Now we shall show that Zorn’s Lemma implies Tychonov’s Product Theorem. We need the concept of convergence for filters.
Definition 2.12. We say that a filterF onX converges tox∈X ifU(x)⊆ F. A filter basisB converges toxif the filter generated byB converges tox.
A point to which a filter, respectively, filter basisF converges is also called a
limit pointofF ut
It is immediate that a filter basisBconverges toxiff for each neighborhoodU ofxthere is a memberB ∈ Bsuch thatB⊆U.
A sequence (xn)n∈Nis said to converge to xif for every neighborhoodU of x there is an N ∈ N such that xn ∈ U for all n > N. Once we are given these definitions it is an easy exercise to show that a sequence (xn)n∈Nconverges to x iff the filter basisB=
{xn, xn+1, . . .}:n∈N converges tox.
For an ultratilterU on a topological spaceX, a pointx∈X is a limit point iff for allU ∈U(x) and eachF ∈ U we haveU∩F6=∅ iffx∈T
F∈UF.
Theorem 2.13. (UT)For a topological space(X,O)the following statements are equivalent:
(i) X is compact.
(ii) Every ultrafilter converges.
Proof. (i)⇒(ii): LetU be an ultrafilter. By (i) there is anxsuch thatx∈V for allV ∈ U (see 2.3). This means that U∩V 6=∅ for allU ∈U(x) and all V ∈ U.
Then F def= {F : U ∩V ⊆F, U ∈ U(x), V ∈ U } is a filter containing U. Since
U is maximal among all filters, we have F =U and thusU =F ⊇U(x), i.e., U converges tox.
(ii)⇒(i). (UT) LetBbe a filter basis of closed sets; we must show thatT B 6=∅.
By the Ultrafilter Theorem (UT), the filter basisBis contained in an ultrafilterU which by (ii) converges to some elementx. Let U be a neighborhood ofx. Then U ∈ U. Now let B ∈ B; since B ⊆ U we have B ∈ U, and thus B∩U ∈ U; in particular,B∩U 6=∅. Thereforex∈B=B for allB∈ B. ut The proof of the preceding characterisation theorem for compactness required actually the Ultrafilter Theorem (UT). This axiom allows the proof of other char-acterisation theorems for compactness as is exemplified in the following exercise.
Exercise E2.4. Recall that a subbasis for a topology O is any subset S, such thatO is the smallest topology containingS. This means that for each open set U ∈Oand eachx∈U, there are finitely many subbasic open setsS1, . . . , Snsuch thatx∈S1∩ · · · ∩Sn⊆U.
A setAof closed sets is called a subbasis for the closed sets if{X\A:A∈A}is a subbasis for the topology, that is iff for each closed setAand a pointx /∈Athere are finitely many subbasic closed setsS1, . . . , Snsuch thatA⊆S1∪· · ·∪Sn⊆X\{x}.
Prove the following theorem: Alexander Subbasis Theorem. Let S be a subbasis of the topology of a topological space X. Then X is compact if and only if any open cover taken fromS has a finite subcover.
[Hint. By definition of compactness ifX is a compact space thenanyopen cover has a finite subcover. We have to assume that every open cover ofsubbasicopen sets has a finite subcover and then conclude that any open cover has a finite subcover.
We might just as well assume that for a suitable subbasisS of closed sets every filterbasis generated by subbasic closed sets has a nonempty intersection, and prove that each ultrafilter converges (Theorem 2.13.) So letU be an ultrafilter. We must show thatT
F∈UF 6=∅because this intersection is the set of all limit points ofU. LetF ∈ U andx∈X\F. SinceS is a subbasis for the set of closed sets, there contained in the filterU and therefore has a nonempty intersection; the set of all finite intersections of elements ofF0 is a filter basis Bof closed subbasic sets. It therefore satisfiesTF0=TB 6=∅by hypothesis. We claim
(3) \
F∈F
F =\ F0.
The relation “⊆” is clear (why?). For a proof of the reverse containment, assume x /∈T
F∈UF and show thatx /∈T
F0. Under this assumption there is anF ∈ U such thatx /∈F. Then by (2) there is anS∈ F0such thatx /∈S. Hencex /∈T
F0. This shows “⊇”. Thus (3) holds and the left side is nonempty. This is what we had to show.]
