¨
Orebro universitet
Instutitionen f ¨or naturvetenskap och teknik Matematik C, 76 – 90 h ¨ogskolepo ¨ang
From the Axiom of Choice
to Tychono
ff’s Theorem
Gustav H¨orngren H¨osterminen 2014
Handledare: Holger Schellwat Examinator: Yang Liu
Sj¨alvst¨andigt arbete, 15 hp Matematik, C – nivå, 76 – 90 hp
Abstract
A topological space X, is shown to be compact if and only if every net in X has a cluster point. If s is a net in a product Q
α∈AXα, where each Xα is a compact topological space, then, for every subset B of A, such that the restriction of s to B has a cluster point in the partial productQ
α∈BXα, it is found that the restriction of s to B ∪ {γ} – extending B by one element γ ∈ A \ B – has a cluster point in its respective partial productQ
α∈B∪{γ}Xα, as well.
By invoking Zorn’s lemma, the whole of s can be shown to have a cluster point. It follows that the product of any family of compact topological spaces is compact with respect to the product topology. This is Tychonoff’s theorem.
The aim of this text is to set forth a self contained presentation of this proof. Extra attention is given to highlight the deep dependency on the axiom of choice.
Contents
1 Introduction 5
1.1 Background . . . 5
1.2 Chapter outline . . . 5
2 Sets 7 2.1 Basic notation and terminology . . . 7
2.2 Relations and Mappings . . . 9
2.3 Generalized unions, intersections and cartesian products . . . 11
2.4 The Axiom of choice . . . 14
2.5 Orderings and Zorn’s lemma . . . 16
3 Point Set Topology 19 3.1 Basics . . . 19
3.2 Compactness . . . 23
3.3 The product topology . . . 24
4 Nets 27 4.1 Directed sets . . . 27
4.2 Nets and Subnets . . . 28
4.3 Nets and Compactness . . . 32
4.3.1 A proof of Tychonoff’s theorem . . . 34
A A proof of Zorn’s lemma 41 A.1 The Bourbaki fixed-point theorem . . . 41
A.1.1 A proof of the Bourbaki fixed-point theorem . . . 41
Chapter 1
Introduction
1.1
Background
Tychonoff’s theorem states that the generalized cartesian product of any family of compact topological spaces is compact with respect to the induced product topol-ogy. The proof was first stated and proved in 1930 by Andrey N. Tychonoff.
In 1992 American mathematician Paul Chernoff, published an article [1] in which he introduced a new way to prove the theorem. According to his survey, there are now – counting his own contribution – four different proofs of the theo-rem.
Chernoff’s proof uses the theory of nets (a generalization of the theory of se-quences) and how compactness of topological spaces finds a description in terms of the theory of nets. Although this way of describing compactness had been known for many decades, its application to a proof of Tychonoff’s theorem was yet un-known.
The aim of this essay is to present a detailed version of Chernoff’s proof and the mathematics that it depends upon, thus making the article [1] available to readers with no or little prior knowledge of set theory and point set topology. In fact, with the exception of a couple of illustrating examples, the prerequisites for reading this essay – besides the capability of reading mathematical discourse – are none.
1.2
Chapter outline
Chapter 2 Here we develop the basic notions of the set theoretical language re-quired for the developments in later chapters. In particular many results there are dependent on the axiom of choice, which is why we have treated it ex-tensively. Finally Zorn’s lemma is stated; a proof is supplied in Appendix A.
Chapter 3 The point of this chapter is to define the product topology, which is a topology on the generalized cartesian product of a family of topological spaces, and to find a base for it.
Chapter 4 Here we develop the theory of nets to the point where we can prove that the definition of compactness for a topological space is equivalent to every net in that space having a cluster point. The chapter is concluded by giving a reworked, and more detailed version of the proof in [1], of Tychonoff’s theorem.
The set theory draws mainly from [4], except the proof of Zorn’s lemma, which is based on [5, pages 878–884]. The topology in Chapter 3 is mainly from from [2, chapter 4] and the material about nets draws from [1], [2, chapter 4] and [3, chap-ter 2]. Finally, besides [1], the main proof is also found in [2, pages 136–137].
Chapter 2
Sets
The aim of this chapter is to present the bits of set-theoretical language relevant to later chapters.
2.1
Basic notation and terminology
A set is a collection of objects (or things). If x is an object that belong to the set A, then x is said to be a member of A (‘x is contained in A’ or ‘A contains x,’ are synonymous). We write this as
x ∈ A.
If x is an object that is not a member of A, we may write this as
x < A.
If A and B are sets such that whenever x ∈ B, we have x ∈ A, then B is said to be a subset of A. This we write as
B ⊂ A(or A ⊃ B).
Notice that every set is a subset of itself.
The empty set, denoted by ∅, is defined as the set with no members. The empty set is subset of every set. Indeed, for a set A to not have ∅ as a subset the statement: whenever x ∈ ∅, we have x ∈ A (2.1) must be false. But this is impossible. In fact, the negation of it reads:
But, by definition, it does not exist any x ∈ ∅, so (2.2) is clearly false. Hence (2.1) is true.
Furthermore, we will write
‘A= B’ for ‘A ⊂ B and B ⊂ A,’
with the meaning of ‘for’ taken in the sense of ‘as an abbreviation of’ or ‘instead of.’ In other words, two sets A and B are defined as equal if they are subsets of each other. Thus, in general, if we want to prove that two sets A and B, are equal, we need to prove that A is a subset of B and that B is a subset of A.
As a special case, we have sets whose members can be listed as finite or infinite arrays of characters (each character representing a member of the set). For instance, we have the set of the objects x1, . . . , xnor the set of all positive integers (1 , 2 , 3 ,
and so on). The former we will write as
{ x1, . . . , xn}, (2.3)
and the latter as
{ 1 , 2 , 3 , . . . }. (2.4) An object y, is a member of the set (2.3) if and only if y is equal to at least one of the listed xj. Thus, for the listing in (2.3) and (2.4); order is irrelevant.
Also note that duplicates may be yanked without changing the set, that is, if two different characters represent the same object, we can safely remove one of them and the shorter list of characters will still represent the set it did before the removal; to illustrate we have
{ x1, . . . , xn, a } = { x1, . . . , xn}
whenever a= xk for some k= 1 , . . . , n (the same holds in the infinite case).
A set is said to be finite if it can be written as in (2.3), otherwise it is said to be infinite.
We may introduce sets by a phrase of the form:
the set of all x, such that . . . x . . . (2.5)
where for any given object x (as soon as we “plug it in” to ‘. . . .’) the expression ‘. . . x . . .’ represents a statement about x , that if true, qualifies x as a member and otherwise as a non-member of the set (2.5).
As a convenient abbreviation for (2.5) we have the piece of notation
{ x : . . . x . . . }
Clearly y ∈ { x : . . . x . . . } if and only if the statement ‘. . . y . . .’ is true.
As a tweak of the { x : . . . x . . . }-notation, it is common to write part of the qualifying statement before the ‘ : ’-sign in . For example, we may write the set of all odd, positive integers as
{ x ∈ Z+: x is odd }, where of course Z+represents the set { 1 , 2 , 3 , . . . }.
If A is any set, then the power set of A, denoted byP(A), is defined as the set of all subsets of A. Or in short:
P(A) = { B : B ⊂ A }.
We round up this section by defining three binary operations on the class of sets. Given a pair of sets A and B, we define the union of A and B, denoted A ∪ B, as
A ∪ B= { x : x ∈ A or x ∈ B }, the intersection of A and B, denoted A ∩ B, as
A ∩ B= { x : x ∈ A and x ∈ B }, and the difference between A and B, denoted A \ B, as
A \ B= { x : x ∈ A and x < B }. If A ∩ B= ∅, we say that A and B are mutually disjoint.
2.2
Relations and Mappings
Let A and B be sets. If a ∈ A and b ∈ B then we can form a new kind of object (a, b), called the ordered pair of a and b. Two ordered pairs (a, b) and (a0, b0) are defined as equal if and only if a= a0and b= b0.
The cartesian product A × B of A and B, is defined as
A × B= { (a, b) : a ∈ A and b ∈ B }.
Let A and B be sets. A relation from A to B is a subset of A × B. Let R be a relation from A to B. It is common to write
‘xRy’ for ‘(x, y) ∈ R’,
as in ‘3 ≤ 4’ where ≤ is regarded as the usual less than or equal to relation, from Z+to itself.
The set
{ x ∈ A : xRy for at least one y ∈ B } is called the domain of R and the set
{ y ∈ B : xRy for at least one x ∈ A } is called the range of R.1
A relation R ⊂ A × B, is said to be a mapping from A into B if: (i) the domain of R includes the whole of A,
(ii) whenever xRy and xRy0, then y= y0.
In other words, R is a mapping if for every x ∈ A, there exists a unique y ∈ B such that xRy.
