Metrizability is one of the main ingredients of descriptive set theory. As we noticed in Section 2.4, since generalized Baire space is not metrizable, all those notions from descriptive set theory that depend on metrizability cannot be easily generalized to uncountable cardinals. In this section we will useRκ to fill this
gap between classical and generalized descriptive set theory. In particular we will define a generalization of metrizability toκwhich can be used to study generalized Baire space from a metric point of view. Then we will show that, as in the classical case,κ-Borel sets form a hierarchy.
Note that in this section we will focus our attention on topologies rather thanκ-topologies. This is due to the following facts:
We want to useRκ to generalize results from classical descriptive set theory to generalized Baire space.
In descriptive set theory the topology overκκplays a central role, the use ofκ-topologies would then make
our work harder and sometimes impossible.
The results at the end of this section will be very important in Chapter 4 where we will be forced to consider the standard topology overκκ.
Definition 3.5.1 (κ-metric Space). A κ-metric spaceis a pair(X, d)where X is a set andd:X →Rκ is
a function such that: • d(x, y)≥0.
• d(x, y) =d(y, x).
• d(x, y) +d(y, z)≥d(x, z).
We will calldaκ-metric. Moreover aκ-metricdis called aκ-ultrametric if the following hold:
d(x, z)≤max{d(x, y), d(y, z)}.
Given a κ-metric space (X, d),x∈X andr∈Rκ, we define
B(x, r) ={y|d(x, y)< r} and
B[x, r] ={y|d(x, y)≤r}.
We will call B(x, r)the open ball of centre xand radius r, andB[x, r] the closed ball of centrexand radius
r.
Note that the absolute value is definable in the language of real closed fields as follows:
|x|=y⇔(x≥0→y=x)∧(x <0→y=−x).
Now, sinceRκ is a real closed field and the absolute value overRis a metric, the absolute value overRκ is
aκ-metric (note that all the properties required by metrics are sentences in (+,·,0,1, <)).
Definition 3.5.2 (κ-metrizability). Let τ be a topology overX. Thenτ is said to be κ-metrizableif there is aκ-metricd:X →Rκ such that
B={B(x, r)|x∈X andr∈Rκ}
is a base forτ. Moreover, we say that τ is completely κ-metrizable if every Cauchy sequence in X induced byd has a limit inX.
As we have just seen, the classical metric over Rextends naturally to aκ-metric forRκ. It is not hard
to see that the absolute value induces the interval topology overRκ.
Theorem 3.5.3. The interval topology overRκ isκ-metrizable.
Proof. We want to show that |·|: Rκ →Rκ induces the interval topology overRκ. We will show that the
set of open ballsBis equal to the set of open intervals with end points inRκ. SinceRκ is a real closed field,
the proof is the same as in the classical case.
On the one hand let (r1, r2) be an open interval inRκ. Then
B(r2+r1 2 , r2−r1 2 ) = (r1, r2). Indeed, we have: x∈B(r2+r1 2 , r2−r1 2 )⇔ r2+r1 2 − r2−r1 2 < x < r2+r1 2 + r2−r1 2 ⇔r1< x < r2 ⇔x∈(r1, r2).
On the other hand, let B(x, r) be an open ball, then trivially B(x, r) = (x−r, x+r).
In conclusion B is the set of open sets with end points in Rκ, hence B generates the interval topology
overRκ
Proof. Letdbe aκ-metric which inducesτ and Ca closed set. Define:
Cα={x| ∃c∈C. d(x, c)<
1
α}.
Note that for everyα∈κwe have thatCα=Sc∈CB(c,
1
α) is open. We claimC=
T
α∈κCα.
On the one hand, assumex∈C. Thend(x, x) = 0 andx∈Cαfor everyα∈κ. Thereforex∈Tα∈κCα.
On the other hand if x∈ T
α∈κCα then for allα∈κthere exists c ∈Cα such thatd(x, c)< α1. For each
αchoose cα ∈ Cα∩C. Since limα∈κα1 = 0, we have limα∈κd(x, cα) = 0 and since C is closed, x∈ C as
desired.
Analogously to the classical case, where ωω was completely metrizable, generalized Baire space can be
proved to be completelyκ-metrizable.
Theorem 3.5.5. The generalized Baire spaceκκ is completely κ-metrizable.
Proof. First we define the followingκ-metric:
d(x, y) = ( 0 iffx=y, 1 α+1 αminimal s.t. x(α)6=y(α), forx, y∈κκ.
