Order Topology and Others

Zhou Qi

March 2, 2016

Abstract

This thesis is a summary of order topology, ordinal spaces and some exotictopological spaces. All conclusions in it are from the reference books and articles.

Given any set with a linear order, there will be a natural order topology onthe set. In the first chapter, I’ll introduce general properties of linearly orderedtopological spaces. In the second chapter I will introduce well-ordered sets, anddefined ordinals and ordinal spaces. By introducing some large ordinals, a fewtopological counterexamples will be given in the third chapter.

Assume that readers have some knowledge of basic set theory and mathe-matical analysis, so terms like ordered pair or supremum will not be explainedin this thesis. Also, general topology will be briefly reviewed. On that basis,this thesis is consistent, which means every other term will be introduced in adefinition or a footnote.

Contents

1 Order Topology 21.1 Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Order Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Ordinal Space 92.1 Well-Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Ordinal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Ordinal Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Counterexamples 163.1 Long Ray . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.2 Tychonoff Plank . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

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Chapter 1

Order Topology

1.1 Order

Definition 1.1.1. A binary relation between two sets A and B is a subset ofthe Cartesian product A×B, denoted by R. In common, 〈x, y〉 ∈ R is denotedby xRy.

Especially, A binary relation on a set A is a subset of A×A. In other words,it is a collection of ordered pairs of elements of A.

Definition 1.1.2. A partial order 6 is a binary relation on a set A, whichsatisfies:

1. a 6 a;

2. if a 6 b and b 6 a then a = b;

3. if a 6 b and b 6 c then a 6 c;

∀a, b, c ∈ A.

Note. 6 is not only a symbol but also a subset of A2, according to defini-tion 1.1.1.

Definition 1.1.3. In A, if a 6 b and a 6= b, we say a < b. In other words,

<=6 \{〈a, a〉 : ∀a ∈ A}

We call < a strict partial order on A.

Proposition 1.1.1. Strict partial order < has the following properties:

1. a 6< a;

2. a < b and b < a can’t be true together;

3. if a < b and b < c then a < c;

∀a, b, c ∈ A.

Proof. Trivial.

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Let B be a subset of a partially ordered set 〈A,6 〉, then there is a partialorder inherits naturally.

Definition 1.1.4. The induced partial order on B 6B is defined as 6 ∩B×B.

Definition 1.1.5. Two partially ordered sets A and B are called isomorphicor having the same order type, if there exists a bijection f : A→ B, so that

a1 6 a2 ⇐⇒ f(a1) 6 f(a2)

The bijection f is called an isomorphism.

Definition 1.1.6. A linear order is a certain partial order on A, which satisfies:∀a, b ∈ A, either a 6 b or b 6 a is true. In other words every two elements arecomparable.

The definition of strict linear order and induced linear order are just likethose of partial order, no further introduction.

Given two linear ordered sets A and B, we can define linear orders on theirdisjoint union1 A+B and their Cartesian product A×B.

Definition 1.1.7. A canonical linear order on A+B can be defined as:

6A+B =6A ∪ 6B ∪A×B

That means in addition to their original orders, a < b, ∀a ∈ A, b ∈ B.

Example 1.1.1. N+N, here N is the set of all natural numbers with its standardorder. Denote the numbers in the second set with a bar to distinguish them, sowe have 0 < 1, 0̄ < 1̄ and 1 < 0̄.

Definition 1.1.8. A canonical linear order on A×B can be defined as:

〈a1, b1〉 6 〈a2, b2〉 ⇐⇒ b1 6 b2 or (b1 = b2 and a1 6 a2)

∀〈a1, b1〉, 〈a2, b2〉 ∈ A×B.

Example 1.1.2. N×2. We have 〈0, 0〉 < 〈1, 0〉, 〈0, 1〉 < 〈1, 1〉 and 〈1, 0〉 < 〈0, 1〉.

Actually, N× 2 is isomorphic to N + N.In mathematical analysis, we use Dedekind cuts of rational numbers to con-

struct real numbers. In general, a linearly ordered set also has cuts and Dedekindcuts.

Definition 1.1.9. A cut of a linearly ordered set A is an ordered pair ofnonempty subsets 〈Z,W 〉, satisfying:

1. Z ∩W = ∅;

2. Z ∪W = A;

3. ∀z ∈ Z,w ∈W, z < w.

There are four possible types of a cut:

1Informally, in disjoint union, even the same elements in different sets are regarded asdifferent elements, in comparison with ‘common’ union.

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1. Z has a greatest element, W has a least element;

2. Z doesn’t have a greatest element, W doesn’t have a least element;

3. Z has a greatest element, W doesn’t have a least element;

4. Z doesn’t have a greatest element, W has a least element.

A cut with type (3) or (4) is called a Dedekind cut .

1.2 Topology

Definition 1.2.1 (Axioms of Open Sets). A topology is a collection of subsetsof a set X, denoted by τ and satisfying the following axioms:

1. ∅ ∈ τ and X ∈ τ .

2. The union of any sets in τ is still in τ .

3. The intersection of any finite number of sets in τ is still in τ .

Sets in τ are called open sets. The complement of an open set is called aclosed set . Together with τ , X is called a topological space.

Definition 1.2.2. A collection B of some subsets of a set X is called a base, ifB is a topology of X, here B denotes the collection of unions of any elementsof B. We say B generates a topology.

Proposition 1.2.1. A given B is a base of X if and only if:

1.⋃B∈BB = X;

2. ∀B1, B2 ∈ B, B1 ∩B2 ∈ B.

Proof. Necessity is apparent.Now prove the sufficiency.