If R is a mapping and xRy, then we say that y is the value of the argument x, or that y is the value of R at x.
Terminology. It is common to refer to mappings as functions or transformations, but we will almost exclusively use the word ‘mapping.’
Notation. There are some common pieces of notation used to distinguish those characters that represent mappings. In particular, we will denote the set of all mappings from A into B by BA. Hence, we can write f ∈ BA to state that f is a mapping from A to B. The same may also be conveyed by writing
f: A → B.
If f ∈ BA, it is custom, and often convenient, to denote its value at α by f (α). If, however, the associated value of an argument α is denoted by, say βα, we may draw attention to this by writing
f: A → B such that α 7→ βα.
In this case, we don’t even have to tie a mapping to any particular name, such as f , but we are perfectly alright referring to it by phrases like:
the mapping α 7→ βαin BA or,
the mapping A → B : α 7→ βα
thus refraining completely from representing it by a special character, and instead concentrate on the association that makes up the mapping. In fact, this will be the default throughout this text.
The compound formulæ‘α 7→ βα’ may be read as ‘α maps to βα.’
Let f ∈ BA and let E ⊂ A. Then we we define the image – denoted by f (E) – of E under f as
f(E)= { f (x) : x ∈ E }. The set f (A), is simply called the image of f .
Now let f ∈ BA, as above, but this time let E ⊂ B. Then we define the inverse image– denoted by f−(E) – of E under f as
f−(E)= { x ∈ A : f (x) ∈ E }.
If α 7→ βα is a mapping in BA and β 7→ γβ is a mapping in CB, then the association
α 7→ βα7→γβα
is a mapping in CA. A chain of mappings like this is called a composite mapping. If f ∈ BAand g ∈ CB, their composition α 7→ f (α) 7→ g( f (α)) is denoted by
g ◦ f. That is, g ◦ f (α)= g( f (α)) for all α ∈ A.
Let f : A → X and B ⊂ A. Then the mapping B → X such that x 7→ f (x) is denoted by f |B and is called the restriction of f to B.
The following definition provides a key tool to our proof of Tychonoff’s theo-rem.
2.1 Definition. Let A be a subset of B. And let f ∈ XA be a mapping from A into some set X. Then a mapping g ∈ XBis said to be an extension of f if g|A= f .
2.3
Generalized unions, intersections and cartesian
prod-ucts
Often it is helpful to distinguish sets whose members are also sets, by referring to them as families or collections (of sets).
2.2 Definition. LetF be a family of sets. Then we define [ F = { x : x ∈ F for at least one F ∈ F }, \ F = { x : x ∈ F for all F ∈ F },
Note that the union A ∪ B, in the binary sense, of two sets A and B, is the union of the family {A , B} (similarly for binary intersections). Hence it is justified to speak of the unions and intersections in Definition 2.2 as generalized versions of the binary operations with the same name.
Besides ”regular” families, we will find it useful to talk about so called indexed families.
2.3 Definition. LetF be a family and let A be any set. A mapping
A →F such that α 7→ Xα (2.6) is called an indexed family. We will denote the indexed family (2.6) by {Xα}α∈A.
Notice the difference between writing {Xα}α∈A and { Xα : α ∈ A }. The latter is
actually the image of the former.
Often we will refer to indexed families as, simply, families, but the subscript notation {Xα}α∈A will clarify that we talk about an indexed family (whenever this makes any difference.)
2.4 Definition. We define the unionS
α∈AXα, of an indexed family {Xα}α∈A, as the union of its image. That is
[
α∈A
Xα=
[
{ Xα: α ∈ A }.
Similarly we define the intersectionT
α∈AXα, of {Xα}α∈Aas \ α∈A Xα= \ { Xα: α ∈ A }.
Notation. It is often convenient to denote the union and intersection of a non-indexed familyF as SF∈FF and
T
F∈FF, respectively. When doing so, we are
actually viewingF as the indexed family
F → F such that F 7→ F, that is {F}F∈F.
Strictly speaking, F and {F}F∈F are not equal (in fact {F}F∈F is a subset of
F × F), but their respective unions and intersections are equal, which accounts for most practical purposes.
2.5 Proposition (De Morgan’s laws). Let X be a set and letF ⊂ P(X). Then we have:
(a) S F∈F X \ F = X \ T F∈FF, (b) T F∈F X \ F = X \ S F∈FF.
Proof. The proof is straight-forward. [
F∈F
X \ F= { x : x ∈ X \ F for at least one F ∈ F } = { x ∈ X : x < F for at least one F ∈ F } = x ∈ X : x < \ F∈F F = X \ \ F∈F F, which is (a). Similarly for (b) we have
\ F∈F X \ F= { x : x ∈ X \ F for all F ∈ F } = { x ∈ X : x ∈ F for no F ∈ F } = x ∈ X : x < [ F∈F F = X \ [ F∈F F,
which concludes the proof.
2.6 Definition. We define the generalized cartesian product Q
α∈AXα, of an in-dexed family {Xα}α∈A, as
Y α∈A Xα = f ∈ [ α∈A Xα A : f (α) ∈ Xαfor all α ∈ A .
Notice that if Xα= X for all α in Definition 2.6, the we have Qα∈AXα= XA. The generalized cartesian product is central to Tychonoff’s theorem, which says something about the generalized cartesian products of families of topological spaces, satisfying certain conditions.
A few things are worth noting: For instance, how do the generalized cartesian product of the family {X1, X2} relate to the (non-generalized) cartesian product
X1× X2, of its members? We have
Y
k∈{1 ,2}
Indeed, for f to be a mapping in (X1∪ X2){1, 2} such that f (1) ∈ X1and f (2) ∈ X2,
f must have the form {(1, x1), (2, x2)} where x1 ∈ X1 and x2 ∈ X2. So although
Q
k∈{1,2}Xkand X1× X2 aren’t equal, they do stand in a one-to-one relationship to
each other.
A mapping f ∈ Q
α∈AXα is sometimes called a choice function on the family
{Xα}α∈A (because its image f (A), is what we might get if we were to choose one element from each of the Xα. Every choice function thus corresponds to one way of making such a choice) and for any α ∈ A, it’s value f (α), at α, is called the α-coordinate of f .
Next, note that if {Xα}α∈A is any indexed family, then for every β ∈ A, there exists a unique mapping
Y
α∈A
Xα→ Xβ: f 7→ f (β),
that projects every f ∈Q
α∈AXαon its β-coordinate. These mappings are important and go under a special name.
2.7 Definition. Let {Xα}α∈Abe an indexed family. For all β ∈ A define
πβ:Y
α∈A
Xα→ Xβsuch that f 7→ f (β).
The family {πα}α∈A, of mappings are referred to as the coordinate mappings on {Xα}α∈A.
Thus we may write the α-coordinate of some f ∈Q
α∈AXαas πα( f ).
If Uα ⊂ Xα, then the inverse image π−α(Uα), of the respective coordinate map-ping, is the set of choice functions on {Xα}α∈Awhose α-coordinate is in Uα.
To every family {Xα}α∈A of non-empty sets, there exists at least one choice-function. This is, in fact, a version of the axiom of choice, to which the following section is devoted.
2.4
The Axiom of choice
2.8 Definition. LetA be a non-empty collection of non-empty, mutually disjoint sets. A set B ⊂SA is said to be a selection on A if B contains exactly one member from each A ∈A (that is, for all A ∈ A, B ∩ A is a one-member set).
Hence, if we have a family of non-empty, mutually disjoint sets, a selection on this family is obtained by simply choosing exactly one member of each set. It ap-pears, thus, that whenever we have a non-empty familyA, of non-empty, mutually
disjoint sets, we can always get a selection S onA. Indeed, we could just go ahead and choose one member from each set inA (after all, the sets are non-empty) and the set of members thus chosen is a selection onA. Most literature in algebra and analysis – including [2, 3, 5] – accepts this as a valid move, the formal statement of which is the so called axiom of choice.
The Axiom of Choice. If A is a non-empty collection of non-empty, mutually disjoint sets, then there exists a selection onA.
In certain pathological cases the axiom of choice can cause the existence of some rather strange mathematical objects and phenomena. A discussion of these cases is beyond the scope of this text. One relatively simple example can be found in [2, pages 19–21].
Problematic results, like the one in [2], only arise whenA is infinite, and seems to have something to do with the fact that there isn’t a good answer to the question: How, exactly, are we to choose one member from each set in an infinite family? Psychologically speaking, the bringing about of a selection from an infinite family of mutually disjoint sets, without constructing it in a “reasonably explicit fashion,” [2, page 17] feels a bit artificial, as opposed to whenA is finite; then, at least, we can somehow imagine the choosing in question from a finite array of characters representingA’s members.
There are some nuances of opinion as to what extent and under what conditions it is permissible to make the “choice move” on A and receive a selection to pass on. See [6, pages 490–492] for a brief introduction to this discussion.