Notice that the Alexander Subbasis Theorem requires the Ultrafilter Theorem (UT) which is secured by the Axiom of choice (AC).
The next theorem is the crucial one. It will prove that AC⇒TPT.
Theorem 2.14. (AC)The product of any family of compact spaces is compact.
Proof. Let (Xj:j ∈J) be a family of compact spaces. LetP def= Q
j∈JXj. If one Xj is empty, thenP =∅ and thusP is compact. Assume now that Xj 6=∅. We prove compactness ofP by considering an ultrafilterU onP and showing that it converges.
For eachj∈J the projection prj(U) is an ultrafilter. LetLj ⊆Xj be the set of points to which it converges. Since Xj is compact, Lj 6=∅. By the Axiom of ChoiceLdef= Q
j∈JLj 6=∅. Let (xj)j∈J∈L.
Now let U be a neighborhood ofx def= (xj)j∈J. We may assume that U is a basic neighborhood of the formU =Q
j∈JUj, whereUj =Xj for allj∈J\F for some finite subset ofJ. Then we find a memberM ∈ U such that prj(M)⊆Ujfor j ∈F. Thus M ⊆pr−1j (Uj) and so pr−1j (Uj)∈ U. Accordingly U =Q
j∈JUj = T
j∈Fpr−1j (Uj)∈ U. ut
Notice that we have used the Axiom of Choice by applying the Ultrafilter Theoremandby selecting (xj)j∈J.
Exercise E2.5. Prove:
In a Hausdorff space, a filterF converges to at most one point..
Thus in a Hausdorff space a converging filter converges to exactly one point x, called thelimit pointand writtenx= limF.
Corollary 2.15. (UT)The product of a family of compact Hausdorff spaces is a
compact Hausdorff space. ut
The Ultrafilter Theorem (UT) indeed suffices for a proof of this theorem.
Example. (Cubes) LetIdenote the unit interval [0,1] andDthe complex unit disc. For each setJ the productsIJ andDJ are compact spaces. ut
We have made good use of the concept of a filter and its convergence. In passing we mention the concept of a Cauchy-filter on a metric space. Let us first
recall that in a metric space (X, d) a subsetB⊆X isboundedif there is a number C such that d(b, c) ≤ C for all b, c ∈ B. For a bounded subset B, the number sup{d(b, c) :b, c∈B} exists and is called the diameter ofB. When we speak of the diameter of a subset, we imply that we assume that the subset is bounded.
Definition 2.16. A filterF on a metric space (X, d) is called aCauchy-filterif for eachε >0 it contains a set of diameter less thanε. A filter basis is aCauchy-filter basisif it contains a set of diameter less thanε. ut Clearly, a filter basis is a Cauchy-filter basis if and only if the filter of all super sets of its members is a Cauchy-filter.
Exercise E2.6. Show that a sequence (xn)n∈N is a a Cauchy-sequence iff the filter basis of all{xn, xn+1, . . .},n∈Nis a Cauchy filter basis.
Lemma 2.17. (i) Let F be a Cauchy-filter in metric space. Then there is a countable Cauchy-filter basisC,C1⊇C2⊇ · · ·such that the diameter ofCn is less than n1 andC ⊆ F .
(ii) If F ⊆ G are two filters in a metric space such that F is a Cauchy-filter
andG converges toxthen F converges to x. ut
IfCconverges, and thus the filterhCiof all supersets of theCn converges, that is, contains some neighborhood filterU(x), then the given filterFconverges. If now we select in each setCn an elementcn, then (cn)n∈N is a Cauchy-sequence. Then {cn, cn+1, . . .} : n ∈ N} is a Cauchy-filter basis B which converges iff (cn)n∈N converges. Moreover,hCi ⊆ hBi.
Proposition 2.18. A metric space(X, d)is complete if and only if every
Cauchy-filter converges. ut
For a given ε, a precompact metric space is covered by finite number of open ε-balls. Thus any ultrafilter contains one of them. Hence every ultrafilter on a precompact spaces is a Cauchy-filter. Thus on a complete precompact metric space every ultrafilter converges. This is an alternative proof that a metric space is compact iff it is complete and precompact. This approach has the potential of being generalized beyond the metric situation.
Exercise E2.7. Fill in the details of this argument.