We claim thatdis a κ-metric:
Trivially d(x, y)≥0, moreover d(x, y) = 0 if and only ifx=y andd(x, y) =d(y, x). We want to prove the following:
∀x∀y∀z.d(x, y) +d(y, z)≥d(x, z).
If d(x, y) = 0 or d(y, z) = 0 the statement follows trivially. Assumed(x, y) = α1+1, d(y, z) = β+11 and
d(x, z)6= 0, in particular d(x, z) = γ+11 . First assumeα≤β. Sinceyβ =z β and α≤β, we have that
d(x, z) =d(x, y)≤d(x, y) +d(y, z) as desired. Now assume β < α. Sincexα=y αandβ < α, we have thatd(x, z) =d(y, z)≤d(x, y) +d(y, z) as desired. Note that we actually proved thatdis a κ-ultrametric.
Now we want to show that the set of open balls induced bydis a base for generalized Baire space. On the one hand letw∈κ<κ. We claim that
[w] = B(p, 1 |w|+ 1) wherew⊂p. By definition we have p0∈[w]⇔w⊆p0⇔d(p, p0)< 1 |w|+ 1 ⇔p 0∈B(p, 1 |w|+ 1)
as desired. On the other hand let B(p, r) withr∈Rκbe an open ball. Note that by definition ofd, we have
B(p, r) = B(p,α1
r+1) withαr greatest ordinal such that
1
αr+1 ≤r. Now let w=pαr as before. We have
p0 ∈[w]⇔w⊂p0⇔d(p, p0)< 1 |w|+ 1 = 1 αr+ 1 ⇔p0 ∈B(p, 1 αr+ 1 ) = B(p, r) as desired.
Finally we want to prove that κκ is complete w.r.t. d. Let (p
α)α∈κa Cauchy sequence in κκ. Note that
∀ε∈R+κ∃γ < κ∀α, α
0≥γ. d(p
α, pα0)< ε. Now, define the following sequence
p`0 =hi,
p`α=pγα γsmallest s.t. ∀α, α0≥γ. d(pα, pα0)< 1
Since the starting sequence was Cauchy,pγ always exists. We claim that (p`α)α∈κis monotone. Letα≤β.
Ifp`α=p`β then triviallyp`α⊆p`β. Hence assumep`α6=hi 6=p`β. Letγandγ0 be such that
p`α=pγα p`β=pγ0β.
Without loss of generality assume γ0 ≥γ a similar proof works forγ0 < γ. By the definition we have
d(pγ, pγ0)< 1
α+ 1, and therefore
p`α=pγα⊆pγ0β=p`β as desired. Then the sequencep`=Sα∈κp`αis well defined.
Now we want to prove that limα∈κpα=p`, that is
∀ε∈R+
κ∃γ < κ∀β≥γ. d(pβ, p`)< ε.
Fixε∈R+κ. Letαsuch that
1
α+1 ≤ε. We have that
p`α=pγα,
withγ smallest ordinal such that
∀β, β0≥γ. d(pβ, pβ0)< 1
α+ 1 In particular, we have that
∀β≥γ. d(pβ, pγ)<
1
α+ 1. Then, sincep`α=pγα, we have that
∀β ≥γ. d(pβ, p`)<
1
α+ 1 ≤ε, therefore limα∈κpα=p` as desired.
Definition 3.5.6 (κ-separability). Let τ be a topological space over X. Thenτ is κ-separableiff it has a dense subset of cardinality κ.
Since in the previous section we have proved that the set of intervals with end points in Qκ is a base for
the interval topology overRκ, we have the following:
Proposition 3.5.7. The interval topology over Rκ isκ-separable.
Recall that we assumedκ<κ=κ, therefore the standard base ofκκ is of cardinalityκ. In particular we have:
Proposition 3.5.8. The generalized Baire space κκ isκ-separable.
Polish spaces have a central role in descriptive set theory, it is natural then to generalize them to uncountable cardinals.
Definition 3.5.9 (κ-Polish Space). Let τ be a topological space over X. Then, τ is κ-Polish iff it is
κ-separable and completelyκ-metrizable.
By what we have proved before, we have:
Corollary 3.5.10. The generalized Baire space κκ isκ-Polish.