We will use the axiom of choice several times in the sequel. In fact, several of the results that we will prove are not possible without it. But we will always be explicit about using the axiom of choice when doing so. Moreover we will use it in a different form.
2.9 Proposition (The multiplicative axiom). If {Xα}α∈A is a non-empty family of non-empty sets, thenQ
α∈AXα, ∅. Proof. Put
A = {α} × Xα: α ∈ A .
Note that for any pair of α , β ∈ A, if α= β then obviously {α}× Xα= {β}× Xβ; else, if α , β, then {α} × Xαand {β} × Xβcannot have any element in common. Indeed, for all members (ξ, x) ∈ {α} × Xα, there is only one choice for ξ (that is, ξ= α) so
we can not have (ξ, x) ∈ {β} × Xβ, since this would render ξ= β, and thereby α = β. Thus it is clear thatA is a non-empty collection of non-empty, mutually disjoint sets.
It turns out that S ∈Q
α∈AXα. In fact, note that S ⊂ A × S
α∈AXα. Thus it
only remains to verify that for each α ∈ A there exists xα ∈ Xαsuch that (α, xα) ∈ S ,
and if x ∈S
α∈AXαwith (α, x) ∈ S then x= xα.
Accordingly, let α ∈ A. Since S is a selection onA, S ∩ {α} × Xα is a
one-member set. Hence there exists xα ∈ Xα such that (α, xα) ∈ S . Now, suppose
x ∈ S
α∈AXα such that (α, x) ∈ S . By (α, x) ∈ S ⊂ SA, we get some β ∈ A such that (α, x) ∈ {β} × Xβ. Clearly β = α, so (α, x) ∈ S ∩ {α} × Xα, which is
a one-member set that already contains (α, xα). Hence (α, xα) = (α, x), rendering x= xα, which concludes the proof. Bertrand Russel used Proposition 2.9 instead of the axiom of choice and called it the multiplicative axiom [7, page 536]. In practice Russel’s version of the axiom of choice is certainly more useful (although less primitive). Clearly it implies the axiom of choice. Hence the multiplicative axiom and the axiom of choice mutually equivalent.
2.5
Orderings and Zorn’s lemma
2.10 Definition. Let S be a set. A relation6 on S is called a partial ordering on S if
(i) x6 x, for all x ∈ S ,
(ii) If x6 y and y 6 x, then x = y, for all x, y ∈ S , (iii) If x6 y and y 6 z, then x 6 z, for all x, y, z ∈ S . We also say that S is partially ordered (by6).
Terminology. The three listed properties (i), (ii) and (iii) that make up a partial ordering in Definition 2.10, are commonly called, the reflexive, anti-symmetric and transitiveproperty, respectively.
We will occasionally adopt the notation: ‘x < y’ for ‘x6 y and x , y’.
2.11 Definition. A partial ordering6 on a set S , is called a total ordering if for all x, y ∈ S , we have either x 6 y or y 6 x. Then S is said to be totally ordered.
Notice that if T is a subset of some partially ordered set S , then T becomes par-tially ordered as well. This by letting the partial ordering on S induce an ordering on T . On top of being partial, the induced ordering may very well be total. Below there will be plenty occasion to speak of totally ordered subsets of some partially ordered set.
2.12 Definition. Let T be a subset of some partially ordered set S . An element b ∈ S is called an upper bound for T in S if for all x ∈ T , we have x 6 b. If also b6 b0for all upper bounds b0, then b is called a least upper bound (for T in S ). 2.13 Definition. Let S be partially ordered by6 and let m ∈ S . We will say that m is a maximal element of S if whenever x ∈ S such that m6 x, then m = x.
Notice that neither upper bounds nor maximal elements necessarily are unique (there may be several), but that least upper bounds are.
2.14 Definition. A partially ordered set S is said to be inductively ordered if every totally ordered subset T of S , has an upper bound b in S . Moreover, we say that S is strictly inductively ordered if b is a least upper bound for T in S .
Zorn’s lemma. If S is a non-empty, inductively ordered set, then S has a maximal element.
Zorn’s lemma is a critical constituent of every known proof of Tychonoff’s theorem as well as many other results in algebra and analysis (notably the Hahn-Banach theorem [2, page 158] or that every vector space has a basis.)
In Appendix A, we have set forth a proof of the statement that Zorn’s lemma follows from the axiom of choice.
Chapter 3
Point Set Topology
3.1
Basics
3.1 Definition. Let X be a non-empty set. A familyO ⊂ P(X) is called a topology on X if
(i) ∅ , X ∈ O,
(ii) SF ∈ O for all F ⊂ O, (iii) TF ∈ O for all finite F ⊂ O.
The ordered pair (X,O) is called a topological space.
Terminology. If (X,O) is a topological space, then the sets E ∈ O are said to be open(in X) and sets of the form X \ E, where E ∈O, are said to be closed (in X.) If the topologyO is understood or unspecified, we may drop reference to it and speak of X, simply, as a topological space.
Let X be a topological space. Notice that the sets ∅ and X are both open and closed in X.
3.2 Example. For every non-empty set X, {∅, X} and P(X) are both topologies on Xcalled the trivial and discrete topology respectively.
3.3 Proposition. Let X be a topological space and let E ⊂ X.
(a) If E is closed in X, then X \ E is open in X.
Proof. If E is closed in X, then there exists an open set F in X, such that E= X \ F, so X \ E= X \ X \ F = { x : x ∈ X and x < X \ F } = { x ∈ X : not (x < F) } = F,
which is open in X. Hence (a).
To prove (b), assume that X \ E is open in X. Then X \ X \ E is closed in X, but with identical reasoning as in the proof of (a), we get
X \ X \ E= E.
Hence (b).
3.4 Proposition. Let X be a topological space. IfF is a family of closed sets then TF is also closed in X. Furthermore, if F also is finite, then S F is closed in X as well.
Proof. We begin by proving the first part of the statement. Accordingly, let F ⊂ P(X) be a family of sets such that each F ∈ F is closed in X.
We want to show thatT
F∈FF is closed in X. By Proposition 3.3.b, this will
follow if we can show that X \ T
F∈FF is open in X. But by Proposition 3.3.a,
X \ F is open in X for all F ∈ F. Hence SF∈F X \ F is open in X. Now by
Proposition 2.5.a we get [
F∈F
X \ F= X \ \
F∈F
F, thus concluding (a).
To prove the second part, assume thatF, in addition, is finite. We want to show that S
F∈FF is closed in X. But by Proposition 3.3.b, this
will follow if we can show that X \ S
F∈FF is open in X. But, since X \ F is open
in X for all F ∈ F and F is finite, we get TF∈F X \ F to be open in X and thus
X \ S F∈FF is open in X, since [ F∈F X \ F= X \ [ F∈F F by Proposition 2.5.b.
(a) Ao= S{ E : E ⊂ A and E is open in X }, (b) A= T{ E : E ⊃ A and E is closed in X }.
Aois called the interior of A and A is called the closure of A.
Let X be a topological space and let A ⊂ X. Note that Aois open in X and that Ais closed in X. In fact, Aois the largest open set that is a subset of A and A is the smallest closed set that has A as a subset.
3.6 Definition. Let X be a topological space and let x ∈ X. A set U ⊂ X is called a neighborhood of x if x ∈ Uo.
3.7 Definition. Let X be a toplogical space and let A ⊂ X. A point x ∈ X is called an accumulation point of A if for every neighborhood U of x, we have A ∩ Uo\ {x}
, ∅. The set of accumulation points of A will be denoted by acc(A).
3.8 Proposition. Let X be a topological space and let A ⊂ X. Then A= A∪acc(A). Proof. First we will show that A ⊂ A ∪ acc(A). Notice that this will follow if we can show that X \ [A ∪ acc(A)] ⊂ X \ A. Consequently, let x ∈ X \ [A ∪ acc(A)]. We want to show that x < A. But since x < acc(A) there must exist an open set U, such that x ∈ U and A ∩ U \ {x}= ∅. Hence, since x < A, we get A ∩ U = ∅. In particular, X \ U is both closed and A ⊂ X \ U, which renders A ⊂ X \ U. Now, since x ∈ U, we get x < X \ U and thus x < A as well.
It remains to show that A ∪ acc(A) ⊂ A. Again we will do this by showing X \ A ⊂ X \[A ∪ acc(A)].
Accordingly, let x ∈ X \ A. We need to show that x < A and x < acc(A). But x < A follows immediately, since we have A ⊃ A and x < A. To show that x < acc(A) it is sufficient to find a neighborhood U of x, such that A ∩ U = ∅. But note that X \ A is open and contains x. Hence it is a neighborhood of x that in addition does not intersect A, which is exactly what we were looking for. 3.9 Proposition. Let O be a family of topologies on some set X. Then T O is a topology on X.
