introTopology

8/4/2019 introTopology

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Notes on Topology

Alex Nelson

May 31, 2009

Contents

1 Introduction 2

I Act I: Topological Spaces 2

2 Properties of Sets 2

3 Topological Spaces 3

4 Basis For Topological Spaces 54.1 Given a Topological Space, Finding a Basis. . . . . . . . . . . . . . . 74.2 Example of Usefulness of Bases. . . . . . . . . . . . . . . . . . . . . . 84.3 A Sub-Basis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

5 Continuous Functions 8

6 Closed Sets 10

II Act II: Construction of Topological Spaces 11

7 The Order Topology 11

8 The Product Topology 12

III Act III: Properties of Topological Spaces 14

9 Connectedness 14

Abstract

Topology is the study of boundaries. We usually think of an open set

as not containing its boundary, and a closed set as one that contains its

boundary. We generalize these notions in topology suitably.

Email: [email protected]

1
mailto:[email protected]:[email protected]


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2 Properties of Sets 2

1 Introduction

Loosely speaking, basic topology is concerned with boundaries. What is an openset? When is a set in one piece? When can we put a set inside a box? Whatis a boundary? What is a neighborhood? And so on.

The basic approach to introducing topology will be done in three acts. The

first act will introduce what is a topology, basic definitions, various operationson sets, neighborhoods, and bases for topologies. The second act will deal withthe construction of topological spaces, specifically subspaces and product spaces.The third will deal with a variety of properties a topological space can have (e.g.compactness, connectedness, countability and seperation axioms, etc.).

Part I

Act I: Topological Spaces

2 Properties of Sets

Just a review of a few properties of sets, and some basic approaches to proofs.

Proposition 1. For any set A

1. The empty set is a subset of A:

A, A.

2. The union of A with the empty set is A

A = A, A.

3. The intersection of A with the empty set is the empty et

A = , A.

4. The Cartesian product of A and the empty set is empty

A = , A.

5. The only subset of the empty set is itself

A A = , A

6. The power set of the empty set is a set containing only the empty set

2

= {}

7. The empty set has cardinality zero (it has zero elements). Moreover, theempty set is finite

|| = 0.


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Given some statementIf P, then Q. (2.1)

its contrapositive is

If Q is not true, then P is not true. (2.2)

Its converse isIf P, then Q. (2.3)

Sometimes its easier to prove the converse than to prove a given proposition.To show that two sets A, B are equal, we need to show A B and B A.

3 Topological Spaces

Problem 2. The fundamental aim of topology, for all practical purposes, is togeneralize real analysis to sets other than R. We begin by generalising the notionof what an open set is. That is, previously we thought of an open set as a setthat doesnt include its boundary, for example

S = {x R : 1 < x < 1} (3.1)

is an open interval since it doesnt include -1 or 1. We wish to generalize this notion,so lets begin by introducing a collection of open subsets of a given set X called atopology on X.

Definition 3. A Topology on a set X is a collection T of subsets of X havingthe following properties:

1. and X are in T

2. The union of the elements of any subcollection of T is in T

3. The intersection of the elements of any finite subcollection of T is in T.

A set X for white a topology has T been specified is called a topological space.

We generalize the notion of an open set to be any subset U X such thatU T. This weakens our previous notion of an open set, i.e. this topologicaldefinition includes our previous intuition about open sets and extends beyond it.We call elements of X points of the space. In fact, we will define open sets aselements of T:

Definition 4. Let (X, T) be a topological space. Then the members of T aredefined as open sets.

Now this definition of a topological space is rather abstract. Lets consider a fewexamples using finite sets.

Example 5. Consider the set X = {a, b} where a, b are arbitrary objects. Thepossible topologies on this are

1. {, {a, b}} This is trivially a topology, since it includes X, . Their unionX = X, their intersection X = are both in the topology. This isdubbed the trivial topology. The only open sets are empty and X.


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a b a b

a b a b

Figure 1: An example of all the possible topologies on the set X = {a, b} corre-sponding to (from top to bottom, left to right) the discrete topology, {, {b}, X},{, {a}, X}, and the trivial topology.

2. {, {a}, {a, b}} This is also a topology, since {a} {a, b} = {a}, {a} X = X,{a} = {a}, {a} = . All of these are in the topology.

3. {, {b}, {a, b}} This is just switching a with b, which is also a topology.

