3.4 Proof of Theorem 1.3.8
3.4.7 Well-behaved ranks on the Baire class ξ functions
In this section we finally show that there actually exist ranks with very nice properties. Two of these ranks will answer Question 1.3.1 and Question 1.3.2.
Throughout the section, let1≤ξ < ω1 be fixed.
First we need a result concerning unboundedness.
Definition 3.4.40. Let A and B be disjoint Π0ξ+1 sets. Then they can be separated by a ∆0ξ+1 set (see e.g. [51, 22.16]). Since every ∆0ξ+1 set is the transfinite difference of Π0ξ sets, A and B can be separated by the transfinite difference of such a sequence. Let αξ(A, B) denote the length of the shortest such sequence.
Definition 3.4.41. Letf be a Baire class ξ function, and p < q ∈Q. Then {f ≤p}and{f ≥q}are disjointΠ0ξ+1sets. Let theseparation rank off be
αξ(f) = sup
p<q p,q∈Q
αξ({f ≤p},{f ≥q}).
Note that this really extends the definition of α1.
Theorem 3.4.42. For every1≤ξ < ω1 the rankαξ is unbounded inω1on the characteristic Baire classξ functions.
Proof. Let U ∈ Π0ξ(2ω×X) be a universal set forΠ0ξ(X) sets, that is, for every F ⊂ X, F ∈ Π0ξ(X)there exists a y ∈ 2ω such that Uy =F. For the existence of such a set see [51, 22.3]. Let us use the notation Γζ(X) for the the family of sets H ⊂X satisfyingαξ(H, Hc)< ζ. From [51, 22.27] we have Γζ(X)⊂∆0ξ+1(X). We will show that there exists a∆0ξ+1 set for everyζ < ω1
which is universal for the family of Γζ sets. Since X is uncountable, there is a continuous embedding of2ω into X ([51, 6.5]), hence no universal set exists in 2ω×X for the family of ∆0ξ+1(X) sets (easy corollary of [51, 22.7]). This implies for everyζ < ω1that Γζ 6=∆0ξ+1, hence the rank is really unbounded.
Letp:ζ×N→Nbe a bijection. Forη < ζ andy∈2ωwe defineφ(y)η∈2ω by φ(y)η(n) = y(p(η, n)). First we check that for a fixed η < ζ the map y 7→ φ(y)η is continuous. Let U = {x∈ 2ω : x(0) = i0, . . . , x(n) = in} be a set from the usual basis of 2ω. The preimage of U is the set {y ∈2ω : ∀k ≤ n φ(y)η(k) =ik}={y∈2ω:∀k≤n y(p(η, k)) =ik}, which is a basic open set, too. NowUη ={(y, x) : (φ(y)η, x)∈ U } is a continuous preimage of aΠ0ξ set, henceUη∈Π0ξ(2ω×X)(see [51, 22.1]). Let
U0={(y, x)∈2ω×X :the smallest ordinalη such that(y, x)6∈ Uη is odd, if such anη exists, or no suchη exists andζ is odd}.
Now we check thatU0∈∆0ξ+1(2ω×X). LetVη=T
θ<ηUθ, then these sets form a continuous decreasing sequence of Π0ξ sets and it is easy to see that U0c is the transfinite difference of the sequence(Vη)η<ζ+1, henceU0c∈∆0ξ+1, proving thatU0∈∆0ξ+1, since the family of∆0ξ+1sets is closed under complements (see [51, 22.1]).
Now we show that U0 is universal. For a setH ∈Γζ(X)there is a sequence (zη)η<ζ in 2ω, such that H is the transfinite difference of the sets Uzη. For every sequence (zη)η<ζ we can find y ∈ 2ω such that φ(y)η = zη. Namely y:p(η, n)7→zη(n)makes sense (sincepis a bijection), and works. Consequently, for H there is y ∈ 2ω, such that H is the transfinite difference of the sets Uzη =Uφ(y)η= (Uη)y. It is easy to see that ifH is the transfinite difference of the sequence((Uη)y)η<ζ then
H ={x∈X:the smallest ordinalηsuch that x6∈(Uη)y is odd, if such anη exists, or no suchη exists and ζis odd},
henceH=U0y. Corollary 3.4.43. For every1 ≤ξ < ω1, every nonempty perfect set P ⊂X and every ordinal ζ < ω1 there is a characteristic function χA ∈ Bξ(X) with A⊂P andαξ(χA)≥ζ.