Proof. Since ∅ , X ∈ O for all O ∈ O, we have ∅ , X ∈ T O and ifF ⊂ T O then F ⊂ O for all O ∈ O. Hence S F ∈ O for all O ∈ O, which is to to say that SF ∈ T O. In the same way we get T F ∈ T O for all finite F ⊂ T O. Thus O is
a topology on X.
3.10 Definition. Let X be any non-empty set and letE ⊂ P(X). Then the topology \
{O : O ⊃ E and O is a topology on X } is called the topology on X generated byE.
Let X be a non-empty set and letE ⊂ P(X). Note that – if nothing else – E is always a subset of the discrete topology. Hence, the intersection in Definition 3.10 is always non-empty, so the topology on X generated byE always exists.
3.11 Definition. LetO be a topology on a set X. A family B ⊂ O is said to be a baseforO if for all U ∈ O \ {∅} there exists F ⊂ B, with U = S F.
3.12 Theorem. Let X be a non-empty set and letE ⊂ P(X). Then the set B = \ F : F ⊂ E and F is finite ∪ {X}
is a base for the topology on X generated byE.
Proof. Denote the topology on X generated byE by O(E) and put O = { U ⊂ X : for all x ∈ U there exists V ∈ B with x ∈ V ⊂ U }. If we can prove the following:
(a) O is a topology on X, (b) B is a base for O, (c) E ⊂ O ⊂ O(E),
in that order, we’re done. Indeed, if (c) thenO = O(E) (provided (a)).
It is immediately clear that ∅ , X ∈ O. Also note thatSF ∈ O for all F ⊂ O. Indeed, if x ∈ SF, then x ∈ F for some F ∈ F, but this F is a member of O, so x ∈ V ⊂ F ⊂SF for some V ∈ B. Thus, to complete the proof of (a), we need to show thatO is closed under finite intersections.
Accordingly let U1, . . . , Un ∈O and let x ∈ U1∩ · · · ∩ Un. We need to show
that there exists V ∈B with x ∈ V ⊂ U1∩ · · · ∩ Un. But for each j= 1 , . . . , n we
have some Vj ∈B with x ∈ Vj ⊂ Uj. Now, since each Vj is a finite intersection of
sets inE, the set V = V1∩ · · · ∩ Vnmust be a finite intersection of sets inE as well,
so V ∈B. Moreover, we have x ∈ V ⊂ U1∩ · · · ∩ Un, and (a) follows.
To prove (b), first note thatB ⊂ O. Indeed, if V ∈ B, then for all x ∈ V, we have x ∈ V ⊂ V. Now, let U ∈O \ {∅}. We need to show that there exists F ⊂ O with U= S F. But note that for every x ∈ U the sets
Ax= { V ∈ B : x ∈ V ⊂ U }
are non-empty. Hence by Proposition 2.9 there exists Γ ∈Y
x∈U
so {Γ(x) : x ∈ U } ⊂ B, and U = Sx∈UΓ(x). Which is (b).
In order to prove (c), first note that if E ∈E then, by {E} being a finite subset ofE, we get T{E} = E ∈ B. Hence we have E ⊂ B ⊂ O. It thus remains to show thatO ⊂ O(E). Accordingly, let U ∈ O \ {∅ , X} (it is clear that ∅ , X ∈ O(E)). We want to show that U ∈ O(E). But since B is a base for O, there exists F ⊂ B such that U = S F.
If we can show that every F ∈ F is a member of O(E), we’re done (since O(E) is closed under unions). But each F ∈ F is a finite intersection of sets in E (recall thatE ⊂ O(E) and that the latter, by being a topology, is closed under finite intersections). Hence F ⊂ O(E), which concludes the proof of (c) and, with that,
the proof.
3.2
Compactness
3.13 Definition. Let X be a topological space and letF be a family of open sets of XwithSF = X, then F is called an open cover of X.
3.14 Definition. Let X be a topological space. X is said to be compact If for every open coverF of X there exists a finite E ⊂ F such that E also is a cover of X. Terminology. The literature on the subject often refers toE in Definition 3.14 as a finite subcoverand defines a topological space X as compact if and only if every open cover of X has a finite subcover.
Allthough, the term ‘subcover’ might sometimes suggest itself to be ‘a cover of a subset of X,’ rather than ‘a proper cover of X that is a subset of the original cover,’ it is always the latter meaning that’s in play.
There are other ways to describe compactness for topological spaces, that turns out to be equivalent to the description put forth in Definition 3.14. We will present two additional descriptions: one will follow just below – Proposition 3.16 – and the other will be presented as Theorem 4.17 on page 33.
3.15 Definition. Let X be any non-empty set. A familyF ⊂ P(X) is said to have the finite intersection property ifTE , ∅ for all finite E ⊂ F.
3.16 Proposition. Let X be a topological space. Then X is compact if and only if for every family F of closed sets with the finite intersection property, we have TF , ∅.
Proof. Starting with the only if -part. Assume that X is compact and let F be a family of closed sets such that for all finiteE ⊂ F, we have T E , ∅.
We want to show thatTF , ∅. But suppose this is not the case. Then we get X \\ F = X \ \ F∈F F= [ F∈F X \ F= X,
where the third identity is due to Proposition 2.5.a. Notice that since all F ∈ F are closed, we have X \ F open for all F ∈ F. Thus we get { X \ F : F ∈ F } to be an open cover of X and since X is compact this cover has a finite subcover, so there existsE ⊂ F such that E is finite and SF∈E X \ F = X. But then – using
Proposition 2.5.a again – we get [
F∈E
X \ F= X \ \
F∈E
F= X \\ E = X,
renderingTE = ∅ contradicting F having the finite intersection property. Hence we can’t haveTF = ∅. Hence T F , ∅ follows, which concludes the only if -part of the proof.
To prove the if -part, suppose that wheneverF is a family of closed sets with the finite intersection property, we haveTF , ∅.
We want to show that X is compact in sense of Definition 3.14. Accordingly, letF be an open cover of X. We are done if we can show that there exists some finiteE ⊂ F such that S E = X. But, by Proposition 2.5.b, we get
X \[ F = X \ [
F∈F
F= \
F∈F
X \ F, and sinceSF = X, it follows that T
F∈F X\F
= ∅. Also note that { X\F : F ∈ F } is a family of closed sets. Hence, by our initial assumption, { X \ F : F ∈F } can’t have the finite intersection property. This means that there exists a finiteE ⊂ F such thatT F∈E X \ F = ∅. By Proposition 2.5.b, we have \ F∈E X \ F= X \ [ F∈E F= X \[ E.
HenceSE = X, which concludes the if -part of the proof and with that the proof.
3.3
The product topology
3.17 Definition. Let {Xα}α∈A be a non-empty family of topological spaces, and let {πα}α∈A be the coordinate-mappings on Q
α∈AXα. Then the topology on X = Q
α∈AXα, generated by the set π−
α(Uα) : α ∈ A and Uαis open in Xα
At this point it is possible to formulate our main theorem.
3.18 Tychonoff’s theorem. Let {Xα}α∈A be a non-empty family of compact topo-logical spaces. Then X= Qα∈AXαis compact with respect to the product topology. We are not ready to give our proof of the theorem just yet. The proof we will give (starting on page 34) is based on the theory of nets (to be treated the next chapter).
The next proposition is key to our proof of Tychonoff’s theorem. Before stating it, however, we will find it convenient to introduce a new piece of terminology. Terminology. If A is a set such that a statementSis true for all but a finite number of α ∈ A, then we will say thatSis true for almost all α ∈ A.
3.19 Proposition. Let {Xα}α∈Abe a non-empty family of topological spaces. Then the set
B = Y
α∈A
Uα: Uαis open in Xαfor allα and Uα= Xαfor almost allα
is a base for the product topology on X = Qα∈AXα.
Proof. Since the product topology on X is generated by the set
E = π−
α(Uα) : α ∈ A and Uαis open in Xα ,
where {πα} is the coordinate mappings on X, we get – by Theorem 3.12 – that the set
B0= \ F : F ⊂ E and F is finite ∪ {X}
is a base for X. Moreover, since X ∈ E (X = π−α(Xα) for any α ∈ A) we get T{X}= X ∈ B0. Hence,
B0= \ F : F ⊂ E and F is finite .
If we can show thatB0= B, we’re done.
Starting with B0 ⊂ B: Let U ∈ B0. Then there exists a finite number of elements α1 . . . , αn ∈ A associated with sets Uαj open in the respective Xαj, such
that U= n \ j=1 π− αj(Uαj) .
Now, for each α ∈ A \ {α1, . . . , αn}, put Uα = Xα ,and notice that we getQα∈AUα to be a member ofB, as well as Y α∈A Uα = f ∈ X : f (αj) ∈ Uαj for all j= 1 , . . . , n = n \ j=1 f ∈ X : παj( f ) ∈ Uαj = n \ j=1 π− αj(Uαj) = U. Hence U ∈B, so B0 ⊂B.