Definition 3.5.11 (κ-Borel Sets). The collection B(X) of κ-Borel sets over a topological space (X, τ) is the smallest containing the topology and closed under unions of size κand complementation.
Since we have generalized metrizability toRκ, we can now show that theκ-Borels form a hierarchy as in
the classical case.
Definition 3.5.12(κ-Borel hierarchy). Let (X, τ)be aκ-metrizable space. Then for1< α < κ+ we define:
Σ(1κ,0)(X) = open sets, Π(1κ,0)(X) = closed sets, Σ(ακ,0)(X) ={[ β∈κ Aβ | ∀β < κ∃γ < α. Aβ∈Πγ(κ,0)(X)}, Π(ακ,0)(X) ={\ β∈κ Aβ | ∀β < κ∃γ < α. Aβ∈Σγ(κ,0)(X)}. As usual ∆(ακ,0)(X) =Σ (κ,0) α (X)∩Π (κ,0) α (X).
Proposition 3.5.13. If (X, τ)isκ-metrizable, then for everyα∈Onwe have Σα(κ,0)(X)⊆Σ(ακ,+10)(X)and
Π(ακ,0)(X)⊆Π
(κ,0)
α+1(X).
Proof. We will proceed by induction overα:
Assume α = 1. Let A ∈ Π1(0,κ)(X). Then A is closed in X. By Proposition 3.5.4 we have that
A=T
β∈κAβ withAβ open for everyβ ∈κ. But thenA∈Π
(κ,0)
2 (X). A similar proof works forΣ (κ,0) 1 (X).
Assumeα >1. LetA∈Π(ακ,0)(X). By definitionA=Tβ∈κAβ withAβ∈Sγ∈αΣ
(κ,0)
γ (X) for allβ < κ.
By inductive hypothesis, for all β < κ, Aβ ∈ Σ
(κ,0)
α (X) and by definition A =Tβ∈κAβ ∈ Π
(κ,0)
α+1(X). A
similar proof works forΣ(ακ,0)(X).
Then we trivially have:
Corollary 3.5.14. If (X, τ)isκ-metrizable, then for everyα∈Onwe have Σ(ακ,0)(X)∪Π(ακ,0)(X)⊆∆(ακ,+10)(X).
Proof. It is enough to proveΣ(ακ,0)(X)⊆Π(ακ,+10)(X) and Π (κ,0)
α (X)⊆Σ(ακ,+10)(X). IfA∈Σ (κ,0)
α (X), then A
is trivially the intersection ofκcopies of itself thereforeA∈Π(ακ,+10)(X). A similar proof showsΠ(ακ,0)(X)⊆
Σ(ακ,+10)(X).
As in the classical case if X isκ-metrizable, one can show that the hierarchy over X contains all and onlyκ-Borel sets.
Theorem 3.5.15. Let (X, τ)be a κ-metrizable space. Then we have B(X) = [ α<κ+ Π(ακ,0)(X) = [ α<κ+ Σ(ακ,0)(X) = [ α<κ+ ∆(ακ,0)(X).
Proof. Note that
[ α<κ+ Π(ακ,0)(X) = [ α<κ+ Σ(ακ,0)(X) = [ α<κ+ ∆(ακ,0)(X)
follows trivially from what we have just proved. Now we will prove
B(X) = [
α<κ+
First note that S
α<κ+Σ
(κ,0)
α (X) is closed under unions of size κ and complementation. Let{Aα}α∈κ be
a family of elements in S
α<κ+Σ
(κ,0)
α (X). By regularity of κ+ there isβ < κ+ such that for allα < κ we
have Aα ∈ Σ (κ,0) β (X). But then S α∈κAα ∈ Π (κ,0) α+1(X) ⊆Σ (κ,0) α+2(X). Then S α∈κAα ∈ S α<κ+Σ (κ,0) α (X). If A ∈S α<κ+Σ (κ,0)
α (X), then there is β such that A ∈ Σβ(κ,0)(X). Therefore, the complement of A is in
Πβ(κ,0)(X)⊆Σβ(κ,+10)(X)⊆S α<κ+Σ (κ,0) α (X). ThenB(X)⊆Sα<κ+Σ (κ,0) α (X).
Now we want to show that S
α<κ+Σ
(κ,0)
α (X) ⊆ B(X). We will prove that Σ
(κ,0)
α (X) ⊂ B(X) and
Π(ακ,0)(X)⊂B(X) for everyα∈κ+. We will proceed by induction overα.
