a completely regular space is completely regular.Any product of completely regular spaces is completely regular.

Proof. The first part is easily proved (and we used it above). The second part follows from (28.2)

because (15.9) the product of embeddings is an embedding.

Corollary 28.4. Any locally compact Hausdorff space is completely regular.

Proof. Locally compact Hausdorff spaces are open subspaces of compact spaces (23.4). Completelyregular spaces are subspaces of compact spaces (28.2).

Example 28.5 (A normal space with a non-normal subspace). Take any completely regular butnot normal space (for instance (25.5) R×R) and embed it into [0, 1]J for some set J (28.2). Thenyou have an example of a normal (even compact Hausdorff) space with a non-normal subspace.

28.6. The Stone–Cech construction. For any topological space X let C(X ) denote the set of continuous maps j : X → I of X to the unit interval I = [0, 1] and let

∆ : X →j∈C(X) I = map(C(X ), I )

be the continuous evaluation map given by ∆(x)( j) = j (x) or πj∆ = j for all j ∈ C(X ). (The spacemap(C(X ), I ) is a kind of double-dual of X ). This construction is natural: For any continuousmap f : X → Y of X into a space Y , there is an induced map C(X ) ← C(Y ) : f ∗ of sets and yetan induced continuous map

(28.7) map(C(X ), I ) =

j∈C(X)

I f ∗∗−−→

k∈C(Y )

I = map(C(Y ), I )

given by πkf ∗∗ = πf ∗(k) = πkf for all k ∈ C(Y ) (a kind of projection map). The map f ∗∗ is anextension of f in the sense that the diagram

(28.8) X f

∆X

Y

∆Y

j∈C(X) I

f ∗∗

k∈C(Y ) I



60 GENERAL TOPOLOGY

is commutative: πkf ∗∗∆X = πkf ∆X = kf = πk∆Y f .

Put βX = ∆(X ) and define ∆: X → βX to be the corestriction of ∆: X → j∈C(X) I to βX .

Then βX is a compact Hausdorff space (it is a closed subspace of the compact (20.17) Hausdorff space I ) and ∆ is, by design, a continuous map with a dense image ∆ X in βX . This construction

is natural: For any continuous map f : X → Y , the induced map f ∗∗ (28.7) takes β X into β Y for

f ∗∗(βX ) = f ∗∗(∆XX ) ⊂ f ∗∗∆XX (28.8)

= ∆Y f (X ) ⊂ ∆Y Y = βY

and thus we obtain from (28.8) a new commutative diagram of continuous maps

(28.9) X f

∆X

Y

∆Y

βX

βf βY

where βf : β X → βY is the corestriction to βY of the restriction of f ∗∗ to βX . (This makes β into

a functor with a natural transformation from the identity functor from the category of topologicalspaces to the category of compact Hausdorff spaces.)The next result says that X → βX is the universal map of X into a compact Hausdorff space.

Theorem 28.10. Let X be a topological space.

(1) The map X → βX is a continuous map of X into a compact Hausdorff space.(2) For any continuous map f : X → Y of X into a compact Hausdorff space Y there exists a

unique continuous map f : β X → X such that the diagram

(28.11) X

f Y

βX f

commutes. (We say that f : X → Y factors uniquely through βX .)(3) Suppose X → αX is a map of X to a compact Hausdorff space αX such that any map

of X to a compact Hausdorff space factors uniquely through αX . Then there exists a homeomorphism αX → βX such that

X

αX βX

commutes.(4) If X is completely regular then X → βX is an embedding. If X is compact Hausdorff then

X → βX is a homeomorphism.

Proof. (4): Since the corestriction of an embedding is an embedding (15.9) and X → I is an

embedding when X is completely regular (28.2), also X → βX is an embedding. If X is compactHausdorff, X is normal (26.9), hence completely regular, so we have just seen that ∆: X → βX isan embedding. The image of this embedding is closed (20.9.(1)) and dense. Thus the embeddingis bijective so it is a homeomorphism.(1) and (2): If Y is compact Hausdorff then by item (4) ∆Y is a homeomorphism in diagram (28.9)and hence f = ∆−1

Y βf is a possibility in (28.11). It is the only possibility [6, Ex 18.13], for theimage of X is dense in β X .(3) Let X → αX be a map to a compact Hausdorff space satisfying the above universal property.

Then there exist continuous maps αX βX that, by uniqueness, are inverse to each other.