To proveB ⊂ B0, let U ∈ B. Then U = Qα∈AUα for some family {Uα}α∈A
where Uαis open in Xαfor all α and Uα = Xαfor almost all α. Put
B= { α ∈ A : Uα, Xα} and notice that
\
β∈B
π−
β(Uβ) ∈B0, since B is finite, as well as
\ β∈B π− β(Uβ)= f ∈ X : f (β) ∈ Uβfor all β ∈ B =Y α∈A Uα = U.
Chapter 4
Nets
4.1
Directed sets
The theory of nets begin with the notion ‘directed set’. The class of directed sets is a subclass of the class of partially ordered sets where the former is separated from the latter as follows:
4.1 Definition. A partially ordered, non-empty set (A,4), is called a directed set if for any choice of α , β ∈ A, there exists γ ∈ A with α4 γ and β 4 γ.
Terminology. If (A,4) is a directed set we also say that that A is directed by 4. Whenever4 is understood or unspecified, we may drop reference to it and speak of A, simply, as a directed set.
As with any partial ordering we may write ‘β< α’ for ‘α 4 β’.
4.2 Example. Let B be any non-empty set and putA = { A ⊂ B : A is finite }. Note that for each pair of A , A0 ∈ A we have A ∪ A0 ∈ A and A , A0 ⊂ A ∪ A0. Hence (A, ⊂) is a directed set.
4.3 Proposition. Let X be a topological space and x a point in X. LetNxbe the set
of neighborhoods of x.1If we take ‘4’ to designate the relation on Nxdefined by:
U4 V if and only if U ⊃ V, thenNxis directed by4.
Proof. We need to verify the following:
(a) 4 is a partial ordering on Nx,
1N
(b) If U , V ∈Nxthen there exists W ∈Nxsuch that U4 W and V 4 W.
Clearly4 is reflexive and anti-symmetric, and whenever we have U 4 V and V 4 W then U ⊃ V ⊃ W, so U ⊃ W which is to say U 4 W. Hence 4 is transitive which concludes (a).
To prove (b), note that whenever U , V ∈ Nx then also x ∈ Uo∩ Vo which is
open in X and thus a member ofNx, so we get both U 4 Uo∩ Voand V 4 Uo∩ Vo
( U ⊃ Uoand V ⊃ Vo, by Definition 3.5.)
Following the litterature on the subject, we will say thatNx is directed by
re-verse inclusion.
4.4 Proposition. Let (A,4) and (A0, 40) be directed sets, and let400be a relation
with respect to A × A0defined by:
(α, α0)400 (β, β0) if and only if α 4 β and α0 40β0. Then A × A0is directed by400.
Proof. Clearly400is a partial-ordering on A × A0, so it remains to show that given (α, α0) , (β, β0) ∈ A × A0 there exists (γ, γ0) ∈ A × A0 such that (α, α0) , (β, β0) 400 (γ, γ0). But note that there exists γ ∈ A with α , β 4 γ as well as a γ0 ∈ A0 with
α0, β0
40γ0, so (γ, γ0) works.
4.2
Nets and Subnets
4.5 Definition. Let X be any set and let A be a directed set. Any mapping in XAis called a net in X.
Notation. For a convenient way to denote a net of the form
A → X: α 7→ xα
we will adopt the notation hxαiα∈Aor simply hxαi (if A is understood). We will also
say that nets of this form are based on A.
4.6 Example. The set of positive integers is directed by the usual5. Hence every sequence is a net. Indeed, consider an arbitrary sequence {xn} in some set X, and
recall that this is in fact a short-hand notation for the mapping
4.7 Example. Let X be a non-empty set. Then any subsetM of P(X) with X ∈ M is directed by ⊂. Suppose that M is a σ-algebra on X. Then every measure µ : M → [0, ∞] is a net in [0, ∞].
4.8 Definition. Let (A,4) and (B, 40) be directed sets. A mapping β 7→ αβ in AB
is called cofinal if for each α0 ∈ A there is at least one β0 ∈ B such that whenever
β <0β
0we have αβ < α0.
4.9 Example. Increasing mappings2 from some directed set A into itself are al-ways cofinal. Indeed, let A be a set directed by4 and let φ ∈ AA be an increasing mapping (i.e. α4 φ(α) for all α ∈ A). Then choose any α0 ∈ A. To be able to say
that φ is cofinal we now need to find a β0 ∈ A such that whenever β < β0, we have
φ(β) < α0, but note that in this case we can simply choose α0itself to work as such
a β0.
Staying with the case where φ ∈ AAand A being directed by4, we’ve just seen (in Example 4.9) that if φ is increasing, then φ is cofinal. The converse, however, isn’t necessarily the case. In fact, there are cofinal mappings that map a directed set into itself, but aren’t increasing (cf. Example 4.10).
Likewise, it might also – under these circumstances – be tempting to make the conjecture that every cofinal mapping in AA is monotonically increasing, that is, whenever α 4 α0 we have φ(α) 4 φ(α0). But as the following example will illustrate, this is not the case either. Worth mentioning is also the fact that there are plenty of monotonically increasing mappings that are not cofinal, among which the mapping
x 7→arctan(x)
in RR is one example. Indeed, simply pick α
0 = π2. There are no β0 ∈ R such that
arctan(β0) ≥ π2.
4.10 Example. Consider the mapping
φ : Z+→ Z+: α 7→ (α+ 3) − (α mod 3) − (α mod 5).
First we will show that φ is cofinal. But notice that for any choice of α ∈ Z+, we have φ(α) ≥ α − 4, or equivalently: φ(α+ 4) ≥ α. Hence, if we pick an α0 ∈ Z+
we can simply put β0= α0+ 4 so that whenever we have β ∈ Z+such that β ≥ β0,
we get φ(β) = φ(β + 4 − 4) ≥β − 4 ≥β0− 4 = (α0+ 4) − 4 = α0,
which is to say that φ is cofinal.
As we run through the first 10 arguments of φ we get the associated values according to the following table:
x 1 2 3 4 5 6 7 8 9 10 · · · φ(x) 2 1 3 2 6 8 7 6 8 12 · · ·
This shows clearly that φ is neither increasing (in the sense of α ≤ φ(α) for all α) nor monotonically increasing (in the sense of having φ(α) ≤ φ(α0) whenever α ≤ α0). This holds even when we cut of the domain of φ to exclude all integers that
aren’t sufficiently large. In fact, notice that for any α ∈ Z+of the form α= 3n5m we have φ(α)= α + 3, but since
(3n5m+ 1) mod 3 > 0 and
(3n5m+ 1) mod 5 > 0, we get
φ(α + 1) ≤ α + 2 < φ(α).
Neither will φ start to become increasing beyond any sufficiently large point. Indeed, suppose α0were to be such a point. Then we cannot have any α ∈ Z+such
that α > α0and α mod 5= 4 (we would get φ(α) = α − 1 − (α mod 3) < α).
In preparation for the next definition (which is somewhat intricate), it is helpful to take note of the following:
Let hxαiα∈Abe a net in some set X. Notice that if φ ∈ ABand B is any directed set besides A, then the composite mapping
β 7→ φ(β) 7→ xφ(β)
4.11 Definition. Let A and B be directed sets and hxαiα∈Aa net in some set X. If φ is a cofinal mapping in AB, then the net hx
φ(β)iβ∈Bis called a subnet of hxαiα∈A.
4.12 Proposition. Let s, s0and s00be nets in the same set X. If s00is a subnet of s0 and s0is a subnet of s, then s00is a subnet of s.
Proof. Assume s , s0and s00to be based on (A,4) , (B, 40) and (C,400) respectively and take
hxαiα∈A
to denote s. That s0is a subnet of s is to say that there is a cofinal mapping φ ∈ AB such that
s0= hxφ(β)iβ∈B.
And that s00is a subnet of s0is in equal fashion to say that there is a cofinal mapping ψ ∈ BC such that
s00 = hxφ(ψ(γ))iγ∈C = hxφ◦ψ(γ)iγ∈C .
Hence, if we can show that the composite mapping φ ◦ ψ in AC is cofinal we’re done. But let α0 ∈ A. We want to show that there exists γ0∈ C such that φ ◦ ψ(γ) <
α0for all γ<00 γ0.
By φ being cofinal, we get β0 ∈ B such that φ(β) < α0 for all β <0 β0, and
by ψ being cofinal, there exists γ0 ∈ C such that ψ(γ) <0 β0 for all γ <00 γ0. It is
thus clear that for all γ<00 γ0, we have φ ◦ ψ(γ)< α0. Hence φ ◦ ψ is cofinal, thus
concluding the proof.