Proof. Since P is nonempty perfect, it is an uncountable Polish space with the subspace topology, hence the rank αξ is unbounded on the characteristic Baire class ξ functions defined on P by the previous theorem. Hence we can take a characteristic functionf0 ∈ Bξ(P)withαξ(f0)≥ζ, and set
f(x) =
f0(x) ifx∈P 0 ifx∈X\P .
It is easy to see thatf ∈ Bξ(X), hence it is enough to prove thatαξ(f)≥ζ.
For this, it is enough to prove that αξ({f0 ≤ p},{f0 ≥ q}) ≤ αξ({f ≤ p},{f ≥q})for every pair of rational numbersp < q. For this, letH∈∆0ξ+1(X) where {f ≤p} ⊂H ⊂ {f ≥q}c andH is the transfinite difference of the sets (Fη)η<λwithλ=αξ({f ≤p},{f ≥q})andFη ∈Π0ξ(X)for everyη < λ.
Let H0 =P ∩H and for every η < λ let Fη0 = P ∩Fη. It is easy to see thatH0 separates the level sets{f0 ≤p}and{f0≥q}andH0 is the transfinite difference of the sets (Fη0)η<λ. And since H0 ∈∆0ξ+1(P)and Fη0 ∈ Π0ξ(P) for every η < λ ([51, 22.A]), we have the desired inequality αξ({f0 ≤ p},{f0 ≥ q})≤αξ({f ≤p},{f ≥q}). Thus the proof is complete.
Let againf be of Baire classξ. Let
Tf,ξ ={τ0 :τ0 ⊇τ Polish, τ0⊂Σ0ξ(τ), f ∈ B1(τ0)}.
So Tf,ξ is the set of those Polish refinements of the original topology that are subsets of the Σ0ξ sets turningf into a Baire class1 function.
Remark 3.4.44. Clearly,Tf,1={τ} for every Baire class 1 functionf. In order to show that the ranks we are about to construct are well-defined, we need the following proposition.
Proposition 3.4.45. Tf,ξ6=∅ for every Baire class ξfunction f.
Proof. By the previous remark we may assume ξ ≥2. For every rational p the level sets {f ≤ p} and {f ≥p} are Π0ξ+1 sets, hence they are countable intersections ofΣ0ξ sets. In turn, theseΣ0ξ sets are countable unions of sets from S
η<ξΠ0η(τ). Clearly, S
η<ξΠ0η(τ)⊂∆0ξ for ξ≥2. By Kuratowski’s theorem [51, 22.18], there exists a Polish refinementτ0⊂Σ0ξ(τ)ofτ for which all these countable many∆0ξ sets are in∆01(τ0). Then for every rationalpthe level sets are nowΠ02(τ0)sets, and the same holds for irrational numbers too, since these level sets can be written as countable intersection of rational level sets, proving
Tf,ξ6=∅.
We now define a rank on the Baire classξfunctions starting from an arbitrary rank on the Baire class 1 functions.
Definition 3.4.46. Letρbe a rank on the Baire class 1 functions. Then for a Baire classξfunctionf let
ρ∗ξ(f) = min
τ0∈Tf,ξ
ρτ0(f), (3.4.18)
whereρτ0(f)is theρ-rank off in the topologyτ0.
Remark 3.4.47. From Remark 3.4.44 it is clear thatρ∗1=ρfor every ρ.
Proposition 3.4.48. Let ρand η be ranks on the Baire class 1 functions. If ρ= η, or ρ ≤η, or ρ ≈η, or ρ. η then ρ∗ξ =ηξ∗, or ρ∗ξ ≤ηξ∗, or ρ∗ξ ≈ηξ∗, or ρ∗ξ . ηξ∗, respectively. Moreover, the same implications hold relative to the bounded Baire class 1 functions.