Definition 28.12. A compactification of the space X is an embedding c : X → cX of X into a compact Hausdorff space cX such that c(X ) is dense in cX .



GENERAL TOPOLOGY 61

If X is completely regular then X → βX is a compactification (28.10.(4)) and it is called theStone–Cech compactification of X . Conversely, if X has a compactification then X is homeo-morphic to a subspace of a compact space and therefore (28.2) completely regular. We state theseobservations in

Lemma 28.13. X has a compactification if and only if X is completely regular.

The Stone–Cech compactification β X of a completely regular space X is the maximal compac-tification in the sense that for any other compactification X → cX there is a closed quotient mapβX → cX map such that

X

βX cX

commutes. The map βX → cX is closed because it is continuous map of a compact space into aHausdorff space (20.9.(1)) and surjective because its image is closed and dense.

The Alexandroff compactification ωX (23.2) of a noncompact (!) locally compact Hausdorff

space X is the minimal compactification in the sense that ωX = cX/(cX − X ) for any othercompactification X → cX . Indeed, there is a closed quotient map cX → ω X , taking cX − X toω, such that

X

cX ωX

commutes. The fact that X is open (23.9) in cX implies (as in the proof of 23.2) that this map iscontinuous. (What is the minimal compactification of a compact Hausdorff space?)

For instance, there are quotient maps

[0, 1]

β R

ωR = S 1

C

of compactifications of R. Here

C = 0 × [−1, 1] ∪ (x, sin 1

x) | 0 < x ≤ π−1 ∪ L

is the Warsaw circle, a compactification of R with remainder C −R = [−1, 1], which is obtained byclosing up the closed topologist’s sine curve S by (a piece-wise linear) path L from (0, 0) to (π−1, 0).In particular, C/[−1, 1] = S 1. More generally, there are quotient maps β Rn → [0, 1]n → ωRn = S n

of compactifications of euclidian n-space Rn.Investigations of the Stone–Cech compactification of the integers β Z raise several issues of a

very fundamental nature [9, 10, 8] whose answers depend on your chosen model of set theory.

29. Manifolds

Definition 29.1. A manifold is a locally euclidian second countable Hausdorff space.

Manifolds are locally compact and locally path connected since they are locally euclidian. Theyare also metrizable, hence they satisfy all countability axioms (24.1) and all separation axioms(25.1):

manifold =⇒

locally compact Hausdorff 2nd countable

28.4=⇒

completely regular

2nd countable

27.6,24.2=⇒

metrizableLindelof

2nd countablecountable dense subset

=

⇒ metrizable

26.10=⇒

normal



62 GENERAL TOPOLOGY

The line with two zeroes is an example of a locally euclidian space that is not Hausdorff. Thelong line [6, Ex 24.12] is a locally euclidian space that is not 2nd countable.

The n–sphere S n and the real projective n–space RP n (16.13) are examples of manifolds. S n

lies embedded in Rn+1 from birth but what about RP n? Does RP n embed in some euclidian

space?

Theorem 29.2 (Embeddings of compact manifolds). Any compact manifold embeds in RN for some N .

The support of a continuous function φ : X → R is the closed set

supp φ = x ∈ X | f (x) > 0The function φ is constant with value 0 outside its support.

Theorem 29.3 (Partition of unity). Let X = U 1 ∪ · · · ∪U k be a finite open covering of the normal space X . The there exist continuous functions φi : X → [0, 1], 1 ≤ i ≤ k, such that

(1) supp φi ⊂ U i(2) k

i=1 φi(x) = 1 for all x

∈ X .

Proof. We show first that there is an open covering V i of X such that V i ⊂ U i. Since the closedset X − (U 2 ∪ · · · ∪ U k) is contained in the open set U 1 there is an open set V 1 such that

X − (U 2 ∪ · · · ∪ U k) ⊂ V 1 ⊂ V 1 ⊂ U 1

by normality (25.2). Now V 1 ∪ U 2 ∪ · · · ∪ U k is an open covering of X . Apply this procedure onceagain to find an open set V 2 such that V 2 ⊂ U 2 and V 1 ∪ V 2 ∪ U 3 ∪ · · · ∪ U k is still an open coveringof X . After finitely many steps we have an open covering V i such that V i ⊂ U i for all i.