4.13 Example. Let {xn} be a sequence in some non-empty set X. A subsequence,
usually denoted by {xnj} of {xn} is a composite mapping
N → N → {xn: n ∈ N} : j 7→ nj 7→ xnj
such that n1 < n2< . . .
Recall that {xn} is a net in X (cf Example 4.6). Moreover, by Example 4.9, the
mapping j 7→ njis cofinal. Hence every subsequence is also a subnet.
Since every sequence {xn} qualifies as a net and every subsequence {xnj} to {xn}
qualifies as a subnet to {xn} (when the latter is viewed as a net), it is fair to say
(to this extent at least) that the theory of nets and subnets is a generalization of the theory of sequences and subsequences. Many more analogies do exist and we will look at a few of them in the next section.
4.3
Nets and Compactness
It turns out that the notion of compactness in topological spaces can be described completely in terms of the theory of nets. This section gives the details.
4.14 Definition. Let hxαiα∈Abe a net in X and let U ⊂ X. Then we say that: (a) hxαi is eventually in U if there exists α0∈ A with xα ∈ U for all α < α0,
(b) hxαi is frequently in U if for all α ∈ A, there exists β < α with xβ∈ U.
The definition above puts no demands on the set X. But note that the following definition only deals with cases in which the set X is a topological space.
4.15 Definition. Let X be a topological space and let hxαiα∈Abe a net in X. Then we say that:
(a) hxαi converge to a point x ∈ X if for all neighborhoods U of x, hxαi is
eventually in U,
(b) A point x ∈ X is a cluster point of hxαi if for all neighborhoods U of x, hxαi is frequently in U.
4.16 Proposition. Let hxαiα∈A be a net in a topological space X and let x ∈ X. Then x is a cluster-point of hxαi if and only if hxαi has a subnet converging to x. Proof. Starting with the if -part: Assume that hxαβiβ∈B is a subnet converging to x
(we may take4 and 40to designate the relations that directs A and B respectively). We need to show that for any neighborhood U of x, hxαi is frequently in U. Hence let U be an arbitrary neighborhood of x and let α0 ∈ A. If we can find an
α < α0 with xα ∈ U, then the if -part follows. But β 7→ αβbeing cofinal, gives us
a β1 ∈ B with αβ < α0for all β <0 β1, and hxαβi converging to x gives us a β2∈ B
with xαβ ∈ U for all β <0 β2. Now, since B is directed by40 we get a β ∈ B with
β1, β2 40 β. Note that αβ < α0and xαβ ∈ U.
It remains to show the only if -part. To do this assume that x is a cluster-point of hxαi.
Our aim is to find a subnet converging to x.
LetN be the set of neighborhoods of x in X. By Proposition 4.3, N is directed by reverse inclusion. Hence, by Proposition 4.4,N × A is directed by 400, defined by:
(U, γ)400(U0, γ0) if and only if U ⊃ U0and γ4 γ0. Next, for all (U, γ) ∈N × A we put
Notice that x being cluster-point of hxαi, togheter with U being a neighborhood of x, renders each X(U,γ) non-empty. Indeed, given some (U, γ) ∈ N × A, we get
hxαi frequently in U, thus there exists α < γ with xα ∈ U. This α is an element of X(U,γ).
Now by Proposition 2.9 we get Y
(U,γ)∈N×A
X(U,γ), ∅.
Hence there exists a mapping (U, γ) 7→ α(U,γ)in AN×Asuch that α(U,γ) ∈ X(U,γ)
for all (U, γ) ∈N × A.
This mapping is cofinal. Indeed, letting α0 ∈ A we may simply choose any
V0 ∈N and consider elements (U, γ) <00 (V0, α0) to see directly that α(U,γ) < γ <
α0.
We thus get hxα(U,γ)i to be a subnet of hxαi.
To conclude the proof it remains only to show that hxα(U,γ)i converge to x.
Ac-cordingly, let V0be a neighborhood of x. We need to find some (U0, γ0) ∈N × A
such that whenever (U, γ)<00 (U0, γ0) we have xα(U,γ) ∈ V0. But (V0, γ0) works for
any choice of γ0 ∈ A. Indeed, let (U, γ) <00 (V0, γ0). Then it follows immidiately
that xα(U,γ) ∈ U ⊂ V0.
4.17 Theorem. Let X be a topological space. Then X is compact if and only if every net in X has a convergent subnet.
Proof. Starting with the only if -part: Assume X is compact and let hxαi be a net, based on (A,<), in X . We need to show that hxαi has a subnet that converges to some x ∈ X. Start by putting, for all α ∈ A
Eα= { xβ : β< α }.
Note that the family {Eα}α∈A has the finite intersection property (desribed in Defi-nition 3.15). 3
Since for all α, Eαis a subset of its closure Eα, the family {Eα}α∈Aalso has the finite intersection property, and since X is compact it follows from Proposition 3.16 thatT
α∈AEαis non-empty, so there exists x ∈T
α∈AEα.
We will now show that this x is a cluster-point of hxαiα∈A, which by Proposi-tion 4.16, would guarantee the existence of a subnet of hxαi, that converge to x. But
3This follows from the induction-principle. Indeed, ifT
γ∈BEγ , ∅ (trivially true whenever |B|= 1) for all subsets B of A of size n, then for any subset B of A of size n + 1, we can yank some β0 from B and getTγ∈B\{β0}Eγ, ∅, because B \ {β0} is of size n. Thus we get some β1 ∈ A with
xβ1 ∈ Eγfor all γ ∈ B \ {β0}, which means that β1 < γ for all γ ∈ B \ {β0}. Now by Definition 4.1
there exists β2∈ A with β0, β14 β2, thus xβ2∈ Eγfor all γ ∈ B, rendering
T
let U be a neighborhood of x in X and let α0∈ A, then we have x ∈ Eα0, which, in
particular, renders U ∩ Eα0 , ∅.
4
Hence there exists β < α0 with xβ ∈ U, so x is a cluster-point of hxαi, which
concludes the only if -part of the proof.
To prove the if -part: assume that every net in X has a convergent subnet. We want to prove that X is compact.
First note, however, that by Proposition 4.16, we have just assumed that every net in X has at least one cluster-point. Now, suppose X is not compact. Our plan is to prove this to be impossible by using this assumption to find a net in X that doesn’t have any cluster-points.
Since X is not compact we get an open cover {Uβ}β∈B of X that does not have any finite sub-covers. Put
A = { A ⊂ B : A is finite }.
Note that (A, ⊂) is a directed set (cf. Example 4.2) and for each A ∈ A, X\Sα∈AUα
is non-empty. Hence by Proposition 2.9 we have Y
A∈A
(X \[
α∈A
Uα) , ∅.
Thus we get a mapping A 7→ xA in XAsuch that xA ∈ X \Sα∈AUαfor all A ∈A.
This mapping is clearly a net in X. Denote it by hxAi.
Now we claim that hxAi does not have any cluster-points. Indeed, let x be any
element of X. Since {Uβ}β∈Bis an open cover of X we get x ∈ Uβ for some β ∈ B.
In particular Uβis a neighborood of x. Note that {β} ∈ A and that for all A ⊃ {β} we have
xA< [
α∈A
Uα ⊃ Uβ,
that is, xA < Uβ for all A ⊃ {β}, which is to say that hxAiA∈A is not frequently in
Uβ, so x is not a cluster-point of hxAi, which concludes the if -part of the proof and
with that the proof.
4.3.1 A proof of Tychonoff’s theorem
Since our proof of Tychonoff’s theorem is quite involved, we will – for readability-purposes – make use of a full section to outline it.
4If x ∈ E
α0then x itself is an element in U ∩ Eα0. Else, if x < Eα0then we have x ∈ acc(Eα0), since by Proposition 3.8, Eα0= Eα0∪ acc(Eα0), so U ∩ Eα0, ∅ follows.
Throughout this section, let {Xα}α∈A be a non-empty family of compact topo-logical spaces. To prove Tychonoff’s theorem (page 25), which is the statement that Q
α∈AXαis compact with respect to the product topology, it is – by Theorem 4.17 – sufficient to show that every net in Qα∈AXαhas a cluster point. Accordingly, let h fiii∈I be a net inQα∈AXα.
Our aim is to show that h fiii∈Ihas a cluster point.
To begin with, note that for each subset B of A, the restrictions fi|B, of the
mappings fi, are all members of the productQα∈BXα. Hence fi|B i∈I is a net in
Q
α∈BXα.5 Sets of this form may be called subproducts ofQ
α∈AXαand they come with their respective product topology. With this in mind put
P = [
B∈P(A)
n
p ∈Y
α∈B
Xα : p is a cluster point of h fi|Bii∈Io,
and define a relation6 on P as:
p6 q if and only if q is an extension of p, with the meaning of ‘extension’ taken in the sense of Definition 2.1.
Notice that (P, 6) is a partially ordered set. Another way to view P is as a subset to A × S
αXα. In fact the ordering 6 is nothing but the standard subset
ordering ⊂ on A × S
αXα, only restricted toP, which in particular means that if
pand q are members ofP and Bpand Bqare their respective domains, then to say
that p 6 q, is simply to say that we have Bp ⊂ Bq and that q(α) = p(α) for all
α ∈ Bp.
Our plan is to show thatP is non-empty and inductively ordered by 6 and thus, by Zorn’s lemma, we will get a maximal element p∗∈P. If the domain of p∗turns out to be A (which it will,) then we’re done.
4.18 Lemma. Letγ ∈ A. Then the net h fi|{γ}ii∈Ihas a cluster point inQα∈{γ}Xα.
Proof. Since Xγis compact the net h fi(γ)i in Xγ, has – by Theorem 4.17 – a cluster
point p in Xγ. Notice that
Y
α∈{γ}
Xα= X{γ}γ = {(γ, x)} : x ∈ Xγ .
We will proceed to show that {(γ, p)} is a cluster point of h fi|{γ}ii∈I.
Accord-ingly, let U ⊂ Xγ{γ} be neighborhood of {(γ, p)}. We need to show that h fi|{γ}i is
frequently in U. But by Proposition 3.19, the set n Y
α∈{γ}
Uα: Uαis open in Xγo = { U{γ}γ : Uγis open in Xγ}
5Chernoff [1], denotes this net by { f
is a base for the product topology on Xγ{γ}. Thus there exists an open set Uγ in Xγ
such that {(γ, p)} ∈ Uγ{γ}⊂ U.
Now clearly, h fi(γ)i is frequently in Uγ. Hence, h fi|{γ}i is frequently in Uγ{γ}.
Indeed, fi|{γ} = γ, fi(γ) for all i ∈ I and Uγ{γ} = { {(γ, x)} : x ∈ Uγ}, so we have fi|{γ} ∈ Uγ{γ}whenever fi(γ) ∈ Uγ, which concludes the proof.
In particular, Lemma 4.18 shows that P is non-empty. Indeed, there exists a mapping {γ} → Xγ that is a member ofP (namely the cluster point to h fi|{γ}ii∈I)
for all γ ∈ A. Next, we will turn to show thatP is inductively ordered by 6. 4.19 Lemma. LetT be a non-empty, totally ordered subset of (P, 6). Then there exists p∗∈P such that p∗is an upper bound forT.
Proof. Put
p∗=[ T,
and for each p ∈ T, denote the domain of p by Bp. We will show that p∗ is an
upper bound forT in P.
We’re done if we can show the following three items:
(a) p∗∈Q
α∈B∗Xα, where B∗= Sp∈TBp,
(b) p∗is an extension of every p ∈T, (c) p∗is a cluster point of the net h fi|B∗i.
Indeed, if (a) and (c) holds, then p∗∈P, and if in addition (b) holds, then p∗is also an upper bound forT.
Starting with (a): Clearly p∗⊂ B∗× S
α∈B∗Xα, thus – for (a) – it only remains
to show that for every α ∈ B∗, two conditions holds, namely
(i) there exists x ∈ Xαsuch that (α, x) ∈ p∗,
(ii) if x0∈S
α∈B∗Xαwith (α, x0) ∈ p∗, then x= x0.
But let α ∈ B∗. By the fact that α ∈ Bpfor some p ∈T, we get α, p(α) ∈ p ⊂ p∗,
which is (i). Now suppose x ∈ S
α∈B∗Xα such that (α, x) ∈ p∗. Then there exists
q ∈ T with x = q(α) and since T is totally ordered by extension, we have either pbeing an extension of q or q being an extension of p. Either way, p(α) = q(α) holds. Hence (ii), and (a) is clear.
Note that (b) now follows immediately. In fact, if p ∈ T and α ∈ Bp then
(α, p(α)) ∈ p ⊂ p∗, and by p∗being a mapping, there can be no x other that p(α) such that (α, x) ∈ p∗, which is to say that p(α)= p∗(α), so p∗is an extension of p.
Thus it only remains to prove (c).
Accordingly, let U be a neighborhood of p∗inQ
α∈B∗Xα. We need to show that
h fi|B∗i is frequently in U. But by Proposition 3.19, the set
n Y
α∈B∗
Uα : Uαis open in Xαfor all α and Uα = Xαfor almost all αo is a base for the product topology on Q
α∈B∗Xα. Hence, there exists a family
{Uα}α∈B∗ such that p∗ ∈ Qα∈B∗Uα ⊂ U, all Uα are open in the respective Xα
and Uα = Xαfor almost every α ∈ B∗.6 Thus, we’re done if we can show that the net h fi|B∗i is frequently inQα∈B∗Uα.
Accordingly, let i0 ∈ I. We need to find some i < i0 with fi|B∗ ∈ Qα∈B∗Uα.
But, let α1, . . . , αnbe those α ∈ B∗for which Uα , Xα (they’re finite in number) and note that if we can find some i < i0 with fi(αj) ∈ Uαj for all j = 1 , . . . , n,
we’re done. Indeed, for α ∈ B∗ \ {α1, . . . , αn}, we have fi(α) ∈ Uα trivially,
since then Uα = Xα. But the family of domains {Bp}p∈T for the members ofT, is
totally ordered by ⊂. Hence there exists a p ∈ T with {α1, . . . , αn} ⊂ Bp. Clearly
p ∈Q
α∈BpUα, since otherwise p
∗wouldn’t extend p (which it does).
Again by Proposition 3.19, we getQ
α∈BpUα to be a neighborhood of p, and
since p is a cluster point to the net h fi|Bpii∈I, this net is frequently inQα∈BpUα.
Thus there exists an i< i0with fi|Bp ∈Qα∈BpUα. This, in particular, is to say that
fi(αj) ∈ Uαj for all j= 1 , . . . , n, which concludes the proof.
Recall that if we can show thatP has a member whose domain is A, Tychonoff’s theorem will follow. But by Lemma 4.19, P is inductively ordered. And since P is also non-empty, it follows by Zorn’s lemma that P has a maximal element p∗. Denote the domain of p∗by B∗.
4.20 Lemma. B∗= A
Proof. Suppose that B∗ , A. Then there exists γ ∈ A \ B∗. Since p∗ is a cluster point of h fi|B∗i, it follows from Proposition 4.16 that this net has a subnet
fi( j)|B∗ j∈J (4.1)
converging to p∗. Recall that J is some directed set and j 7→ i( j) is some cofinal mapping in IJ.
Now, consider the net
h fi( j)|{γ}ij∈J (4.2)
6In [1] Chernoff refers to such a set as a basic neighborhood of p∗
inQ
α∈{γ}Xα. By Lemma 4.18, there exists pγ ∈Q
α∈{γ}Xαsuch that pγis a cluster point to the net (4.2). Hence, by Proposition 4.16, the net (4.2) has a subnet that converges to pγ. Notice how this subnet must be of the form:
fi( j(k))|{γ} k∈K (4.3)
where K is some directed set and the mapping k 7→ j(k) is some cofinal mapping in JK.
Put q= p∗∪ pγ. Clearly q ∈Q
α∈B∗∪{γ}Xα. Moreover, we have q > p∗. Thus
we can’t have q ∈P, since this contradicts the maximality of p∗. Hence, if we can show that q ∈P, using the assumption B∗, A, we’re done.
But notice that, by Proposition 4.12, the net
fi( j(k))|B∗∪ {γ} k∈K (4.4)
inQ
α∈B∗∪{γ}Xαis a subnet of the net
fi|B∗∪ {γ} i∈I (4.5)
Hence, if the net (4.4) converges to q, then by Proposition 4.16, q is a cluster point of the net (4.5), which is to say that q ∈P.
Before proceeding to show this it is worthwhile to take note of the fact that there are three different directed sets in play here, that doesn’t necessarily have anything to do with each other, other than that they furnish a foundation for a pair of cofinal mappings to connect them according to
K → J → I.
The point is that they cannot be assumed to be directed by the same ordering (as if they all where subsets of, say N.) Hence we can but assume I , J , K to be directed by4 , 40, 400respectively.
To show that the net (4.4) converges to q, let U be a neighborhood of q in Q
α∈B∗∪{γ}Xα. We need to find k0∈ K such that whenever k <00 k0, we have
fi( j(k))|B∗∪ {γ} ∈ U.
But by Proposition 3.19, U has at least one subset of the form Y
α∈B∗∪{γ}
Uα (4.6)
that contains q and where Uα is open in Xα for all α and Uα = Xα for almost all
α. Clearly p∗ ∈ Q
p∗and pγ. Hence by Proposition 3.19, these sets are neighborhoods of p∗and pγ
respectively.
Since the net (4.1) converges to p∗ there exists j0 ∈ J such that whenever
j<0 j0, we have
fi( j)|B∗∈ Y
α∈B∗
Uα.