⋃B∈BB = X implies axiom of open sets (1), and

definition 1.2.2 implies axiom of open sets (2).To explain how the conditions imply axiom of open sets (3), just select

arbitrary U, U ′ ∈ B,

U =⋃α

Bα, U′ =

⋃β

B′β , here Bα, B′β ∈ B, ∀α, β

ThenU ∩ U ′ =

⋃α,β

(Bα ∩B′β)

According to condition (2), Bα ∩B′β ∈ B, ∀α, β. Then U ∩ U ′ ∈ B.

Example 1.2.1. The Euclidean topology τe on R is generated by the collectionof all open intervals.

Definition 1.2.3. Given a subset Y of a topological space 〈X, τ〉, the subspacetopology τY is defined as {Y ∩ U : U ∈ τ}.

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Apparently, the definition satisfies the three axioms of open sets. Besidesthe definition, there is another way to introduce the subspace topology.

Definition 1.2.4. The subbase of Y , denoted by BY is defined as {Y ∩ B :∀B ∈ B}.

Easy to verify that the subbase generates the subspace topology.

Definition 1.2.5. Given two topological spaces 〈X, τX〉, 〈Y, τY 〉, we use thebase τX × τY to generate the product topology on X × Y .

Some Topological Properties

Now briefly explain some important topological properties which will be men-tioned later. Detailed and further properties can be found on Munkres’ bookTopology: A First Course.

Definition 1.2.6. Let X be a topological space. A neighborhood of an elementx (or a subset) is a subset Y of X that includes an open set U containing x (orthe subset). Correspondingly, x is called a interior point of Y . The collectionof all interior points of Y is called the interior of Y .

Definition 1.2.7. Let Y be a subset of X, x ∈ X. If the intersection betweenY \{x} and every neighborhood of x is nonempty, then x is called a limit point ofY . The collection of all limit points of Y is called the derived set of Y , denotedby Y ′.

Definition 1.2.8 (Separation Axioms).

• T1: two different elements of X each has a neighborhood not containingthe other element.

• T2: two different elements of X have disjoint neighborhoods.

• T3: a closed subset of X and an element not in it have disjoint neighbor-hoods.

• T4: two disjoint closed sets of X have disjoint neighborhoods.

• completely normal : every subspace of X is T4 ⇐⇒ every two separatedsets2 can be separated by neighborhoods.

• T5: T1 and completely normal.

Proposition 1.2.2. T2 implies T1; T1 along with T4 implies T2 and T3.

Proof. The first statement is trivial.T1 is equivalent to that every singleton3 is closed, which is quite easy to

verify. Then the second statement becomes apparent.

Definition 1.2.9. X is connected if X can’t be the union of two disjointnonempty open subsets.

2Each is disjoint from the other’s closure (A’s closure = A ∪A′).3A set which contains only one element.

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Definition 1.2.10. X is path-connected if there exists a path joining any twoelements in X, i.e. ∀x, y ∈ X, there exists a continuous function f : [0, 1] →X, f(0) = x, f(1) = y.

Definition 1.2.11. X is compact if every open cover of X has a finite subcover.

Definition 1.2.12. X is sequentially compact if every sequence of X has aconverging subsequence.

Definition 1.2.13 (Countability Axioms).

• C1: each element of X has a countable neighborhood basis (∀U,∃B ⊆ U ,here U is a neighborhood of x and B is in the neighborhood basis).

• C2: X has a countable base.

Apparently, C2 implies C1.

1.3 Order Topology

In R, we can choose the collection of all open intervals as a base to generatethe Euclidean topology. Similarly, we can generate the order topology on anylinearly ordered set A. We define open intervals on A as these:

• (a, b) = {c : a < c < b};

• (a,→) = {c : a < c};

• (←, b) = {c : c < b};

∀a, b, c ∈ A.

Definition 1.3.1. These kinds of sets are called open intervals on A.

Definition 1.3.2. The order topology on a linearly ordered set containing morethan one element is generated by the collection of all open intervals. A spacewhose topology is an order topology is called a linearly ordered topological space(LOTS).

The definition is proper, because it’s quite easy to verify that the collectionof all open intervals satisfies both conditions in proposition 1.2.1.

Note. Whenever a LOTS is mentioned, we always assume it contains morethan one element to avoid being loaded down with trivial details.

The next two propositions show that any LOTS is T5.

Proposition 1.3.1. Any LOTS is T2.

Proof. Let X denote a LOTS. ∀a, b ∈ X, a 6= b, either a < b or b < a due todefinition 1.1.3 and 1.1.6. Without loss of generality, we assume that a < b.

If there exists a c so that a < c < b then (←, c) and (c,→) are the requiredneighborhoods. If not, namely (a,→) ∩ (←, b) = (a, b) = ∅, therefore (←, b)and (a,→) are disjoint neighborhoods of a and b.

Proposition 1.3.2. Any LOTS is completely normal.

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Proof. Let A and B be separated subsets of X. For each a ∈ A there is an openinterval Ua such that a ∈ Ua ⊆ X \ B, and for each a ∈ B there is an openinterval Ub such that b ∈ Ub ⊆ X \ A. Let UA =

⋃a∈A Ua and UB =

⋃b∈B Ua;

clearly A ⊆ UA, B ⊆ UB , and UA ∩ UB ⊆ X \ (A ∪B).Let U = UA ∩UB . If U = ∅, then the proof is done, so assume that U 6= ∅.

Define a relation ∼ on U : p ∼ q ⇐⇒ [min{p, q},max{p, q}] ⊆ U ; it’s easilyverified that ∼ is an equivalence relation.