4. {, {a}, {b}, {a, b}} This is also a topology, its all possible subsets ofX. Thisis given a special name known as the power set or discrete topology.Here, every subset is open.

We can see a diagrammatic scheme of these topologies in fig (1).

Example 6. This is a nonexample. Consider the set X = {a,b,c}. The collection

T = {, {a, b}, {b, c}, X} . (3.2)

Observe that{a, b} {b, c} = {b} / T (3.3)

So its not closed under finite intersections, so T isnt a topology.

Proposition 7. If we have some topological space (X, T), and for each x X wehave {x} T, then T is the discrete topology.

Proof. We want to show that T is the power set of X. We know that arbitraryunions of subsets of X in T is also in T, and by hypothesis all singletons areelements of T. It follows that any arbitrary subset of U X can be written as anarbitrary union of singletons

U =xU

{x}. (3.4)

Since this is any arbitrary subset of X, it follows that every subset of X is in T.Thus T is the discrete topology.

Definition 8. Let T, T be two topologies on the same set X. We say that

T T

T is finer than T

T is coarser than T.(3.5)

Remark 9. This notion of finer and coarser topologies are worthy of explana-tion. The typical explanation is to think of sand paper, or gravel. Munkres [3] givesthe example of having a topology be visualized by a collection of large rocks. If webreak it up into finer gravel, we can union them together to get back the coarsertopology and it takes up less volume. Intuitively, this is a finer configuration ofpebbles.


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4 Basis For Topological Spaces 5

4 Basis For Topological Spaces

Problem 10. Oftentimes, a topology is cumbersome to work with, so we wantto make life easier. In linear algebra, we dont work with vectors aribtrarily, wework with linear combinations of basis vectors. We use an analogous concept fortopological spaces. We work with basis elements when discussing topological prop-

erties, it simplifies life a bit. For us, however, our basis elements are not vectors!Instead, they are a selected collection of open subsets of a given topological spaceX with some extra properties. They are the basic building blocks for a topology,so intuitively if all the building blocks share a property, the topology should haveit too.

Definition 11. If X is a set, a basis for a topology on X is a collection B ofsubsets of X (or basis elements) such that

1. For each x X, there is at least one basis element B containing x.

2. If x belongs to the intersection of two basis elements B1, B2, then there is abasis element B3 such that

x B3 B1 B2. (4.1)

IfB satisfies these two conditions, then we define the topology generated bythe basis B by first saying an open set U X is in the topology generated byB iff x U, there is a B B such that x B U.

Remark 12. We can alternatively say that if an open set U in the topologygenerated by B, then

U =J

B (4.2)

where J is some indexing set, B B are basis elements.

Remark 13. A few notes on the properties of the basis for a topology.

1. The first property is basically saying the union of all our basis elements coversX. There are no holes in our topology.

2. The second property is a bit more abstract. We want intersections to becovered by some number of basis elements. If we didnt have this property,the intersection of two basis elements wouldnt be an open set, which impliesthat basis elements are not open sets. This is bad, it would imply that thetopology generated by a basis isnt really a topology!

Problem 14. Note that the definition of the topology generated is not yet shownto be a topology, we need to prove that it is a topology.

Theorem 15. Let B be a basis for a topology on X. Then T, the topologygenerated by B, is a topology.

Proof. We need to prove that the topology generated by a given basis satisfiesthe properties of a topology. We will do this property by property. Let B be thebasis for a topology on X, T by the topology generated by B.

1. (Contains , X)

(a) We see that T is vacuously true.


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(b) We see that for each x X there is a Bx B such that x Bx. Itfollows that

X =xX

{x} xX

Bx (4.3)

Since each Bx X, it follows that

xX

Bx X (4.4)

Thus the union of all basis elements is X.

So both T and X T.

2. (Closure under arbitrary unions) We see that if {U}J is a collection ofopen sets in T and J is some index set, then we can write

I

B = U (4.5)

by virtue of the definition of an open set in the topology generated by B.

IfI =

J

I (4.6)

then I

B =J

I

B

=

J

U (4.7)

is also, by definition, a set in T.

3. (Closure under finite intersections) Let {U}n=1 be some finite collection of

open sets in T. We see that by definition of an open set in the topologygenerated by B

U = I

B (4.8)

for some index set I, and basis elements B B. We see then that thefinite intersection is then

n=1

U =n

=1

I

B

. (4.9)

Let

I =n

=1

I (4.10)

thenn

=1

U =I

B . (4.11)

But this is by definition an open set in the topology generated by B.

