1.7 Uniformization
Choice principles in the presence of a universal set are problematic. By Theorem 6, for ex-ample, V ∈ V implies that there is a perfect set and in particular that not every set is well-orderable. And in [FH96a, FH98, Ess00], M. Forti, F. Honsell and O. Esser identified plenty of choice principles as inconsistent with positive set theory. On the other hand, many topo-logical arguments rely on some kind of choice. The followinguniformization axiomturns out to be consistent and yet have plenty of convenient topological implications, in particular with regard to compactness.
A uniformization of a relation R ⊆ V2 is a function F ⊆ R with dom(F) = dom(R). The uniformization axiom states that we can simultaneously choose elements from a family of classes as long as it is indexed by a discrete set:
Uniformization If dom(R)is a discrete set,Rhas a uniformization.
Unless the relation is empty, its uniformization will be a set by the additivity axiom. There-fore the uniformization axiom can be expressed with at most one universal and no existential quantification over classes, and therefore still be equivalently formulated in a first-order way, using axiom schemes. Let us denote byESUrespectivelyTSUessential respectively topologi-cal set theory with uniformization.
In these theories, at least all discrete sets are well-orderable. The following proof goes back to S. Fujii and T. Nogura ([FN99]). We callf:a→ aachoice functioniff(b) ∈bfor every b ∈a.
Proposition 17(ESU). A setais well-orderable iff it is Hausdorff and there exists a contin-uous choice functionf:a→ a, such thatb\ {f(b)}is closed for allb.
In particular, every discrete set is well-orderable and in bijection toκ⊕for some cardinalκ.
Proof. Ifais well-ordered, we only have to defineF(b) = min(b). In a well-order, the minimal element is always isolated, sob\F(b)is in fact closed. To show that Fis a set, letc ⊆ abe closed. Then the preimage ofcconsists of all nonempty subsets ofawhose minimal element is inc. Assumeb /∈F−1[c], that isF(b)∈/ c. Then
(a ∩ ♦((−∞,F(b)]∩c)) ∪ [F(b) +1,∞)
is a closed superset ofF−1[c]omittingb, where byF(b) +1 we denote the successor ofF(b), and ifF(b)is the maximal element, we consider the right part of the union to be empty. Hence F−1[c]is in fact closed, proving thatFis continuous and a set.
For the converse, assume now that f is a continuous choice function. A set p ⊆ a is an approximationif:
• a ∈p
• pis well-ordered by reverse inclusion⊇.
• For every nonempty proper initial segmentQ⊂ p, we haveT Q∈ p.
• For every non-maximalb∈p, we haveb\ {f(b)} ∈p.
We show that two approximationspandqare always initial segments of one another, so they are well-ordered by inclusion: LetQbe the initial segment they have in common. Since both containb = T
Q, that intersection must be inQand hence the maximal element ofQ. Ifb is not the maximum of eitherporq, both containb\ {f(b)}, which is a contradiction because that is not inQ.
Thus the union P of all approximations is well-ordered. Assume T
P has more than one element. Then P∪{T
P,T
P\f(T
P)} were an approximation strictly larger thanP. ThusT P is empty or a singleton. Since there is no infinite descending chain, and for every bounded ascending chainQ ⊆ P, we haveT
Q ∈ P,P is closed, soP ∈ V. Also,⊇is a set-well-order onP. Thusais also set-well-orderable, becausef Pis a continuous bijection ontoa:
Firstly, it is injective, because after the first b with f(b) = x, x is omitted. Secondly, it is surjective, because ifb∈P is the first element not containingx, it cannot be the intersection of its predecessors and thus has to be of the form b = c\ {f(c)}. Hence x = f(c). If xis a member of every element ofP, thenT
P = {x} ∈Pandx= f({x}).
Now letabe a discrete set. We only have to prove that a continuous choice functionf:a → a exists. In fact, any choice function will do, sinceais discrete and hence every function on ais continuous. And the existence of such a function follows from the uniformization axiom, applied to the relationR⊆d×ddefined by:xRyiffy∈x.