Assume α= 1. Then by definitionΣ(ακ,0)(X)⊂B(X) andΠ
(κ,0)
α (X)⊂B(X).
Assume α > 1. Let A ∈ Σ(ακ,0)(X). Then A = Sβ∈κAβ, where for all β ∈ κ we have Aβ ∈
S
γ∈αΠ
(κ,0)
γ (X). Then by inductive hypothesis Aβ ∈ B(X) for every β ∈ κ. Now since B(X) is closed
under union of sizeκ,A∈B(X) as desired.
We will now follow Kechris [17] to prove that the hierarchy over κκ does not collapse.
Definition 3.5.16 (Universal Sets). Let(X, τ)be aκ-metrizable space. Moreover let
Γ(X)∈ {Σ(κ,0)
α (X),Π
(κ,0)
α (X)|α∈κ
+}
andY be a set. We will call U a Y-universal set for Γ(X) ifU ∈Γ(Y ×X)and
Γ(X) ={Uy |y∈Y}
where
Uy={x| hy, xi ∈U}.
Theorem 3.5.17. Let (X, τ) be a κ-metrizable and κ-separable space. Then for every α ∈ On, the sets Σα(κ,0)(X) andΠα(κ,0)(X)haveκκ-universal sets.
Proof. We will only prove the theorem forΣ(ακ,0)(X). We will proceed by induction overα.
Assume α= 1. Since (X, τ) is κ-separable, it has a baseB of cardinality at most κ. Let (Bβ)β∈κ be a
listing ofB. Define
hy, xi ∈U1⇔y∈κκ∧x∈X∧x∈ [
y(α)=1
Bα.
ThereforeU1 is trivially open inκκ×X. Moreover, letA∈Σ(κ,0)
1 (X). SinceA is open,A=
S
β∈IBβ for
someI⊆κ. Now define
y(β) = 1⇔β∈I. We have x∈Uy1⇔ hy, xi ∈U1⇔x∈ [ β∈I Bβ⇔x∈A.
ThereforeU1isκκ-universal forΣ(κ,0)
1 (X). A similar proof works forΠ (κ,0) 1 (X).
Assume α >1. Letη:κ→αbe a monotone function such that sup{η(β) + 1|β∈κ}=α
(note that every ordinalκ < α < κ+ has cardinality κ). For everyy∈κκ andβ∈κdefine
(y)β(γ) =y(dβ, γe)
Note that the functiony7→(y)β is continuous. Now, let w∈κ<κ. We have
Takem= sup{dβ, w(γ)e |γ <|w|}. We claim
(−)−β1[w] = [
p∈P
[pm] whereP ={p∈κκ| ∀γ <|w|. p(dβ, w(γ)e) =w(β)}.
On the one hand assume p∈(−)β−1[w] thenp(dβ, w(γ)e) =w(γ) for everyβ < |w|. Hence p∈ P and
p∈S
p0∈P[p0m].
On the other hand, ifp∈S
p0∈P[p0m], then there isp0∈P such thatpm=p0m, hence
∀γ <|w|. p(dβ, w(γ)e) =w(γ),
thereforep0 ∈(−)−β1[w] as desired.
By inductive hypothesis for every β < αwe have thatΠ(βκ,0)(X) has aκκ-universal setUβ. Now define
hy, xi ∈Uα⇔ ∃βh(y)β, xi ∈Uη(β).
Since every (−)β is continuous, and everyUη(β)is κκ-universal for Σ
(κ,0)
η(β)(X), then U
α is inΣ(κ,0)
α (X).
Moreover, let A ∈ Σ(ακ,0)(X). Then A is a union of at most κ elements in Sβ<αΠ
(κ,0)
β (X), call them
{Aβ}β∈κ. Now for every Aβ letγβ be the smallest such thatAβ∈Π
(κ,0)
η(γβ)(X) and define
yA(dβ, γe) =yβ(γ) whereU γβ
yβ =Aβ.
ThenUyαA =A. Now we have
x∈ [ β∈κ Aβ⇔ ∃β ∈κ. x∈Aβ ⇔x∈Uγβ yβ ⇔ ∀γ. yA(dβ, γe) =yβ(γ) ⇔ h(yA)β, xi ∈Uγβ ⇒ hy, xi ∈Uα. Moreover, we have hy, xi ∈Uα⇔ ∃β. h(yA)β, xi ∈Uγβ ⇔x∈Uγβ (yA)β ⇒x∈Aβ ⇔x∈ [ β∈κ Aβ.