Thus the topology generated T byB satisfies all the properties of a topology.


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Lemma 16. Let X be a set, let B be a basis for a topology T on X. Then Tequals the collection of all unions of elements ofB.

Proof. Given any arbitrary U T, we can write it as

U =xU

Bx (4.12)

where x Bx U is a basis element. We know such an element exists by property1 of a basis element, and we know it is contained in U by property 2 of a basiselement.

We have that any element of T can be written as an arbitrary union of basiselements, we need to show that it contains every union of basis elements. We knowby definition, every basis element is open. We also know that arbitrary union ofopen sets is open. By definition, open sets are elements of a specified topology.Bases are defined for a given topology, which implies the arbitrary union of basiselements is in the given topology. So every union of basis elements is in the topologyT.

Remark 17. This lemma basically says that any open set U can be written asa union of basis elements. Note that it does not specify this union is necessarily

unique. This stands in stark contrast to linear algebra, where a vector is a uniquelinear combination of basis elements. It is an unfortunate fact of life that topologistschoose poor definitions.

Remark 18. So, just to review, if we have a basis B for the topological space(X, T), if U X, and if for each x U there is a Bx B such that x Bx U,then U is a union of basis elements.

4.1 Given a Topological Space, Finding a Basis.

Problem 19. But now that we have some notion of a basis for a topology, thequestion is How can we find a basis for a given topology?

Lemma 20. Let (X, T) be a topological space. Suppose that C is a collection of

open sets of X such that for each open set U of X and each x U, there is anelement C C such that x C U. Then C is a basis of X.

Proof. We need to do two things. First we need to show that C is a basis. Second,we need to show that the topology generated by C, call it T, is really just T.

1. We see that given each x X, there is at least one C C containing it. Thisis, by hypothesis, true. We need to show that if x C1 C2, then there isa C3 C such that x C3 C1 C2. We see that C1 C2 is open, so byhypothesis there is a C3 C1 C2 such that x C3. This means that C isa basis.

2. We need to show that T = T.

(a) We see that any element U T is such that for some x U, there is a

C C

such that x C U so U T

by definition.(b) We also see that any element V T is such that it is the union of basis

elements in C, by lemma (16). But each C C is such that C T.This means that V (as a union of elements of T) is also in T.

This implies that T = T, as desired.

Thus C is a basis for the topology T on X.


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4.2 Example of Usefulness of Bases.

Problem 21. In the beginning of this section, we mentioned that it is useful to usebases instead of topologies since they allow us to prove properties faster. That is, ifwe can prove some topological property holds on each basis element (or equivalently,an arbitrary basis element), then it holds for the topology. We will illustrate theuse of bases in making such a claim.

Lemma 22. Let B, B be bases for the topologies T, T (respectively) on X.Then the following are equivalent:

1. T is finer than T.

2. For each x X and each basis element B B containing x, there is a basiselement B B such that x B B.

Proof. (2) (1). We see that an element U T can be written as the union ofbasis elements {B}J. But since, for each x U, there is a B

B

such thatx B B. Then B

U, which implies U T

.(1) (2). Given x X, and a B B that contains x. We see that B T. But by(1), T T so B T. So consider B as an open set that coincidentally contains a

point x. Then by definition ofB

as a basis, there is a basis element B

B

suchthat x B B.

Remark 23. The gist of this lemma boils down to this: given two bases on a givenset, and one is finer than the other, the topology generated by the finer basisis finer than the topology generated by the coarser basis.

4.3 A Sub-Basis.

Definition 24. Let (X, T) be a topological space. A Subbasis for the topologyon X is the collection of subsets of X whose union equals X. The TopologyGenerated by a Subbasis is defined as the collection T of all unions of finiteintersections of subbasis elements.

Remark 25. In the authors opinion, the previous definition is a bad one. Itappeals too much to linear algebraic intuition of a subbasis as a basis of a subspace,and that is not the case here. Instead, it should really have the intuition of aprebasis. That is, after a bit of manipulation we can get a basis. But it is not asthough anyone can change this definition now!

5 Continuous Functions

Problem 26. One of the interesting properties we had with functions in real anal-ysis was the notion of continuity. Is there a way to generalize this notion to atopological setting?