It follows that every discrete set ais well-orderable. Therefore, it is comparable in length to On. If an initial segment ofawere in bijection toOn, then as the image of a discrete set,On would be a set. Henceamust be in bijection to a proper initial segmentα⊕ofOn. Ifκis the cardinality of α, there is a bijection betweenκ⊕ andα⊕. Composing these bijections proves the claim.
It follows that there exists an infinite discrete set iffω ∈ On. The uniformization axiom also allows us to define for every infinite cardinal κ a cardinal 2κ, namely the least ordinal in bijection toκ⊕. Proposition 17 then shows that, just as inZFC,Onis not only a weak but even a strong limit.
Like the axiom of choice, the uniformization axiom could be stated in terms of products.
Of course, it only speaks of products ofD-few factors at first, but surprisingly it even has implications for larger products as long as the factors are indexed by a D-compact well-ordered set. D-compactness for a well-ordered set just means that no subclass of cofinality
> Onis closed.
Proposition 18 (ESU+T3+Union). Let w be a D-compact well-ordered set, a ∈ V and ax ⊆anonempty for everyx∈wI. Then the productQ
x∈wIaxis nonempty.
1.7. UNIFORMIZATION 33
Proof. Recall that the product is defined as:
Y
x∈wI
ax =
F∪(w0×a) | F :wI → V, ∀x F(x)∈ ax
We do induction on the length ofwand we have to distinguish three cases:
Case 1: Ifw has no greatest element, its cofinality must beD-small or else it would not be D-compact Hausdorff. So lethyα|α < κibe a cofinal strictly increasing sequence. Using the induction hypothesis and the uniformization axiom, choose for everyα < κan element
fα ∈ Y
x∈]yα,yα+1]I
ax
Then the union of thefαis an element ofQ
x∈wIax.
Case 2: Assume that w has a greatest element pand thatw\ {p} is a set. Then this is still D-compact Hausdorff and hence the induction hypothesis applies, so there is an element f:w\ {p}→ aof the product missing the last dimension. For anyy∈ap, the setf∪ hp,yiis inQ
x∈wIax.
Case 3: Finally assume thatwhas a greatest elementpand thatw\ {p} is not a set. By the induction hypothesis,
Py = Y
x∈[−∞,y]I
ax
is a nonempty set for everyy < p. The unionQ = S
y<pPy is not a set, because otherwise its domain dom(S
Q) = w\ {p}would also be a set. But since Q ⊆ (w×a), it does have a closure which is a set, and this closure must have an elementgwithp ∈ dom(g). We will show thatf =g∪(w0×a)witnesses the claim, that is,f ∈Q
x∈wIax. If z ∈ wI, then g is not in the closure of S
y<zPy, because that is a subclass of the set ((−∞,z]×a). Thusgis in the closure ofS
z6y<pPy, which is a subclass of:
Mz = (w×a) ∩ {r| r∩({z}×a) ∈61az}
If we can show thatMzis closed, we can deduce thatg ∈Mzfor everyz∈wI and therefore g wI = f wIis a function fromwItoawithf(x)∈ axfor allx∈wI. Thusfis indeed an element of the product.
To prove thatMzis closed in(w×a)∩♦({z}×az), assumeris an element of the latter but not of the former. Then there are distinctx1,x2 ∈ az, such thathz,x1i,hz,x2i ∈z. Sinceaz is Hausdorff, there areu1andu2, such thatx1 ∈/ u1,x2 ∈/ u2andu1∪u2 = a, and
((w\ {z})×a ∪ {z}×u1) ∪ ((w\ {z})×a ∪ {z}×u2) is a closed superset ofMzomittingr.
Some of the models of topological set theory we will encounter are ultrametrizable, which in the presence of the uniformization axiom is a very strong topological property. A setais ultrametrizableif there is a decreasing sequenceh∼α |α∈Oni of equivalence relations on a such thatT
α ∼α= ∆aand theα-balls[x]α = {y | x ∼α y} forx ∈ aandα ∈ Onare a base
of the natural topology onain the sense of open classes, that is, the relatively open classes U ⊆ aare exactly the unions of balls. If that is the case, theα-balls partitiona into clopen sets for everyα.
Proposition 19(ESU). Every ultrametrizable set is aD-compact linearly orderable set.
Proof. For everyα ∈On, the classCαof allα-balls is a subclass ofa. Ifb∈ aandx∈ b, then♦[x]αis a neighborhood ofbinawhich contains only one element ofCα, namely[x]α. HenceCα has no accumulation points and is therefore a discrete set. That means there are onlyD-fewα-balls for everyα ∈On.
Now letA ⊆ aandT
A = ∅. For eachα, letBαbe the union of allα-balls which intersect every element ofA. ThenT
αBα = ∅and everyBαis closed.
Assume that allBα are nonempty. Then for everyαall but D-few members of the sequence hBα|α∈Oni are elements of the closed set Bα, so every accumulation point must be in T
α∈OnBα, which is empty. Thus{Bα | α∈On}has no accumulation point and is a discrete subset ofB0. Hence it isD-small, which means that the sequencehBα|α∈Oniis eventually constant, a contradiction.
Therefore there is a Bα which is empty, and by definition everyα-ball is disjoint from some element of A. Since there are only D-few α-balls, the uniformization axiom allows us to choose for everyα-ball[x]αan elementc[x]
α ∈Adisjoint from[x]α. The set of thesec[x]
α is discrete and has an empty intersection. This concludes the proof of theD-compactness.
Since it is discrete, the set Cα can be linearly ordered and there are onlyD-few such linear orders for every α. If L is a linear order on Cα, let RL be the partial order relation on a defined byxRLyiff [x]αL[y]α. RL is a set because it is the union ofD-few sets of the form [x]α×[y]α. LetSαbe the set of all suchRL. The sequencehSα|α∈Onican only be eventually constant if a is discrete, in which case it is linearly orderable anyway. If a is not discrete, however, S = S
αSα must beD-large and therefore have an accumulation point 6ina2. Because eachSαisD-small,6is in the closure of everyS
β>αSβ. Forx,y∈a, let tα,x,y = (a2\([y]α×[x]α)) ∩ ♦{hx,yi}.
We will show that6is a linear order ona:
Assumex 6= y. Then there is anαsuch thatx α y. Every element ofS
β>αSβ assigns an order to[x]αand[y]α, so it is in exactly one of the disjoint closed setstα,x,yandtα,y,x. There-fore the same must be true of6, so we havex 6 yiff noty 6 x. This proves antisymmetry and totality.
Ifx6 y6zandx,y,zare distinct, then there is anαsuch thatx α y α zα x. Then6 is in the closure of neithertα,y,xnortα,z,y, and must therefore be in the closure of
[
β>α
Sβ ∩ tα,x,y ∩ tα,y,z,
which is a subset oftα,x,z, because every element ofSis transitive. It follows that6is also in tα,x,z and thusx6 z, proving transitivity.
1.7. UNIFORMIZATION 35
Finally,6is reflexive because for everyx∈a, all ofSlies in the seta2 ∩ ♦{hx,xi}. Another consequence of the uniformization axiom is the following law of distributivity:
Lemma 20(ESU). Ifdis discrete and for eachi ∈d,Jiis a nonempty class, then [
i∈d
\
j∈Ji
j = \
f∈Q
i∈dJi
[
i∈I
f(i).
Proof. Ifxis in the set on the left, there exists an i ∈ dsuch that xis an element of every j ∈ Ji. Thus for every functionfin the product,x ∈ f(i). Hencexis an element of the right hand side.
Conversely, assume thatxis not in the set on the left, that is, for everyi ∈d, there is aj ∈Ji such that x /∈j. Letfbe a uniformization of the relationR= {hi,ji | i∈ d, x /∈j ∈Ji}. Then x /∈S
i∈If(i).
It implies that we can work with subbases in the familiar way. Let us callKregularif every union ofK-fewK-small sets isK-small again. Then in particularDis regular.
Lemma 21(ESU). LetK ⊆ Dand letBbe aK-subbase of a topologyT such that the union ofK-few elements ofBalways is an intersection of elements ofB. ThenBis a base ofT.
Proof. We only have to prove that the intersections of elements ofBare closed with respect to K-small unions and therefore constitute aK-topology. But ifIisK-small, and eachhbi,j|j ∈ Jiiis a family inB, we have
[
i∈I
\
j∈Ji
bi,j = \
f∈Q
i∈IJi
[
i∈I
bi,f(i)
by Lemma 20, and everyK-small unionS
i∈Ibi,f(i)is an element ofBagain.
Thus ifKis regular andSis aK-subbase ofT, the class of allK-small unions of elements ofS is a base ofT. SinceS
i♦ai = ♦S
iai, the sets of the following form constitute a base of the exponentialK-topology:
♦Ta ∪ [
i∈I
Tbi,
whereI isK-small and a,bi ∈ T for alli ∈ I. As that is sometimes more intuitive, we also use open classes in our arguments instead of closed sets. By settingU = {aandVi = {bi, we obtain that every open class is a union of classes of the following form:
TU∩ \
i∈I
♦TVi
That is, these constitute a base in the sense ofopenclasses. SinceU = U∩♦U, the class Ucan always be assumed to be the union of theVi.
Lemma 21 also implies that given a class B, the weak comprehension principle suffices to prove the existence of the topology K-generated by B: A setcis closed iff for everyx ∈ {a, there is a discrete family(bi)i∈I in B, such that c ⊆ S
ibi andx /∈ S
ibi. In particular, the K-topology of ExpK(X)exists (as a class) whenever the topology ofXis a set.
Lemma 22(ESU). LetKbe regular andXaK-topologicalT0-space.
1. IfXisT1, then ExpK(X)isT1(but not necessarily conversely).
2. XisT3iff ExpK(X)isT2. 3. XisT4iff ExpK(X)isT3.
4. D⊆ Xis dense inXiff theK-small subsets ofDare dense in ExpK(X).
Proof. In this proof we useand♦with respect toX, not the universe, so ifT is the topology ofX, we seta= Taand♦a= ♦Ta.
(1): Fora ∈ExpK(X), the singleton{a} = a∩T
x∈a♦{x}is closed in ExpK(X).
(As a counterexample to the converse consider the case whereK = κ is a regular cardinal number andX = (κ+1)⊕, with the κ-topology generated by the singletons{α} for α < κ.
This is not T1, because {κ} is not closed, but it is clearly T0. We show that its exponential κ-topology isT1: Leta ∈ExpK(X). Then eithera⊆ κis small ora= X.
In the first case,{a}= a∩T
x∈a♦{x}is closed. In the second case,{a} = {X} = T
x∈κ♦{x}is also closed.)
(2): (⇒) Let a,b ∈ ExpK(X) be distinct, wlog x ∈ b\a. Then there are disjoint open U,V ⊆Xseparatingxfroma. Hence♦UandV separatebfroma.
(⇐) Firstly, we have to show thatXisT1. Assume that{y} is not closed, so there exists some otherx∈cl({y}), and byT0,yis not in the closure ofx, so cl({x})⊂ cl({y}). The two closures can be separated by open base classes U∩T
i♦Ui and V∩T
j♦Vj of ExpK(X), whose intersection(U∩V)∩T
i♦Ui∩T
j♦Vjis emtpy. Hence there either exists aUidisjoint from V– which is impossible because cl({x}) ∈V∩T
i♦Ui–, or there is aVjdisjoint fromU: But sinceVj∩cl({y}) 6= ∅, we havey∈ Vj. Hencey /∈ U3 x, contradicting the assumption that xis in the closure ofy.
Now letx /∈ a. Then aandb = {x}∪a can be separated by open base classesU∩T
i♦Ui andV∩T
j♦Vjof ExpK(X), whose intersection(U∩V)∩T
i♦Ui∩T
j♦Vjis emtpy. Hence there either exists aUi disjoint fromV – which is impossible becausea ∈ V∩T
i♦Ui–, or there is aVj disjoint fromU: Then Vj andU separatexfrom a, becausebmeets Vj anda does not, sox∈ Vj.
(3): In both directions, theT1property follows from the previous points.