Let T ⊆ U contain exactly one point of each ∼-class.4 If a ∈ A, andp, q ∈ Ua ∩ T with p < q, now prove that p < a < q. Suppose that a < p.Since p ∈ T ⊆ U , there is a b ∈ B such that p ∈ Ub; [a, q] ⊆ Ua ⊆ X \ B, sob /∈ [a, q]. If b < a, then a ∈ [b, p] ⊆ Ub ∩ A = ∅, so a < p < q < b. But then[p, q] ⊆ Ua ∩ Ub ⊆ U , so p ∼ q, contradicting the choice of p and q, thereforep < a. Similarly a < q. Similarly, if a ∈ B and p, q ∈ Ua ∩ T with p < q, thenp < a < q. Therefore |Ua ∩ T | 6 2, ∀a ∈ A ∪B.

Now fix p ∈ T . Let Ap = {a ∈ A : p ∈ Ua} and Bp = {a ∈ B : p ∈ Ua},Ap 6= ∅ 6= Bp since p ∈ U . Suppose that a < p for some a ∈ Ap. If b ∈ Bp andb < p, then either a < b and b ∈ Ua, or b < a and a ∈ Ub, since the sets Uaand Ub are open intervals. So p < b, and since b ∈ Bp was arbitrary, p < Bp.Similarly Ap < p and therefore Ap < p < Bp. If instead p < a for some a ∈ Ap,it shows similarly that Bp < p < Ap.∀a ∈ A ∪B, define Va as these:

Va =

Ua, if Ua ∩ T = ∅Ua ∩ (p,→), if Ua ∩ T = {p} and p < a

Ua ∩ (←, p), if Ua ∩ T = {p} and a < p

Ua ∩ (p, q), if Ua ∩ T = {p, q} and p < a < q

Let VA =⋃a∈A Va, VB =

⋃a∈B Va, apparently VA and VB are open, A ⊆

VA, B ⊆ VB , now prove that VA ∩ VB = ∅.Suppose not, then there exists a ∈ A and b ∈ B such that Va ∩ Vb 6= ∅;

without loss of generality assume that a < b. Fix q ∈ Va ∩ Vb, it’s easy to seethat a < q < b, since Va and Vb are open intervals. Moreover, q ∈ U , so q ∼ pfor a unique p ∈ T . Let I = [min{p, q},max{p, q}]. Then I ⊆ U ⊆ X \ (A∪B),so a, b /∈ I, and therefore a < p < b. If p 6 q, then p ∈ Va ∩ T ⊆ Ua ∩ T , and byconstruction Va ⊆ (←, p), and p /∈ Va, that leads to a contradiction. If q 6 p,then p ∈ Vb ∩ T ⊆ Ub ∩ T , so that Vb ⊆ (p,→), and p /∈ Vb, which is again acontradiction. Therefore VA ∩ VB = ∅, X is completely normal.

Corollary. Any LOTS is T5.

Consider a subset Y of X. There are two topologies on Y : the induced ordertopology τ6Y

and the subspace topology τY . Sometimes they are the same: Qas a subset of 〈R,6, τ〉; sometimes not:

Example 1.3.1. If Y = {−1} ∪ { 1n : n ∈ N+} as a subset of 〈R,6, τ〉, then{−1} is open in τY , but not open in τ6Y

.

In fact, τY is always thinner than τ6Y, which means τ6Y

⊆ τY . Just notethat every set in the base of τ6Y

is also in the base of τY . But when are theythe same? We have the following conclusion.

4The axiom of choice is required here.

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Proposition 1.3.3. τ6Y= τY ⇐⇒ for every Dedekind cut 〈Z,W 〉 of Y , if Z

has a greatest element z, then z is the infimum of W ; if W has a least elementw, then w is the supremum of Z.

Proof. Let’s start with necessity. Suppose there exists a Dedekind cut 〈Z,W 〉 ofY , without loss of generality, assume that Z has a greatest element z while Wdoesn’t have a least element, and z isn’t the infimum of W . So ∃x ∈ X, z < xand x < w,∀w ∈ W . Therefore Z = (←, x) ∩ Y , is an element of τY . However,since W doesn’t have a least element, Z can’t be an element of τ6Y

.Now prove the sufficiency. We only need to prove: τY ⊆ τ6Y

, which isequivalent to that the intersection between any open interval of X and Y is anopen set of τ6Y

.According to axiom of open sets (3), a bounded open interval is the intersec-

tion between two unbounded open intervals, so only unbounded open intervalsneed to be verified: ∀x ∈ X, (←, x) ∩ Y and (x,→) ∩ Y are elements of τ6Y

.If x ∈ Y , then (←, x)∩ Y and (x,→)∩ Y are also open intervals in 〈Y,6Y 〉.

If x 6∈ Y , 〈Z,W 〉 = 〈(←, x) ∩ Y, (x,→) ∩ Y 〉 is a cut of Y (assume Z and Ware both nonempty). Due to the condition, the cut can’t be a Dedekind cut,otherwise x must be in Y . If Z has a greatest element z and W has a leastelement w, then

Z = (←, x) ∩ Y = (←, w)Y

W = (x,→) ∩ Y = (z,→)Y

are elements of τ6Y. If Z doesn’t have a greatest element and W doesn’t have

a least element, then

Z = (←, x) ∩ Y =⋃z∈Z

(←, z)Y

W = (x,→) ∩ Y =⋃w∈W

(w,→)Y

are also elements of τ6Y.

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Chapter 2

Ordinal Space

2.1 Well-Order

Definition 2.1.1. A well-order on a set A is a certain linear order, satisfyingthat every nonempty subset of A has a least element.

Example 2.1.1. Any finite linearly ordered set is well-ordered.

Example 2.1.2. N is well-ordered with its standard order.

Example 2.1.3. N× N is well-ordered with its canonical order.

Proof. For a nonempty subset {〈m,n〉 : m,n ∈ N}, firstly find the least n in thesubset, then find the least m in with the least n. Then we get the least element.We can do that because N is well-ordered.

Corollary. For any finite number of well-ordered sets A1, A2, . . . , An, A1×A2×· · · ×An is well-ordered with its canonical order.

Proposition 2.1.1. Any element a except the greatest element of a well-orderedset A has a successor b, meaning a < b and there doesn’t exist c satisfyinga < c < b.

However an element isn’t necessarily a successor of another element.

Proof. Since a isn’t the greatest element, {x > a : x ∈ A} is nonempty so thereexists a least element b which is the successor.

Consider the well-ordered set N × N, the element 〈0, 1〉 isn’t a successor ofany other element.

If an element isn’t a successor of any other element, then we call it a limitelement . The least element is also a limit element. Correspondingly, an elementis called a predecessor of its successor.

Proposition 2.1.2. A is well-ordered ⇐⇒ there doesn’t exist ai ∈ A,∀i ∈ N,so that a0 > a1 > a2 > · · ·

Proof. If there exists such ai, so {ai : ∀i ∈ N} doesn’t has a least element thenA is not well-ordered. Thus the sufficiency is proved.

If A isn’t well-ordered then there exists a subset B which doesn’t a leastelement. Select an arbitrary element b0 in B. Due to the hypothesis, b0 isn’t

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the least element. So there exists b1 < b0. In the same way, there existsb2 < b1, . . ., then we have the infinite descending {bi,∀i ∈ N}. Thus the necessityis proved.

Proposition 2.1.3 (Transfinite Induction). Let A be a well-ordered set, p(a)be a proposition about element a. For any element a ∈ A, if ∀b < a, p(b) is true=⇒ p(a) is true, then ∀a ∈ A, p(a) is true.

Proof. Let B be the collection of all elements that doesn’t satisfy p. Suppose Bis nonempty, then B has a least element b. However for all elements less thanb, p is true, so p(b) must be true. Therefore B = ∅, ∀a ∈ A, p(a) is true.

Note. The condition already implies p(0) is true, here 0 denotes the least ele-ment.

Definition 2.1.2. Let 〈Z,W 〉 be a cut of a well-ordered set A, we call Z aninitial segment of A. ∅ and A itself are also called initial segments.

Proposition 2.1.4. For any two well-ordered sets A and B, either A is iso-morphic to an initial segment of B, or B is isomorphic to an initial segment ofA.

Proof. Define a function f : A → B, a 7→ the least element in B that doesn’tequal f(a′),∀a′ < a. If {f(a′) : ∀a′ < a} = B, then f(a) will not be defined.This definition ensures f to be strictly increasing.

If f is defined on the whole set A, then the image of f must be an initialsegment of B. That is because ∀b < f(a), b must be in the image of f accordingto the definition of f(a). Therefore f is an isomorphism between A and aninitial segment of B.

If f is defined on an initial segment of A (6= A), then the image of f mustequal B, according to the definition. Therefore f is an isomorphism between aninitial segment of A and B.

Proposition 2.1.5. A can’t be isomorphic to an initial segment of it, unlessthe initial segment is A itself.

Proof. Suppose A is isomorphic to [0, a), a ∈ A, let f denote the isomorphism.So ∃a′ ∈ A, f(a′) < a′. Because f is strictly increasing, we get

a′ > f(a′) > f(f(a′)) > f(f(f(a′))) > · · ·

That contradicts proposition 2.1.2.

Corollary. If A is isomorphic to an initial segment of B and B is isomorphicto an initial segment of A, then A = B.

Proof. Suppose A 6= B and the two statements are both true, we can get ainitial segment (6= A) of A that is isomorphic to A, using the two isomorphisms.That leads to a contradiction.

These propositions allow us to compare two well-ordered sets. For any well-ordered sets A and B, only one of these is true:

• if A is isomorphic to an initial segment of B ( 6= B), then we say A’s ordertype is less than B’s order type;

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• if they are isomorphic, then we say A’s order type is equal to B’s ordertype;

• if an initial segment of A (6= A) is isomorphic toB, then we say A’s ordertype is greater than B’s order type.

2.2 Ordinal

As written before, two isomorphic ordered sets are said to have the same ordertype. Informally speaking, an ordinal is the order type of all isomorphic well-ordered sets.

However the collection of all isomorphic well-ordered sets may be too ‘big’ tobe a set. To avoid the difficulty, we can choose a standard set from every collec-tion of isomorphic well-ordered sets as the ordinal. So we adopt von Neumann’sdefinition:

Definition 2.2.1. An ordinal is the collection of all ordinals less than it; 0 = ∅is the least ordinal.

More strictly, a set α is an ordinal if and only if:

1. the inclusion relation ∈ on α is a well-order;

2. any element of α is also a subset of α.

For example, 1 = {∅} is the successor of 0, 2 = {∅, {∅}} is the successor of1, and ω = {0, 1, 2, . . .} is the least infinite ordinal. N with its standard orderhas the ordinal ω.

With no doubt, an ordinal is a well-ordered set, so we can compare twoordinals. If α is isomorphic (actually equal) to an initial segment of β, then wesay α 6 β.