Definition 27. (Real Analysis Definition of Continuous) Let f : X Y be afunction, X and Y be sets. We say that f is Continuous if for each > 0 thereis a corresponding > 0 such that

|x x0| < |f(x) f(x0)| < (5.1)

Or in other words, for each neighborhood of f(x0), there is a neighborhood ofx0.


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Remark 28. Observe that what we are doing is specifying a neighborhood in therange of a certain size, then demanding there is a corresponding neighborhood inthe domain of a certain size. We can do this in real analysis since the reals aresufficiently nice (they are a metric space which is a really really strong conditionfor a topological space). We want to generalize this for topological spaces so whenwe work with metric spaces we recover our previous notion of continuity.

Definition 29. Let X, Y be topological spaces, f : X Y be a function. Wesay that f is continuous at x0 if for every neighborhood V of f(x), there is aneighborhood U of x such that f(U) V.

Remark 30. This notion of continuity is nearly identical to the notion we previ-ously introduced from real analysis. The difference is that we are not using a metricto keep track of every open neighborhood in the domain and the range.

But observe we demand each open neighborhood V off(x) has a correspondingneighborhood U in the domain such that the image of U is contained in V. This isprecisely what we did in the real analysis case, we specified an neighborhood, anddemanded that for each neighborhood there is a corrsponding neighborhoodin the domain such that the image of the neighborhood is contained in the neighborhood.

Theorem 31. Let X and Y be topological spaces, if f : X Y is such that foreach open set V Y its preimage f1(V) X is open, then f is continuous.

Proof. Trivial.

Remark 32. (Inverse Functions) Note that contrary to appearances continuitydoes not demand the function be invertible! The preimage of a set is not the sameas the functions inverse. Consider f : X Y, and Z Y is some open set. Then

f1(Z) = {x X : f(x) Z} (5.2)

which does not say anything about the existence of an inverse for f.

Definition 33. Let X, Z be topological spaces, and Y X be a subspace. A map

i : Y Xx i(x) = x

(5.3)

is defined as the Inclusion Map. Similarly, a continuous map

r : X Yx r(x) = x if x Y

(5.4)

is defined as the Retraction if the retraction of i to Y is the identity on A.

Remark 34. We use the inclusion map and the retraction to extend and restrictfunctions (respectively) by composing them with the functions of interest. Notethat the restriction of a continuous function (i.e. composing it with a retraction) is

the composition of two continuous functions and thus continuous.Definition 35. Let X, Zbe topological spaces, Y Xbe a subspace. Iff : X Zis a function, we can define the Restriction of f to Y as

f r : X Z (5.5)

the composition of the retraction map (which is just idY on Y) with f.


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6 Closed Sets 10

Remark 36. Observe that the restriction of a function is just composition offunctions.

Lemma 37. (Pasting Lemma) Let X and Y be topological spaces, and U, V beopen sets in X such that X = U V. Suppose f : U Y and g : V Y arecontinuous and

f(x) = g(x) for all x U V (5.6)

Then the function h : X Y defined by

h(x) =

f(x), if x U

g(x), if x V(5.7)

is continuous.

6 Closed Sets

Problem 38. We defined what it means for a set to be open by just demandingit be in a topology on a space. So how do we define a set to be closed? Intu-

itively, its an open set that contains its boundary, but what is a boundarytopologically?

Definition 39. Let X be a topological space, A X be some subset. We say thatA is Closed if X A is open.

Remark 40. This is kind of weaseling out of the problem, just define a set to beclosed if its compliment is open. This doesnt seem intuitive or immediately clearwhy one would want to define it this way, but we will see why later on.

Definition 41. Let X be a topological space, Y X.

1. The Closure of Y (denoted Y or closure(Y)) is the intersection of all closedsets containing Y.

2. The Interior of Y (denoted Y0 or int(Y)) is the union of all open sets con-tained in Y.

Remark 42. Observe that the closure of a set is closed, the interior of a set isopen.

Remark 43. The closure of a set is the smallest closed set containing it, but weconveniently avoided the problem of What do we mean by smallest? by usingintersections of closed sets. Similarly, the interior of a set is the largest open setcontained in it.

Remark 44. A set Y is closed iff Y = Y and it is open iff Y = Y0.

Definition 45. Let X be a topological space, A X be some subset. We define

the Boundary of A (denoted A) to be

A (X A) (6.1)

the intersection of the closure of A with the closure of the compliment of A.