In addition: since the mapping k 7→ j(k) is cofinal, there exists k1 ∈ K such that
j(k)<0 j0for all k<00 k1.
Since the net (4.3) converges to pγ, it is eventually in
Y
α∈{γ}
Uα
so there exists k2∈ K such that whenever k <00 k2we have,
fi( j(k))|{γ} ∈ Y
α∈{γ}
Uα.
Now, since (K,400) is a directed set, there exists k0 ∈ K with k1, k2 400 k0. At
this point we can verify directly that whenever k <00 k0we have
fi( j(k))|B∗∪ {γ} ∈ Y
α∈B∗∪{γ}
Uα
that as indicated by (4.6), is a subset of U, thus concluding the proof. Tychonoff’s theorem follows.
Appendix A
A proof of Zorn’s lemma
A.1
The Bourbaki fixed-point theorem
A.1 Definition. Let (A,6) be a partially ordered set. A mapping f ∈ AA is said to be increasing, if x6 f (x) for all x ∈ A.
We remind the reader at this point that a set S is said to be strictly inductively ordered if whenever T is a non-empty, totally ordered subset of S , then T has a least upper bound in S .
Notice that the only difference between a set S being strictly inductively or-dered and inductively oror-dered, is that the upper bounds for the non-empty, totally ordered subsets are least upper bounds in the former.
A.2 Bourbaki fixed-point theorem. 1 Let A be a non-empty, strictly inductively ordered set and let f be an increasing mapping in AA. Then there exists x0 ∈ A such that f(x0)= x0.
It is fair to say that the proof is rather technical in the sense that it is quite long and, for the sake of readability, has to be broken down into several smaller steps. For these reasons we will let it occupy its own section.
A.1.1 A proof of the Bourbaki fixed-point theorem
Throughout this section let A be a set strictly inductively ordered by 6 and let f ∈ AAbe an increasing mapping on A. If we can show that f has a fixed point, the
1This is essentially the same theorem that is stated in [5, page 881]. There Lang credits the
Bourbaki group (of which he was a member himself.) The naming here, however, is due to the present author and is not generally recognized. A fixed point of a mapping in AAis, of course, a point
Bourbaki fixed-point theorem will follow. We shall conclude this through a series of lemmas.
Before proceeding, fix some a ∈ A (whose existence is guaranteed by A being non-empty) and letA designate the set of all B ⊂ A such that:
(i) a ∈ B,
(ii) f (B) ⊂ B,
(iii) For all non-empty, totally ordered subsets T of B, the least upper bound for T is a member of B.
The point here is that each totally ordered member ofA will contain a fixed point of f . Hence our task can be translated into showing thatA contains such an element. To be precise:
A.3 Lemma. If M ∈ A and M is totally ordered, then there exists b ∈ M with f(b)= b.
Proof. Clearly M is non-empty. So by A being strictly inductively ordered, we get some b ∈ A such that b is the least upper bound for M in A. Moreover b ∈ M (M is a totally ordered subset of itself!) Hence f (b) ∈ M, which renders f (b) 6 b. But since f is increasing, we still have b6 f (b), so b 6 f (b) 6 b which is to say
f(b)= b.
It will be made clear that the setTA, is such an M as mentioned in Lemma A.3. Through remainder of this section, take M to designateTA.
A.4 Lemma. M ∈A
Proof. That M is a subset of A follows immediately from the identity M = T A. Hence it remains to show that M satisfies each of the three terms given in the definition ofA.
First we need to show that a ∈ M, but a ∈ B for all B ∈A, which is to say that a ∈TA, or equally a ∈ M.
Next we need to show that f (M) ⊂ M. Accordingly, let x ∈ M. Since x ∈ B, for every B ∈A, it follows that f (x) ∈ B for those B (that is, f (x) ∈ T A). Hence
f(M) ⊂ M.
Finally, let T be a non-empty, totally ordered subset of M and let b be the least upper bound for T in A. We need to show that b ∈ M, but for all B ∈A we have T ⊂ M ⊂ B, so b ∈ B for every B ∈A, which is to say that b ∈ T A = M, which
Thus it remains to show that M is totally ordered. This will be done through Lemma A.5–A.8.
A.5 Lemma. For all x ∈ M, a6 x.
Proof. Since M is a subset of every member ofA it suffices to show that the set A0= { x ∈ A : a 6 x }
is a member ofA. But clearly a ∈ A0, and if x ∈ A0 then a6 x 6 f (x) (since f is increasing,) so f (A0) ⊂ A0.
Finally, if T is a non-empty, totally ordered subset of A0 and b the least upper bound for T in A, then x 6 b for some x ∈ T (since T is non-empty) and thus, by a 6 x (since T ⊂ A0,) we get a 6 b, which renders b ∈ A0, concluding that
A0∈A.
We now introduce the last two pieces of notation used exclusively for technical purposes within this section.
For each c ∈ M put
Mc= { x ∈ M : x 6 c or f (c) 6 x }.
If c ∈ M such that for all x ∈ M with x < c we have f (x)6 c, then c is called an extreme point of M.
A.6 Lemma. If c is an extreme point of M then Mc= M.
Proof. Let c be an extreme point of M. Note that it is sufficient to prove that Mc∈A.
By Lemma A.5, we get a6 c (since c ∈ M), and thus a ∈ Mc, so Mcsatisfies
criterion (i) for being a member ofA.
Next, we want to show that (ii) holds for Mc, namely that f (Mc) ⊂ Mc.
Ac-cordingly let x ∈ Mc. Then one of x < c or x = c or f (c) 6 x, must hold. And
by x 6 f (x), the third case translates to f (c) 6 f (x) and here f (x) ∈ Mc follows
immediately.
If x < c then, by c being an extreme point of M, we get f (x) 6 c and thus f(x) ∈ Mc. If x = c then, of course, f (c) = f (x) rendering f (x) ∈ Mc. Hence
f(Mc) ⊂ Mc.
Finally, to show that (iii) holds for Mc, let T be a non-empty, totally ordered
subset of Mcand b the least upper bound for T in A. We need to prove that b ∈ Mc.
Note that if there exists x ∈ T with f (c)6 x, then, since x 6 b, by b being an upper bound for T in A, f (c) 6 b follows and thus b ∈ Mc. We may thus assume that
an upper bound for T in A. Hence b 6 c, since b is the least upper bound for T . Thus we get b ∈ Mc, concluding that Mc ∈ A and thereby M ⊂ Mc followed by
M= Mc.
A.7 Lemma. Each element of M is an extreme point of M.
Proof. Let E designate the set of extreme points of M. Since E ⊂ M it is sufficient to prove that M ⊂ E. Again we shall use the fact that M ⊂ B, for all B ∈A and show that E ∈A.
Clearly a is an extreme point of M, because there is no x ∈ M with x < a, by Lemma A.5.
To show that f (E) ⊂ E, let c be an extreme point of M. We need to show that f(c) is an extreme point as well. Accordingly, let x ∈ M with x < f (c) (the task is to show f (x) 6 f (c)). But note that M = Mc, by Lemma A.6, so we get x ∈ Mc
and thus x 6 c or f (c) 6 x, where the latter possibility is ruled out immediately (x < f (c)) so we get either x < c or x= c. If x < c then, by c being an extreme point, we get f (x) 6 c and, since c 6 f (c), f (x) 6 f (c) follows. If x = c then, trivially, f (x)6 f (c). Thus f (E) ⊂ E.
It remains to show that E satisfies the condition under (iii), in the definition of A (then we’re done). Accordingly, let T be a non-empty, totally ordered subset of E and b the least upper bound for T in A. We need to prove that b is an extreme point (of M).
Accordingly, let x ∈ M with x < b. We want to show that f (x) 6 b. But first note that b ∈ M. Indeed, T ⊂ E ⊂ M. So b ∈ M, since M ∈A (by Lemma A.4.)
Second, we couldn’t have the case where x is another upper bound for T (this would render b6 x, contradicting x < b). Still, since every c ∈ T is an extreme point of M, it follows that Mc = M, for every c ∈ T, that is, for each c ∈ T, either
x 6 c or f (c) 6 x. Hence we must have at least one c ∈ T with x 6 c, since otherwise c6 f (c) 6 x for all c ∈ T which would render x to be an upper bound for T , which, as pointed out, is not possible. So we do have some c ∈ T with either x < c or x = c. If x < c then, by c being an extreme point, we get f (x) 6 c, and thus f (x)6 b. If x = c then x is an extreme point of M, so by Lemma A.6, b ∈ Mx.
Hence f (x)6 b in this case also (otherwise b 6 x). Which concludes that f (x) 6 b,
and the proof.
We can now show that M is a non-empty, totally ordered subset ofA. Note that this concludes the proof of The Bourbaki fixed-point theorem (Lemma A.3) and as such this section.