If α < β, then both α ∈ β and α & β are true. In fact,

Proposition 2.2.1. α < β ⇐⇒ α ∈ β ⇐⇒ α & β

Proof. That is easy to see, just give some consideration to the definition ofordinals.

A notable fact is, the collection of all ordinals is not a ordinal, not evena set. If it was, then it must be the greatest ordinal, but any ordinal has asuccessor which is greater than itself. This conclusion is known as the Burali-Forti paradox.

Ordinal Arithmetic

Since the elements of an ordinal are all ordinals, we can say that an ordinalis the successor of another ordinal. Or we can redefine the term ‘successor’ inanother way, due to the specificity of ordinals.

Definition 2.2.2. The successor of an ordinal α is defined as α∪{α}, denotedby α+1. Apparently, α+1 is also an ordinal. An ordinal is called the predecessorof its successor. This definition is consistent with the previous definition.

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Proof. α ∈ α+ 1, so α < α+ 1. Suppose ∃β, α < β < α+ 1, then β ∈ α ∪ {α},so either β ∈ α ⇐⇒ β < α or β ∈ {α} ⇐⇒ β = α. That leads to acontradiction. Therefore the two definitions are consistent.

An ordinal is called a successor ordinal , if it has a predecessor. Else calleda limit ordinal .

Proposition 2.2.2. A successor ordinal has a predecessor, which is preciselyits greatest element in it.

Proof. Let α+ 1 denote a successor ordinal. If ∃β ∈ α+ 1 which is greater thanα, then α < β < α+ 1 so α+ 1 can’t be the successor of α. Therefore α is thegreatest element of α+ 1.

If an ordinal α has a greatest element β, then α > β. If α > β + 1, thenβ can’t be the greatest element because β + 1 is also in α. So β < α 6 β + 1.There doesn’t exist an ordinal between β and β + 1, so α = β + 1. That meansβ is the predecessor of α, making α a successor ordinal.

Proposition 2.2.3. If ∀i ∈ I (here I is an index set), αi is an ordinal, then⋃i αi is also an ordinal, and it’s precisely the supremum of {αi}.

Proof. To prove that⋃i αi is an ordinal, we only need to prove that if α ∈

⋃i αi

then β ∈⋃i αi for any β < α. That is quite apparent because if α is in some

αi then β is in that αi too, for any β < α.⋃i αi is a upper bound of {αi} because ∀i, αi <

⋃i αi. And every upper

bound of {αi} must contain⋃i αi as a subset according to proposition 2.2.1.

Therefore⋃i αi is the least upper bound, namely the supremum of {αi}.

After those preparatory work, we can finally define the sum of ordinals. Wehave defined the canonical order on A + B in section 1.1 , here A and B areboth linearly ordered sets. Naturally, we want the sum of two ordinals α+ β tobe the order type of A+B, here A and B each has the order type α and β.

Definition 2.2.3 (Addition of Ordinals). Since we have known that α + 1 =α ∪ {α}, we can define α+ β for any ordinal β recursively:

1. α+ 0 = α;

2. α+ (β + 1) = (α+ β) + 1;

3. α+ β =⋃γ<β(α+ γ), for limit ordinal β 6= 0.

Proposition 2.2.4. If A and B each has the order type α and β, then A+Bhas the order type α+ β.

Proof. We use the transfinite induction to prove it. Assume ∀γ < β, A+C hasthe order type α+ γ, here γ is the order type of C.

If β = 0 then B = ∅, the proposition becomes trivial.If β = γ + 1 is a successor then B has a greatest element b, B = B \ {b} ∪

{b}. According to the hypothesis, there is an isomorphism between α + γ andA + (B \ {b}). In addition, by mapping γ(as an element of β) to b we can getan isomorphism between α+ β and A+B.

If β is a limit ordinal, then for any initial segment C of B (C 6= B), wehave an isomorphism between α+ γ and A+ C, here γ is the order type of C.All the isomorphisms are consistent so we can paste them together, getting anisomorphism between α+ β and A+B.

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Example 2.2.1. 0 + α = α, 1 + ω = 2 + ω = ω, ω + 1 6= ω.

We can find out two facts by this example: firstly, the addition of twoordinals isn’t commutative; secondly, α + γ may equal β + γ even if α 6= β, sothe subtraction of two ordinals doesn’t always make sense.

We can use the addition of ordinals to define the multiplication:

Definition 2.2.4 (Multiplication of Ordinals).

1. α · 0 = 0;

2. α(β + 1) = αβ + α;

3. αβ =⋃γ<β(αγ), for limit ordinal β 6= 0.

Similar to proposition 2.2.4 both in conclusion and in proof, if A and B eachhas the order type α and β, then A×B has the order type αβ.

Example 2.2.2. 0α = 0, 2ω = 3ω = ω, ω · 2 = ω + ω.

Similarly too, the multiplication of two ordinals isn’t commutative and thedivision of two ordinals doesn’t always make sense.

Finally, we use the multiplication to define the exponentiation:

Definition 2.2.5 (Exponentiation of Ordinals).

1. α0 = 1;

2. αβ+1 = αβ · α;

3. αβ =⋃γ<β(αγ), for limit ordinal β 6= 0.

In the three definitions of ordinal arithmetic, we have only used two basicsteps to create new objects: getting a successor of an ordinal and getting theunion of some ordinals. According to definition 2.2.2 and proposition 2.2.3, wealways get an ordinal after every step.

2.3 Ordinal Space

After spending so much space in set theory, let’s talk about something more‘topological’: ordinals as linearly ordered topological spaces. As a LOTS, anordinal α is often written as [0, α) to emphasize that it’s the space consisting ofall ordinals less than α, and α+ 1 is written as [0, α].

Note. When ‘the topology’ of an ordinal is mentioned, it always refers to theorder topology.

Proposition 2.3.1. Any ordinal space is totally disconnected1.

Proof. Let S be a subset of an ordinal and S has more than one element, thenS has a least element β. Both (←, β + 1) ∩ S = {β} and (β,→) ∩ S = S \ {β}are nonempty and open in S, so S is disconnected. Therefore the ordinal spaceis totally disconnected.

1Any subset having two element or more is disconnected.

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For any finite ordinal, the topology is simply the discrete topology2, so onlyinfinite ordinals can be interesting.

Successor ordinals less than α are isolated3 in α. That is because {β+ 1} =(β, β+2). Meanwhile, limit ordinals (except 0) less than α are not isolated in α.Just note that there doesn’t exist a greatest element less than a limit ordinal, soany open interval containing the limit ordinal must contain some ordinals lessthan it. Therefore the derived set of α is precisely the set consisting of limitordinals (except 0) less than α.

Proposition 2.3.2. [0, α) is compact ⇐⇒ α is a successor ordinal.

Proof. If α is a limit ordinal, then the open cover {[0, β)}β<α can’t have a finitesubcover because any finite subcover has a greatest element (elements of thecover are ordinals) and the greatest element is less than α.

If α is a successor ordinal, then [0, α) has a greatest element α0. Let S bea open cover of [0, α). Choose a open set S0 in the cover that contains α0,then there is a open interval I0 ⊆ S0 that contains α0. The open interval I0looks like (α1,→), so [0, α)\ I0 also has a greatest element α1. Similarly, chooseS1 (S1 may equal S0 but that doesn’t matter) and I1 and find the greatestelement of [0, α) \ (I0 ∪ I1), denoted by α2 . . . According to proposition 2.1.2,this process must end after finite steps. Then {Ii} is a finite cover of [0, α),therefore {Si}i=1,...,n is a finite subcover of S.

Example 2.3.1. [0, ω), [0, ω]

Let’s talk about the least two transfinite ordinals.ω is the ordinal of N with its standard order. ω looks like {0, 1, 2, 3, . . .}.

The topology is also the discrete topology.ω + 1 is the successor of ω, and it looks like {0, 1, 2, 3, . . . ω}. The topology

is not the discrete topology, because {ω} is not open.[0, ω] is compact while [0, ω) isn’t. [0, ω] is the one-point compactification

(adding a greatest element to get a compact space) of [0, ω). In common, forany limit ordinal α 6= 0, [0, α] is the one-point compactification of [0, α).

Example 2.3.2. [0, ω1), [0, ω1]

Let ω1 denote for the least uncountable ordinal. From Zermelo’s Theorem4

we know that there does exist an uncountable ordinal. The least uncountableordinal ω1 consists of all countable or finite ordinals.

Proposition 2.3.3. [0, ω1) is sequentially compact.

Proof. Let {αn}n∈N be a sequence of [0, ω1). We can get an increasing subse-quence by mapping n to the n-th least element of {αn}. Then the supremum(or union) of that subsequence is also countable so it’s an element of [0, ω1).The supremum couldn’t be a successor ordinal, otherwise its predecessor is aupper bound too. If an open neighborhood of the supremum (must containsanother element because the supremum is a limit ordinal) doesn’t contain anyelement in {αn}, then the least element of the neighborhood is also an upperbound. Therefore the supremum is a limit point of {αn}, we have a convergingsubsequence of {αn}.

2With discrete topology, any subset of the space is open.3x is isolated in X means {x} is open in X.4Every set can be well-ordered.

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According to proposition 2.3.2, [0, ω1) is not compact. As a result, [0, ω1) isnot metrizable. And a more interesting fact is,

Proposition 2.3.4. In [0, ω1], ω1 is a limit point of [0, ω1), but any sequenceof [0, ω1) couldn’t converge to ω1.

Proof. ω1 is a limit point of [0, ω1) since {ω1} is not open, and the secondstatement is because the supremum (=union) of any countable subset of [0, ω1)(elements are countable ordinals) is a countable ordinal.

As a corollary, [0, ω1] is not C1 since the only element without a countableneighborhood basis is ω1. That shows the subspace [0, ω1) is C1.

Proposition 2.3.5. Neither [0, ω1) nor [0, ω1] is C2.

Proof. We only need to prove that [0, ω1) isn’t C2, since [0, ω1] is not even C1.Suppose [0, ω1) is C2, then it has a countable base. Let α be the supremum ofthe collection of the least elements from each set in that base. Because the baseis countable, so α ∈ [0, ω1). Then (α,→) is not open because any open set mustcontain an ordinal less than or equal to α, according to the construction of α.That leads to a contradiction.

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Chapter 3

Counterexamples

3.1 Long Ray

The ray [0,+∞) can be defined as [0, 1)×ω, as a product of two linearly orderedsets. With this definition, ∀a ∈ {0}∪R+, a = 〈a−bac, bac〉, here b·c is the floorfunction.

Now replace ω by ω1, we get:

Definition 3.1.1. The long ray L = [0, 1)× ω1.

Proposition 3.1.1. L is sequentially compact.

Proof. Let {〈xn, αn〉}n∈N be a sequence of [0, ω1). Since [0, ω1) is sequentiallycompact, there is a converging subsequence of {αn}, denoted by {αi}i∈N, con-verges to α. [0, 1] is also sequentially compact so {xi} has a converging sub-sequence, denoted by {xj}j∈N, converges to x ∈ [0, 1]. Therefore {〈xj , αj〉}converges to 〈x, α〉 (if x = 1 then 〈1, α〉 = 〈0, α+ 1〉).

Consequently, L isn’t homeomorphic1 to [0, 1) since [0, 1) is not sequentiallycompact.

Proposition 3.1.2. L is path-connected and is therefore connected.

We need a lemma to prove this proposition.

Lemma. For any ordinal α ∈ (0, ω1), [0, 1)× α is homeomorphic to [0, 1), andthe homeomorphism can be chosen to be order-preserving.

Proof. First of all, [0, 1) × 1 = [0, 1). Assume ∀β ∈ (0, α), [0, 1) × β is order-preservingly homeomorphic to [0, 1).

If α = β+ 1 is a successor ordinal, then we have two order-preserving home-omorphisms: f1 : [0, 1) × β → [0, 12 ), f2 : [0, 1) × {β} → [ 12 , 1). By pasting f1and f2 together we can get a homeomorphism between [0, 1)× α and [0, 1).

If α is a limit ordinal, we will find an increasing sequence α0 < α1 < α2 < · · ·that converges to α (∀β < α,∃i, so that αi > β). Because (0, α) is countable, wecan list the ordinals as γ0, γ1, γ2, . . ., in this list the ordinals are not necessarily

1There exists a bijection between two homeomorphic spaces, preserving all topologicalproperties. The bijection is called a homeomorphism.

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ordered. Now let α0 = γ0. If i > 0, let αi be γj , here j is the least index sothat γj > αi−1. Such αi does exist because α is a limit ordinal, and αi willbe greater than any certain ordinal less than α because the γ-list contains allordinals less than α.

For any i, there exists a homeomorphism which maps [0, 1)× αi+1 to [0, 1).This homeomorphism maps 〈0, αi〉 to A ∈ [0, 1), so it maps [〈0, αi〉, 〈0, αi+1〉)to [A, 1). Rescale these homeomorphisms so [〈0, 0〉, 〈0, α0〉) maps to [0, 12 ),[〈0, α0〉, 〈0, α1〉) maps to [ 12 ,

34 ) and [〈0, α1〉, 〈0, α2〉) maps to [ 34 ,

78 ), etc. Paste

them together and we get an order-preserving homeomorphism between [0, 1)×αand [0, 1).

Now we can prove proposition 3.1.2.

Proof. According to the lemma, for any ordinal α ∈ (0, ω1), [0, 1) × α is path-connected because [0, 1) is path-connected. ∀〈a, α〉, 〈b, β〉 ∈ L, a, b are realnumbers and α, β are countable or finite ordinals, so 〈a, α〉 and 〈b, β〉 can beconnected by a path, as elements of [0, 1) × (α ∪ β + 1). Therefore L is path-connected.

By adding a greatest element ∞ to L, namely one-point compactification,we get the extended long ray L∗ = L+ {∞}.

Proposition 3.1.3. L∗ is connected but not path-connected.

Proof. L∗ is connected because L is connected and the singleton {∞} is notopen.

Suppose L∗ is path-connected, then there exists a continuous function f :[0, 1]→ L∗, f(0) = 〈0, 0〉, f(1) =∞.

Now prove that f is a surjection. If not, there exists an element p ∈ L∗ sothat f−1({p}) = ∅. Then f−1((←, p)) and f−1((p,→)) are disjoint open setsin [0, 1], and f−1((←, p)) ∪ f−1((p,→)) = [0, 1]. That contradicts that [0, 1] isconnected.

Therefore for all α < ω1, f−1((0, 1)×{α}) is nonempty. Since (0, 1)×{α} =(〈0, α〉, 〈0, α + 1〉) is open in L, f−1((0, 1) × {α}) is open in [0, 1], so it mustcontain an open interval as a subset. That contradicts that there can’t beuncountably many disjoint open intervals in [0, 1].

There exists a continuous injection from [0, 1) to L, but couldn’t exist acontinuous surjection. Although L and [0, 1) have the same cardinality, L ismuch ‘longer’ than [0, 1). That is why we call it the long ray.

L is C1 while L∗ is not. Neither L nor L∗ is C2. That is quite similar withhow we dealt with [0, ω1) and [0, ω1].

3.2 Tychonoff Plank

Definition 3.2.1. The Tychonoff plank T is defined as [0, ω1] × [0, ω], herethe topology is not the order topology of its canonical order, but the producttopology.

By removing the element 〈ω1, ω〉 of T , we get the deleted Tychonoff plankT− = T \ {〈ω1, ω〉}.

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Proposition 3.2.1. T is a compact, T2 and is therefore T4. However, T− isnot T4. That shows a subspace of a T4 space isn’t necessarily T4.

Proof. According to proposition 1.3.1 and proposition 2.3.2, both [0, ω] and[0, ω1] are T2 and compact. Because both T2 and compactness are multiplicative,T is T2 and compact and therefore T4.

[0, ω) × {ω} and {ω1} × [0, ω) are both closed sets of T−. Consider anyneighborhood of {ω1} × [0, ω) denoted by U , ∀i < ω, ([0, ω1) × {i}) ∩ U mustcontain (αn, ω1) × {i} as a subset. Let α =

⋃i<ω αn, then α < ω1, (α, ω1) ×

[0, ω) ⊆ U . But any open set V containing 〈α+1, ω〉must contain {α+1}×(n, ω)as a subset, here n is some ordinal less than ω. Therefore 〈α+1, n+1〉 ∈ U ∩V ,T− is not T4.

Definition 3.2.2. Let α be a limit ordinal, an α-indexed sequence in X meansa function from α to X.

An ordinal-indexed sequence is the generalization of a sequence. A commonsequence is a ω-indexed sequence.

If X is a topological space, we say that an α-indexed sequence of elementsof X converges to a limit x, when given any neighborhood U of x there is anordinal β < α such that xγ is in U for all γ > β.

Ordinal-indexed sequences are usually more powerful than common sequences.For example, in [0, ω1], ω1 is a limit point of [0, ω1), but it is not a limit of anycommon sequence in ω1, see proposition 2.3.4. However, it is the limit of theω1-indexed sequence mapping any ordinal less than ω1 to itself.

The next counterexample shows that even ordinal-indexed sequences are notpowerful enough to replace the concept of limit points.

Proposition 3.2.2. In T , the corner point 〈ω1, ω〉 is a limit point of [0, ω1)×[0, ω), but it is not a limit of any ordinal-indexed sequence in [0, ω1)× [0, ω).

Proof. Let S denote [0, ω1) × [0, ω). Suppose ∃f : π → S converges to 〈ω1, ω〉,then ∀〈ρ, σ〉 ∈ S, there exists a least α < π, so that ∀β > α, f(β) ∈ (ρ, ω1) ×(σ, ω). Define g : S → π, 〈ρ, σ〉 7→ α. Let’s discuss some properties of g.

1. If ρ1 > ρ0, σ1 > σ0, then g(〈ρ1, σ〉) > g(〈ρ0, σ〉) and g(〈ρ, σ1〉) > g(〈ρ, σ0〉).

2. g(〈ρ, σ〉) = max{g(〈ρ, 0〉), g(〈0, σ〉)}.

Defineg0 : ω → π, σ 7→ g(〈0, σ〉)

g1 : ω1 → π, ρ 7→ g(〈ρ, 0〉)

g0 and g1 actually determine g. Their images g0([0, ω)) and g1([0, ω1)) aresubsets of π, so they are well-ordered sets. In fact, the order type of g0([0, ω))can’t be greater than ω, since g0 is non-strictly increasing; similarly, the ordertype of g1([0, ω1)) can’t be greater than ω1.

Now I claim that the order type of g0([0, ω)) is precisely ω. Suppose not,its order type is a finite ordinal, g0([0, ω)) has a greatest element α. So ∃β <ω, g0([β, ω)) = {α}. Then the α-th element of the sequence couldn’t actuallyexist, since

⋂γ∈[β,ω)[0, ω1)× (γ, ω) = ∅. Similarly, the order type of g1([0, ω1))

is precisely ω1, or there will exist a countable ordinal β1, so that g1([β1, ω1)) isa singleton.

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Claim: ∀α0 ∈ g0([0, ω)),∃α1 ∈ g1([0, ω1)), so that α0 < α1; and ∀β1 ∈g1([0, ω1)),∃β0 ∈ g0([0, ω)), so that β1 < β0. Suppose not, without loss ofgenerality, assume ∃α ∈ g0([0, ω)), α is greater than any element of g1([0, ω1)).Then g([0, ω1) × g−10 (α)) = {α}, so the α-th element of the sequence couldn’texist.

However, the last two assertions can’t be true together. ∀αi ∈ g0([0, ω)),∃βi ∈g1([0, ω1)) so that αi < βi, since the order type of g1([0, ω1)) is ω1, so ∀i <ω, [0, βi) ∩ g1([0, ω1)) is a countable or finite set. Then

⋃i[0, βi) ∩ g1([0, ω1))

is a countable set, ∃β ∈ g1([0, ω1)), so that ∀i, β > βi > αi since g1([0, ω1)) isuncountable.

Therefore there can’t be an ordinal-indexed sequence converging to 〈ω1, ω〉.

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Bibliography

[1] Nikolai Konstantinovich Vereshchagin, Alexander Shen. Basic Set Theory.American Mathematical Society, 2002.

[2] James R. Munkres. Topology: A First Course. Prentice-Hall, 1974.

[3] Stephen Willard. General Topology. Courier Corporation, 2012.

[4] Lynn Arthur Steen, J. Arthur Seebach. Counterexamples in Topology.Courier Corporation, 2013.

[5] Wang Shangzhi. Subspace of Linear Ordered Topological Space and Sub-ordered Space. Journal of Beijing Teachers College: Natural Sciences Edi-tion, Vol. 11, No. 4, 1990.

[6] Richard Koch. The Long Line. November 24, 2005. http://www.math.

ucsd.edu/~nwallach/LongLine[1].pdf

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Index

ω, 11ω1, 14

base, 4subbase, 5

binary relation, 2

C1−2, 6closed set, 4compact, 6

sequentially compact, 6completely normal, 5connected, 5cut, 3

Dedekind cut, 4

derived set, 5disjoint union, 3

homeomorphic, 16homeomorphism, 16

initial segment, 10interior, 5interior point, 5isolated, 14isomorphic, 3isomorphism, 3

limit element, 9limit point, 5long ray, 16LOTS, 6

neighborhood, 5

open interval, 6open set, 4order

induced linear order, 3induced partial order, 3linear order, 3

partial order, 2strict linear order, 3strict partial order, 2well-order, 9

order type, 3ordinal, 11

limit ordinal, 12successor ordinal, 12

ordinal-indexed sequence, 18

path-connected, 6predecessor, 9, 11

separated sets, 5singleton, 5successor, 9, 11

T1−5, 5topological space, 4topology, 4

discrete topology, 14Euclidean topology, 4order topology, 6product topology, 5subspace topology, 4

totally disconnected, 13Tychonoff plank, 17

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