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7 The Order Topology 11

Example 46. Consider the set X = {a,b,c}. Consider the topology

T = {, {a}, {b}, {a, b}, X} (6.2)

We see that {a, c} is closed since its compliment {b} is open, and we also see that{b, c} is closed since its compliment {a} is open. The intersection of these twoclosed sets {c} is closed, since its compliment is open and the intersection of closed

sets is closed.Observe the closure of{a} is given by the intersection of all closed sets containing

a:{a, c} {a,b,c} = {a, c} (6.3)

and the closure of {b, c} (that is, the compliment of {a}) is itself, i.e. {b, c} is aclosed set so its closure is itself. The boundary of {a} is then the intersection ofthese two sets

{a} = {b, c} {a, c} = {c} (6.4)

which we could not have found if we didnt define the boundary in a topologicalway!

Remark 47. Let X be a topological space, A X be a subset. The definition of

the boundary of A is the same as the closure of A minus its interiorA = A A0. (6.5)

How can we see this? Well, the interior of A is the union of all open sets containedin A. Its compliment would be the closure of X A. We see that the intersectionof (X A) with A is just the closure of A intersected with the compliment of itsinterior, i.e.

A (X A) = A

A0C

= A A0. (6.6)

This is by virtue of the property of compliments

(A B)C = A BC (6.7)

where A and B are subsets of some set U.

Definition 48. Let X be a topological space, Y X. A point x X is a LimitPoint of Y if every neighborhood of x intersects Y {x}.

Remark 49. This definition makes more sense given the notion of what a con-tinuous function is, since we defined continuity in the real analysis situation asthe limit as f(x) approaches f(x0). We take this notion, and use the topologicalnotion of continuity to concoct a topological notion of a limit.

Part II

Act II: Construction of Topological

Spaces

7 The Order Topology

Problem 50. Topologies seem more or less abstract. Isnt there some way tosimplify specifying open sets? We are going to introduce the notion of specifying


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open sets using open intervals. How do we specify open intervals? By using anorder relation a < b, we can specify open intervals.

If X is a simply ordered set, there is a standard topology for it defined usingthe order relation. Its called the order topology.

We should probably first generalize the notion of intervals familiar from realanalysis. Since X is a simply ordered set, there is a (simple) order relation a} (7.2a)

(, a) = {x|x < a} (7.2b)

[a, +) = {x|x a} (7.2c)

(, a] = {x|x a} (7.2d)

The first two are open rays, the last two are closed rays.

8 The Product Topology

Problem 54. If we have two topological spaces X and Y, can we glue themtogether? That is, given two topological spaces, we wish to construct a new one

using only what we know of X and Y.

Definition 55. Let X, Y be topological spaces. The Product Topology onX Y is the topology having as basis the collection B of all sets of the form U V,where U is an open subset of X and V is an open subset of Y.


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Remark 56. Being rigorous, we should probably verify that this B beast really isa basis. For any x y X Y, we see that X Y B so the first condition istrivially satisfied. The second condition, let (x y) U1 V1 and (x y) U2 V2.We see that B is the collection of the product of all open subsets, which allows usto see that

(U1 V1) (U2 V2) = (U1 U2) (V1 V2) (8.1)

is also the product of open sets, so its a basis element. Thus the second propertyof a basis is satisfied.

Theorem 57. IfB is a basis for the topology ofX and C is a basis for the topologyof Y, then the collection

D = {B C|B B, C C} (8.2)

is a basis for the topology of X Y.

Proof. We will use lemma 20 to prove this. That is, for some open set W X Yand each x y W there is a basis element U B such that x U and a basiselement V C such that y V, so x y U V W and U V D. Thus Dis a basis.

Definition 58. Let 1 : X Y X be defined by the equation

1(x, y) = x; (8.3)

let 2 : X Y Y be defined by the equation

2(x, y) = y. (8.4)

The maps 1 and 2 are called the projections of X Y onto its first and secondfactors, respectively.

Remark 59. Observe that projections are surjective provided that both X and Yare nonempty. If one is empty, X Y is empty too, and everything becomes trivial.

Remark 60. Note that if U X is open, 11 (U) = U Y. Similarly, if V Yis open, 12 (V) = X V. Their intersection is U V.

Theorem 61. The collection

S = {11 (U)|U is open in X} {12 (V)|V is open in Y} (8.5)

is a subbasis for the product topology on X Y.

Proof. Let T be the topology on X Y, T be the topology generated by oursubbasis. We see that each element W S is an open subset of the producttopology, which means that the topology generated by S is contained in T, i.e.

T T. (8.6)

We need to show that T T. We see given a basis element U V for T that

U V = 11 (U) 12 (V) (8.7)

which is nothing more than a finite intersection of subbasis elements. This impliesthat T T as desired.


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9 Connectedness 14

Part III

Act III: Properties of Topological

Spaces

9 Connectedness

Problem 62. How can we tell if a space is connected or not? That is, if itsdisconnected, then we should be able to break it up into independent pieces.How do we make this notion of disconnectedness rigorous?

Definition 63. Let X be a topological space. A Seperation of X is a pair U, Vof disjoint, nonempty, open sets whose union is X. The space X is said to beConnected if there is no seperation in it.

Remark 64. Observe that we didnt really define connectedness. Instead we de-fined what it means for it to have a seperation, and then proceeded to defineconnected as not seperated. So, to prove a topological space is connected is

equivalent to disproving the existence of a seperation. Consequently, we will useproof by contradiction a lot when dealing with connectedness.

Remark 65. Observe that if X is connected and Y is homeomorphic to X, thenY is necessarily connected.

Remark 66. A space X is connected if and only if the only closed subsets of Xthat are both open and closed are X and . Otherwise if we had some additionalset U that is closed and open, then X U is also closed and open. But U andX U are disjoint, nonempty, open subsets of X whose union is X. This couldnthappen if X were connected, so U needs to be empty or equal to all of X.

Lemma 67. IfY is a subspace ofX, a seperation ofY is a pair of disjoint nonemptysets A and B whose union is Y, neither of which contains a limit point of the other.

The space Y is connected if there exists no seperation of Y.Proof. Let A, B form a seperation of Y. We see that

A = A Y Y (9.1)

andAC = (A Y)C = B, (9.2)

which implies(A Y) B = . (9.3)

We can argue similarly for (B Y) A = .

Proposition 68. Let T, T be two topologies on X. If T T, then

1. the existence of a seperation in T implies existence of a seperation in T;

2. connectedness in T implies connectedness in T.

Proof.

1. Observe that if A, B forms a seperation in the T topology, then A, B T T. So the seperation is in the T topology too.


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9 Connectedness 15

2. Observe that ifX is connected in the T topology, then there is no seperation(no pair A, B that form a seperation of X that live in T). This means thatno such A, B exist in T, which implies there is no seperation of X in the Ttopology. So X is connected in the T topology.

Problem 69. How can we construct connected topological spaces? This seemsespecially daunting since we only can really know if a topological space is discon-nected.

Theorem 70. If the sets C and D form a seperation ofX, and ifY is a connectedsubspace of X, then Y lies entirely in C or entirely in D.

Proof. Assume for contradiction that Y can lie in both C and D. Then we canobserve that C Y and D Y are subsets of Y that are both open and closed, aswell as disjoint and their union would be Y. How can we see this last point? Well,we know since C and D form a seperation of X that

C D = X (9.4)

so it follows that

(C Y) (D Y) = (C D) Y = X Y = Y. (9.5)

We have our contradiction, it implies Y has a seperation. We reject our assumptionand say that Y has to lie entirely in C or entirely in D.

Theorem 71. The union of connected subspaces ofX with a common shared pointis connected.

Proof. Let {A} be our collection of connected subspaces,

Y = A (9.6)be the subspace we are trying to show is connected,

{y0} =

A (9.7)

be the point common to all connected subspaces.Assume for contradiction Y is disconnected. Then there exists a pair C, D of

disjoint, nonempty, open subsets of Y whose union is Y. Since A is connected, byour previous theorem, it must lie entirely in C or D.

Since all A share a common point, it implies that all A lie entirely in C orthey all lie entirely in D. If some lived entirely in C while the rest entirely in D,there would be no shared point y0 since C and D are disjoint.

So it follows that their union lies entirely in one. But this means that eitherC = and D = Y, or C = Y and D = , which is a contradiction of our assumptionthat C = and D = .

Theorem 72. Let A be a connected subspace of X. If A B A, then B isconnected.


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9 Connectedness 16

Proof. Assume for contradiction that B has a seperation. More precisely, there isa pair of nonempty disjoint open sets C, D whose union is B. Since A B, andA is connected, then A is contained entirely in C or it is contained entirely in D.Suppose it is contained entirely in C, then

C D A D A (9.8)

where A is the set of limit points of A. Recall a limit point x is such that

x A {x}. (9.9)

But observe thatx A {x} C {x} C (9.10)

but this implies thatA C B A C (9.11)

or equivalently, D = since we previously proven D C = , which is a contradic-tion.

Theorem 73. The image of a connected space under a continuous mapping is

connected.

Proof. Let f : X Y be continuous, X be connected. Assume for contradictionthere is a seperation (i.e. a pair A, B of nonempty, open, disjoint subsets whoseunion is Y) in Y. Their preimage would be a seperation in X, which contradicts ourhypothesis that X is connected. We reject our assumption that Y has a seperationand conclude it is connected.

Theorem 74. A finite Cartesian product of connected spaces is connected.

Proof. We will do a sort of inductive proof by contradiction.Base Case (n = 2): Lets consider two connected spaces X1 and X2. Let

Y = X1 X2. (9.12)

For contradiction, assume Y is disconnected. Then there is a pair A, B of disjoint,nonempty subsets of Y whose union is Y. This means that either 1(A), 1(B)forms a seperation ofX1 or 2(A), 2(B) forms a seperation ofX2. But we assumedthat Xi (i = 1, 2) was connected. So we have a contradiction, reject our assumptionthat Y is disconnected, and conclude Y is connected.Inductive Hypothesis (arbitrary n): Suppose this works for

Y = X1 Xn. (9.13)

Inductive Cast (n + 1): Lets consider X1, . . ., Xn+1 connected spaces, let

X = X1 Xn (9.14)

andY = X Xn+1. (9.15)

Then this is just the base case, and Y was shown to be connected.

Definition 75. A space is Totally Disconnected if its only connected sub-spaces are one-point sets.


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References 17

Theorem 76. IfX is a topological space equipped with the discrete topology, thenit is totally disconnected.

Proof. Since X has the discrete topology, we see that every subset of X is openand consequently its compliment (being a subset of X) is open, so every subset isalso closed. The only subspaces of X which has the only subsets be both open andclosed would necessarily be the one-point sets.

(Observe that an arbitrary set Y X with more than one element could bepartitioned into two subsets, which are both open and closed. This implies that Yis not connected.)

It then follows by definition that X is totally disconnected.

Proposition 77. Let Y X, let X and Y be connected. If A and B form aseperation of X Y, then A Y and B Y are connected.

Proof. We will prove that ifAY has a seperation, it means that X has a seperationin the sense of a seperation in a subspace (it is more general this way if X isnta subspace, its still a kosher seperation).

1. Let AB = XY, assume for contradiction that Z := AY has a seperationC D = Z.

2. We will show that B, C D form a seperation of X. Observe first thatC D = A Y.

3. We see that B A = B A = since A, B form a seperation of X Y.

4. We see that Y is closed since X Y = A B is the union of open sets (thusopen).

5. We see that Y B = Y B = .

6. We see that B X Y so B Y = .

7. Observe that

Z B = (A Y) B = (A Y) B = (9.16)

and similarlyZ B = (A Y) B = . (9.17)

8. We see that B and Z = A Y form a seperation of X, implying X is discon-nected. This is a contradiction. We have to reject the assumption that Z hasa seperation.

A similar argument holds for B Y being connected.

References

[1] Aisling McCluskey and Brian McMaster. Topology course lecture notes. http://at.yorku.ca/i/a/a/b/23.htm, 1997. Lecture Notes.

[2] Sidney A. Morris. Topology without tears. http://uob-community.ballarat.edu.au/~smorris/topology.htm, 2001. E-Book.

[3] James R. Munkres. Topology. Prentice Hall, second edition, 2000.
http://at.yorku.ca/i/a/a/b/23.htmhttp://at.yorku.ca/i/a/a/b/23.htmhttp://uob-community.ballarat.edu.au/~smorris/topology.htmhttp://uob-community.ballarat.edu.au/~smorris/topology.htmhttp://uob-community.ballarat.edu.au/~smorris/topology.htmhttp://uob-community.ballarat.edu.au/~smorris/topology.htmhttp://uob-community.ballarat.edu.au/~smorris/topology.htmhttp://uob-community.ballarat.edu.au/~smorris/topology.htmhttp://at.yorku.ca/i/a/a/b/23.htmhttp://at.yorku.ca/i/a/a/b/23.htm

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