Do this one more time to obtain an open covering W i such that W i ⊂ W i ⊂ V i ⊂ V i ⊂ U i forall i. Now choose a Urysohn function (26.1) ψi : X → [0, 1] such that ψi(W i) = 1 and ψi(X −V i) =

0. Then supp ψi = x | ψi(x) > 0 ⊂ V i ⊂ U i and ψ(x) =

ψi(x) > 1 for any x ∈ X since x ∈ W ifor some i. Hence φi = ψi

ψ is a well-defined continuous function on M taking values in the unit

interval such that k

i=1 φi(x) = k

i=1

ψi(x)

ψ(x) =

1

ψ(x) k

i=1 ψi(x) =

ψ(x)

ψ(x) = 1.

Proof of 29.2. Let M be a compact manifold. Cover M by finitely many open sets U i, 1 ≤ i ≤ k,homeomorphic to open sets in Rn by embeddings gi : U i → Rn. Let φi be a partition of unityfor the open covering U i. Extend gi to a function g i : M → Rn given by

gi(x) =

φi(x)gi(x) x ∈ U i

0 x ∈ U i

The extension gi is continuous because the restrictions to the open sets U i and X − supp φi arecontinuous (15.5.(5)). We claim that the continuous map

F : M →k

R × · · · × R ×

k

Rn × · · · × Rn = Rk(n+1)

x → (φ1(x), . . . , φk(x), g1(x), . . . , gk(x))is an embedding. Since M is compact and Rk(n+1) is Hausdorff it is enough (20.9.(2)) to showthat F is injective: Suppose that F (x) = F (y). Choose i such that φi(x) > 0. Then φi(y) = φi(x)is also positive so x, y ∈ U i. But g i|U i = φi · gi|U i is injective so x = y.

In particular RP 2 (16.13) embeds into some euclidian space. Which one?There is up to homeomorphism just one compact manifold of dimension 1, the circle. Compact

manifolds of dimension 2 are described by a number, the genus, and orientability. In particular,any simply connected 2-dimensional manifold is homeomorphic to S 2. We do not know (December2003) if there are any simply connected 3-dimensional manifolds besides S 3 (Poincare conjecture).Classification of 4-dimensional compact manifolds is logically impossible.



GENERAL TOPOLOGY 63

References

[1] Colin C. Adams, The knot book , W. H. Freeman and Company, New York, 1994, An elementary introductionto the mathematical theory of knots. MR 94m:57007

[2] Nicolas Bourbaki, General topology. Chapters 1–4, Elements of Mathematics, Springer-Verlag, Berlin, 1998,

Translated from the French, Reprint of the 1989 English translation. MR 2000h:54001a[3] Glen E. Bredon, Topology and geometry , Graduate Texts in Mathematics, vol. 139, Springer-Verlag, New York,

1993. MR 94d:55001[4] Ryszard Engelking, General topology , second ed., Sigma Series in Pure Mathematics, vol. 6, Heldermann Verlag,

Berlin, 1989, Translated from the Polish by the author. MR 91c:54001[5] Ryszard Engelking and Karol Sieklucki, Topology: a geometric approach , Sigma Series in Pure Mathematics,

vol. 4, Heldermann Verlag, Berlin, 1992, Translated from the Polish original by Adam Ostaszewski. MR94d:54001

[6] James R. Munkres, Topology. Second edition , Prentice-Hall Inc., Englewood Cliffs, N.J., 2000. MR 57 #4063[7] Edwin H. Spanier, Algebraic topology , Springer-Verlag, New York, 1981, Corrected reprint. MR 83i:55001[8] Juris Steprans, The autohomeomorphism group of the ˇ Cech-Stone compactification of the integers, Trans. Amer.

Math. Soc. 355 (2003), no. 10, 4223–4240 (electronic). MR 1 990 584[9] Jan van Mill, An introduction to βω, Handbook of set-theoretic topology, North-Holland, Amsterdam, 1984,

pp. 503–567. MR 86f:54027[10] Russell C. Walker, The Stone- ˇ Cech compactification , Springer-Verlag, New York, 1974, Ergebnisse der Math-

ematik und ihrer Grenzgebiete, Band 83. MR 52 #1595[11] W. Hugh Woodin, The continuum hypothesis, Part I , Notices of the American Math. Soc. 48 (2001), no. 6,

567–576.[12] , The continuum hypothesis, Part II , Notices of the American Math. Soc. 48 (2001), no. 7, 681–690.

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