Proof. The statement for = and ≤ is immediate from the definitions, and the case of ≈ obviously follows from the case ., so it suffices to prove this latter case only. So assume ρ . η (or ρ . η on the bounded Baire class 1 functions). Choose an optimal τ0 ∈ Tf,ξ for η, that is, ηξ∗(f) = ητ0(f). Then ρ∗ξ(f)≤ρτ0(f).ητ0(f) =ηξ∗(f), completing the proof.
Then the following two corollaries are immediate from Theorem 3.4.24, and Theorem 3.4.33.
Corollary 3.4.49. α∗ξ ≤β∗ξ ≤γξ∗.
Corollary 3.4.50. α∗ξ(f) ≈ βξ∗(f) ≈ γξ∗(f) for every bounded Baire class ξ function f.
Note that by repeating the argument of Remark 3.4.30 one can show that α∗ξ differs fromβ∗ξ andγξ∗.
Theorem 3.4.51. If X is a Polish group then the ranks α∗ξ, βξ∗ and γξ∗ are translation invariant.
Proof. Note first that for a Baire classξfunctionf andx0∈X the functions f◦Lx0 andf ◦Rx0 are also of Baire classξ. We prove the statement only for the rankα∗ξ, because an analogous argument works for the ranksβξ∗ andγξ∗.
Let f be a Baire class ξ function andx0 ∈X, first we prove that α∗ξ(f)≥ α∗ξ(f◦Rx0). Letτ0 ∈Tf,ξ be arbitrary and consider the topologyτ00={U·x−10 : U ∈ τ0}. The map φ : x7→ x·x−10 is a homeomorphism between the spaces (X, τ0) and (X, τ00), satisfying f(x) = (f ◦Rx0)(φ(x)). From this it is clear that τ00 ∈Tf◦Rx0,ξ and since the definition of the rank αdepends only on the topology of the space, we have ατ0(f) = ατ00(f ◦Rx0). Since τ0 ∈ Tf,ξ was arbitrary, the fact thatαξ∗(f)≥αξ∗(f◦Rx0)easily follows.
Repeating the argument with the functionf◦Rx0 and elementx−10 , we have α∗ξ(f◦Rx0)≥α∗ξ(f◦Rx0◦Rx−1
0
) =α∗ξ(f), henceα∗ξ(f) =α∗ξ(f◦Rx0). For the functionf ◦Lx0 we can do same using the topology τ00 ={x−10 ·U : U ∈ τ0} and the homeomorphism φ: x7→x−10 ·x, yieldingα∗ξ(f) =α∗ξ(f ◦Lx0). This
finishes the proof.
Theorem 3.4.52. If f is a Baire classξ function andF ⊂X is a closed set thenf·χF is of Baire classξ, andα∗ξ(f·χF)≤1 +α∗ξ(f),β∗ξ(f·χF)≤1 +βξ∗(f) andγ∗ξ(f·χF)≤1 +γξ∗(f).
Proof. Examining the level sets of the functionf·χF, it is easy to check that
Proof. Using Proposition 3.4.27, it is enough to show that there exists a topology τ0 ∈ Tf,ξ such that f : (X, τ0) → R is continuous, and this is clear
from [51, 24.5].
Next we prove a useful lemma, and then investigate further properties of the ranksα∗ξ,β∗ξ andγξ∗.
Lemma 3.4.54. For everynletτn be a Polish refinement ofτwithτn⊂Σ0ξ(τ).
Then there exists a common Polish refinement τ0 of the τn’s also satisfying τ0 ⊂Σ0ξ(τ).
As above, by Kuratowski’s theorem [51, 22.18], we have a Polish topology τ0, for which these countably many∆0ξ(τ)sets are in∆01(τ0)satisfyingτ0 ⊂Σ0ξ(τ).
Now recall the definition of the derivative definingβ:
ω(f, x, F) = inf
Let us now fixf andε >0and let us denote the derivativeDf,with respect to the topology τ0 by Dτ0, and with respect to the topologyτ00 by Dτ00. By Proposition 3.4.5 it is enough to prove thatDτ00(F)⊂Dτ0(F)for every closed setF ⊂X.
For this it is enough to show thatωτ00(f, x, F)≤ωτ0(f, x, F)for everyx∈F whereωτ0(f, x, F)is the oscillation with respect to the topologyτ0. And this is clear, since in the case ofτ00, the infimum in the definition goes through more open set containingx, hence the resulting oscillation will be less.
For the rank γ, we proceed similarly. First we recall the definition ofγ:
ω((fn)n∈N, x, F) = inf
Let us fix a sequence (fn)n∈N of τ0-continuous (hence also τ00-continuous) functions converging pointwise to f, and also fix ε > 0. Let us denote the derivativeD(fn)n∈N,ε with respect toτ0 byDτ0 and with respect toτ00 byDτ00. Again, by Proposition 3.4.5 it is enough to prove that Dτ00(F) ⊂ Dτ0(F) for every closed setF ⊂X. And similarly to the previous case it is enough to prove that the oscillationω((fn)n∈N, x, F)with respect to the topologyτ00 is at most the oscillation with respect toτ0, but this is clear, since, as before, the infimum
goes through more open set in the case ofτ00.
Theorem 3.4.56. The ranksβξ∗ andγ∗ξ are essentially linear.
Proof. We only consider βξ∗, since the proof for the rank γξ∗ is completely analogous.
It is easy to see thatβ∗ξ(cf) =βξ∗(f)for everyc∈R\ {0}, hence it suffices to show thatβξ∗ is essentially additive.
Forf and gletτf andτg be such thatβτf(f) =βξ∗(f)andβτg(g) =βξ∗(g).
Using Lemma 3.4.54 we have a common refinementτ0 of τf and τg with τ0 ⊂ Σ0ξ(τ). Now f, g ∈ B1(τ0), so f +g ∈ B1(τ0), hence τ0 ∈ Tf+g,ξ. Therefore βξ∗(f+g)≤βτ0(f+g). By Lemma 3.4.55 we have thatβτ0(f)≤βτf(f)(in fact equality holds), and similarly forg. But βτ0 is additive by Theorem 3.4.29, so
βξ∗(f+g)≤βτ0(f+g).max{βτ0(f), βτ0(g)} ≤max{βτf(f), βτg(g)}= max{β∗ξ(f), βξ∗(g)}.
Theorem 3.4.57. If f is a Baire classξ function then
α∗ξ(f)≤αξ(f)≤2α∗ξ(f), henceα∗ξ(f)≈αξ(f).
Proof. Forξ= 1the claim is an easy consequence of the definition of the two ranks and Corollary 3.4.14. From now on, we suppose thatξ≥2.
For the first inequality, for every pair of rationals p < q pick a sequence (Fp,qζ )ζ<αξ(f) ⊂ Π0ξ(X), whose transfinite difference separates the level sets {f ≤p}and{f ≥q}.
EveryΠ0ξ(X)set is the intersection of countably many∆0ξ sets, henceFp,qζ = T
nHp,q,nζ , withHp,q,nζ ∈∆0ξ. By Kuratowski’s theorem [51, 22.18], there is a finer Polish topologyτ0⊂Σ0ξ(τ), for whichHp,q,nζ ∈∆01(τ0)for everyp, q, nand ζ < αξ(f), henceFp,qζ ∈Π01(τ0).
This means that the level sets off can be separated by transfinite differences of closed sets with respect toτ0, hence they can be separated by sets in∆02(τ0).
Then it is easy to see that for everyc∈Rthe level sets{f ≤c}and{f ≥c}are countable intersections of∆02(τ0)sets, hence they areΠ02(τ0)sets, proving that f ∈ B1(τ0). Moreover, α1,τ0(f) ≤ αξ(f) easily follows from the construction (hereα1,τ0 is the rankα1 with respect toτ0). And by Corollary 3.4.14 we have α∗ξ ≤ατ0(f)≤α1,τ0(f)≤αξ(f), proving the first inequality of the theorem.
For the second inequality, take a topology τ0 with ατ0(f) =α∗ξ(f). Again, by Corollary 3.4.14, we haveα1,τ0(f)≤2ατ0(f) = 2α∗ξ(f).
It remains to prove that αξ(f) ≤ α1,τ0(f). A τ0-closed set is Π0ξ with re-spect to τ. Therefore, if (Fη)η<ζ is a decreasing continuous sequence of τ0 -closed sets whose transfinite difference separates{f ≤p}and{f ≥q} then the
same sequence is a decreasing continuous sequence of sets fromΠ0ξ(τ), proving
αξ(f)≤α1,τ0(f).
Corollary 3.4.58. αξ and α∗ξ are essentially linear for bounded functions for every ξ.
Proof. αξ ≈α∗ξ by the previous theorem,α∗ξ ≈βξ∗ for bounded functions by Corollary 3.4.50, andβξ∗ is essentially linear by Theorem 3.4.56.
From Corollary 3.4.37 we can obtain the appropriate statement for the ranks α∗ξ, β∗ξ andγξ∗.
Proposition 3.4.59. Iff =Pn
i=1ciχAi, where theAi’s are disjoint∆0ξ+1 sets coveringX and the ci’s are distinct then
α∗ξ(f)≈max
i {α∗ξ(χAi)}, and similarly for βξ∗ andγξ∗.
Proof. The additivity ofα∗ξ impliesα∗ξ(f).maxi{α∗ξ(χAi)}. For the other inequality letτ0be a topology for whichf is Baire class1. Then the characteris-tic functionsχAi are also Baire class1, and hence by Corollary 3.4.37 we obtain ατ0(f) ≈ maxi{ατ0(χAi)}. But by the definition of α∗ξ for every such topol-ogyα∗ξ(χAi)≤ατ0(χAi), thereforemaxi{α∗ξ(χAi)} ≤maxi{ατ0(χAi)} ≈ατ0(f).
Then choosingτ0 so thatατ0(f) =α∗ξ(f)the proof is complete.
Theorem 3.4.60. The ranksα∗ξ,βξ∗andγ∗ξ are unbounded inω1. Moreover, for every nonempty perfect set P⊂X and ordinal ζ < ω1 there exists a character-istic functionχA∈ Bξ(X) withA⊂P such that α∗ξ(χA), βξ∗(χA),γξ∗(χA)≥ζ.
Proof. In order to prove the theorem, by Corollary 3.4.49 it suffices to prove the statement for α∗ξ. Moreover, instead of α∗ξ(χA) ≥ ζ it suffices to obtain α∗ξ(χA)&ζ. And this is clear from Theorem 3.4.57 and Corollary 3.4.43.
Proposition 3.4.61. Iffn, f are Baire classξfunctions andfn→f uniformly thenβξ∗(f)≤supnβξ∗(fn).
Proof. For everynletτn ∈Tfn,ξwithβτn(fn) =βξ∗(fn). Using Lemma 3.4.54, letτ0 be their common refinement satisfyingτ0⊂Σ0ξ(τ), whereτ is the original topology. Note that fn ∈ B1(τ0) for every n, and the Baire class 1 functions are closed under uniform limits [51, 24.4], henceτ0∈Tf,ξ. Then by Proposition 3.4.31 and Lemma 3.4.55 we have
β∗ξ(f)≤βτ0(f)≤sup
n
βτ0(fn)≤sup
n
βτn(fn) = sup
n
β∗ξ(fn).
Proposition 3.4.62. Iffn, f are Baire classξfunctions andfn→f uniformly thenα∗ξ(f).supnα∗ξ(fn)andγξ∗(f).supnγξ∗(fn).
Proof. Repeat the previous argument but apply Proposition 3.4.39 and
Propo-sition 3.4.32 instead of PropoPropo-sition 3.4.31.