HenceUα isκκ-universal forΣ(κ,0)
α (X) as desired.
We can finally prove that, as in the classical case the hierarchy does not collapse.
Theorem 3.5.18. For every α < κ+, we have:
Σ(ακ,0)(κκ)6⊆Π(ακ,0)(κκ) Π(ακ,0)(κκ)6⊆Σ(ακ,0)(κκ).
Proof. We will only prove that
Σ(ακ,0)(κκ)6⊆Π(ακ,0)(κκ),
the fact that
can be proved analogously.
Assume Σ(ακ,0)(κκ)⊆Π(ακ,0)(κκ). LetU beκκ-universal forΣ(ακ,0)(κκ). Define
y∈A⇔ hy, yi∈/U,
thereforeA∈Π(ακ,0)(κκ) =Σ
(κ,0)
α (κκ), and there existsy∈κκsuch thatUy =A. But this is a contradiction
since we havehy, yi ∈U andy∈A.
Note that, now that we have a generalized version of Polish spaces we can ask if, as in the classical case, the Theorem 3.5.18 also holds forκ-Polish spaces of cardinality at least 2κ. In the classical case this is proved by using the Cantor space.
Definition 3.5.19 (Generalized Cantor space). We will call generalized Cantor spacethe set 2κ of binary sequences of lengthκendowed with the topology generated by the base
B={[w]|w∈2<κ},
where2<κ denotes as usual the set of binary sequences of length less than κ.
First note that the proof of Theorem 3.5.17 also works for generalized Cantor space. Then, by using the same argument of Theorem 3.5.18 we have:
Theorem 3.5.20. For every α < κ+, we have:
Σ(ακ,0)(2κ)6⊆Π(ακ,0)(2κ) Π(ακ,0)(2κ)6⊆Σ(ακ,0)(2κ).
From Theorem 3.5.20 we have the following:
Theorem 3.5.21. Let (X, τ)be aκ-Polish spaces. If2κ is a subspace ofX then, for everyα < κ+ we have
Σ(ακ,0)(X)6⊆Π(ακ,0)(X)andΠ(ακ,0)(X)6⊆Σ(ακ,0)(X).
Proof. We will only prove that
Σ(ακ,0)(X)6⊆Π(ακ,0)(X),
the fact that
Π(ακ,0)(X)6⊆Σ
(κ,0)
α (X),
can be proved analogously. Assume Σ(ακ,0)(X)⊆Π
(κ,0)
α (X). Then we have:
Σ(ακ,0)(2κ) ={Y ∩2κ|Y ∈Σ(ακ,0)(X)} ⊆ {Y ∩2κ|Y ∈Π(ακ,0)(X)}=Π(ακ,0)(2κ).
But this contradicts Theorem 3.5.20.
Note that it is still unclear if 2κcan be embedded in everyκ-Polish space of cardinality at least 2κ. The
classical analogous of this fact is usually proved by using a Cantor scheme (see [17, Corollary 6.5]). In particular, in the classical proof given a Polish spaceX one constructs a binary treeT of height ωwith the following properties:
(1) Every node ofT is a nonempty open subset ofX.
(2) Every nodeU ofT and its closureU are a subset of the predecessors ofU inT.
(3) Every branch (Ui)i∈ω ofT starting at the root of the tree is such thatTi∈ωUi is a singleton.
In the generalized case we would have to extend this tree to 2κ(i.e., the generalized tree should have height
κ). While the definitions for base and successor stages ofT can essentially be the same as in the classical proof, it is unclear how to define the tree on limit ordinals. In particular, on limit stages nothing guarantees
Open Question 3.5.22. Does Theorem 3.5.18 generalize toκ-Polish spaces of cardinality at least2κ?
Our generalizations of metrizable and Polish spaces give rise to many open questions. It would be interesting to study if the standard relationship between Polish spaces and the Baire space holds in the generalized case. In particular:
Open Question 3.5.23. Is everyκ-Polish space the continuous image of the generalized Baire space κκ?
We conclude this section proving that, by using κ-metrics, one can prove a generalized version of the Baire Category Theorem.
Theorem 3.5.24 (κ-Baire Category Theorem). Let(X, τ)be a completeκ-metric space such that for every family of nested non empty open balls{Bα}α∈β with β < κwe have: