Brian Semmes, ILLC, University of Amsterdam, Holland (e-mail: [email protected])
A NEW PROOF OF THE JAYNE–ROGERS
THEOREM
Abstract
We give a new simpler proof of a theorem of Jayne–Rogers.
1
Introduction
In this paper we will give a new proof of a theorem of Jayne and Rogers. First recall from [2] the following definitions:
Definition 1. LetX, Y be metric spaces. A functionf:X →Y is said to be
∆0
2-functioniff−1(S)∈Σ02 for everyS∈Σ02(equivalently,f−1(U)∈∆02 for every openU ⊆Y). Sometimes these functions are also calledfirst level Borel functions(see [2]).
The functionf is said to bepiecewise continuousifX can be expressed as the union of an increasing sequenceX0, X1, . . . of closed sets such thatf Xn
is continuous for everyn∈ω.
Observe thatfis piecewise continuous if and only if1there is a∆02-partition
hDn | n∈ωiof X such that f Dn is continuous for everyn ∈ω. For one
direction, if f is piecewise continuous then putting D0 = X0 and Dn+1 = Xn+1\Xn we have the desired partition. Conversely, let Pm,n∈Π01 be such thatDn=Sm∈ωPm,nandPm,n⊆Pm0,nfor everym≤m0 andn∈ω, and let
Xn=Si≤nPn,i. It is easy to check that theXn are increasing and closed, and
thatf Xn is continuous (sincePn,i∩Pn,j=∅wheneveri6=j). Thus, in the
rest of this paper, when we will refer to some piecewise continuous function we will generally have in mind a function with this “partition” property.
Key Words: Baire class 1 functions, piecewise continuous functions, first level Borel functions,∆0
2-functions
Mathematical Reviews subject classification: 03E15, 54H05
1A third equivalent definition is thatX can be covered by a countable familyP 0, P1, . . . of closed sets such thatfPnis continuous for everyn∈ω.
Definition 2. A set S in a metric space is said to be Souslin-F set if it belongs toAΠ0
1, whereA is the usual Souslin operation (see Definition 25.4 in [1]).
A metric spaceX is said to be anabsolute Souslin-F set ifX is a Souslin-F set in the completion ofX under its metric.
Observe that ifX is separable then it is an absolute Souslin-F set if and only if it is Souslin, that is if and only if it is the continuous image of the Baire spaceωω.
Now we are ready to give the statement of the original Theorem.
Theorem 1.1(Jayne–Rogers). IfX is an absolute Souslin-F set, thenf:X →
Y is a∆02-function if and only if it is piecewise continuous.
According to the authors of [2], their proof “even in the case whenX andY are separable, is complicated”. Sixteen years later, S lawomir Solecki provided in [3] a new proof of Theorem 1.1 in the case whenX andY are separable and X is Souslin (in fact he proved a much stronger result which refines Theorem 1.1), but even in that case the proof was quite complicated. Our goal is to provide a simpler proof of Theorem 1.1 that works under the same conditions of Solecki’s version of the theorem — see Corollary 2.2.
We will assumeZF+DC(R) throughout the paper (note that the Jayne–
Rogers’ and Solecki’s proofs are carried out inZFC, but by a simple absolute-ness argument the result must hold also inZF+DC(R)). All spaces considered
are metric. Our notation will be quite standard: the set of the natural
num-bers will be denoted byω, while ifXis any topological space andAis a subset ofX we will denote the closure ofAwith Cl(A). The set of all the binary se-quences of finite length will be denoted by<ω2, andω2 will denote the Cantor space. A functionf:X →Y will be said ofBaire class1 if it is the pointwise limit of a sequence of continuous functionsfn:X →Y. Finally, if (X, d) is
any metric space, a setU ⊆X will be called basic open if it is an open ball ofX, i.e. if U ={x∈ X | d(x, x0)< r} where x0 ∈X and r∈ R+. For all
the other undefined symbols and notions we refer the reader to the standard monograph [1].
2
The proof of the Jayne–Rogers Theorem
The main result of this paper is the following Theorem, from which the Jayne– Rogers Theorem will follow — see Corollary 2.2.
Theorem 2.1. LetX andY be metric spaces withX Polish, and letf:X →
Recall that iff:X →Y is of Baire class 1 then it is alsoΣ02-measurable, i.e.f−1(U)∈Σ02for every open set U ⊆Y, but the converse in general fails. Nevertheless, if we require that X and Y are separable and that X is zero-dimensional thenf is of Baire class 1 just in case it isΣ02-measurable (see e.g. Theorem 24.10 in [1]).
Corollary 2.2. LetX andY be separable metric spaces withX Souslin. Then
f:X →Y is a∆02-function if and only if it is piecewise continuous.
Proof. One direction is trivial. For the other direction, assume toward a
con-tradiction thatf is a∆0
2-function butnot piecewise continuous. LetF be the collection of all the closed setsC of (the completion of) X such that f C is continuous. By Corollary 1 of [4], we have that either there is a countable family of sets in F which cover X, or else there is some Z ⊆ X which is homeomorphic to the Baire spaceωω and such that Z can not be covered by a countable family of sets fromF. Since the first alternative easily implies that f is piecewise continuous, we can assume that the second alternative holds and therefore that f0 =f Z is not piecewise continuous. Note that we can assume also that f0 is of Baire class 1 (otherwise, since Z is Polish and zero-dimensional and Y is separable, we would have that f0 is not even
Σ0
2-measurable and hence not a ∆02-function), and therefore we can apply Theorem 2.1 tof0: this gives the desired contradiction.
Before proving Theorem 2.1 we need a couple of technical lemmas. For the next few results,X0 will be an arbitrary subset of the Polish space X. Given A, B⊆Y we will say thatAandB arestrongly disjoint if Cl(A)∩Cl(B) =∅. Moreover if h:X0 →Y is any function we putAh =h−1(Y \Cl(A)). Note that for everyA, B⊆Y one has (A∪B)h=Ah∩Bh. IfhisΣ0
2-measurable andU, V ⊆Y are strongly disjoint, then we have that ifhUhandh
Vhare
both piecewise continuous then the wholehis piecewise continuous. In fact, UhandVhis a finiteΣ02-covering ofX0(by the strongly disjointness ofU and V), which by the reduction property ofΣ02 can be refined to a ∆02-partition
hD0, D1i of X0 such that D0 ⊆Uh, D1 ⊆Vh, and hence both h D0 and h D1 are piecewise continuous. But if h0:X0 → Y is such that for some
∆0
2-partitionhD0n|n∈ωiofX0 we have thath0D0n is piecewise continuous
for every n, then h0 is piecewise continuous on the whole X0: therefore his
piecewise continuous as well. Now let h: X0 → Y be a Σ0
2-measurable function, x ∈ X0, and A be any subset of Y. We say that xis h-irreducible outside A if for every open neighborhoodV ⊆X0ofxthe functionhAh∩V is not piecewise continuous.
A0. Moreover if X00 ⊆ X0 and x ∈ X00 is h0-irreducible outside A (where h0 =hX00) thenxis alsoh-irreducible outsideA.
Lemma 2.3.Supposeh:X0→Y is aΣ02-measurable function andU0, . . . , Un⊆
Y are basic open sets of Y such thatrange(h)∩Cl(Ui) =∅ for every i≤n.
Thenhis notpiecewise continuous if and only if(∗)there is an x∈X0 and
a basic open setU ⊆Y strongly disjoint from U0, . . . , Un such that h(x)∈U
andxis h-irreducible outsideU.
Proof. Put C = Cl(U0)∪. . .∪Cl(Un). We will prove that h is piecewise
continuous if and only if (∗) does not hold. Ifhis piecewise continuous then the same must hold forh X00 where X00 is any subset of X0, therefore one direction is trivial. For the other direction, assume toward a contradiction that (∗) does not hold, i.e. for every x ∈ X0 and every open set U ⊆ Y strongly disjoint from C such that h(x) ∈ U we have that x is h-reducible outsideU, that is there is some open neighborhood V ⊆ X0 of xsuch that h Uh∩V is piecewise continuous. LetQbe the union of all the open sets W ⊆ X0 such that h W is piecewise continuous (this clearly implies that alsohQis piecewise continuous since X0 is assumed to be separable). We
claim thathX0\Qis continuous (from this easily follows thathis piecewise continuous). To see this we fix anyx∈X0\Qand letU ⊆Y be any open set such thath(x)∈U. We want to show that there is some open neighborhood V ofxsuch that h“(V ∩(X0\Q))⊆U. LetU0 ⊆Y be basic open, strongly disjoint fromC, and such that Cl(U0)⊆U (U0 exists sinceY is metric). Let V ⊆X0 be given by the failure of (∗) on the inputs x and U0, and assume toward a contradiction thath(x0)∈/ Cl(U0) for some x0 ∈ V ∩(X0 \Q). In this case we can clearly find a basic openU00 ⊆Y strongly disjoint fromU0 andC, and such thath(x0)∈U00. Let V0⊆X0 be the open set given by the failure of (∗) on inputsx0 andU00. SinceV andV0 have been chosen in such a way thath(U0)h∩V andh(U00)h∩V0 are piecewise continuous, and since
{(U0)h∩V,(U00)h∩V0}is aΣ02-covering ofV∩V0, by the strong disjointness of U0 andU00we must have thathV∩V0is piecewise continuous, and therefore V ∩V0 ⊆Q: but this means that x0 ∈Q, a contradiction!
Lemma 2.4. Let h:X0 →Y be a Σ0
2-measurable function, x∈X0,A⊆Y,
andU0, . . . , Un be a sequence of strongly disjoint open subsets ofY. Ifxish
-irreducible outsideAthen there is at most onei≤nsuch thatxis h-reducible
outsideA∪Ui.
Proof. Assume that i ≤ n is such that x is h-reducible outside A∪Ui, i.e.
that there is an open neighborhoodV ⊆X0 ofxsuch thath(A∪Ui)h∩V
(A∪Uj)h∩V0 is piecewise continuous. But since Ui and Uj are strongly
disjoint, this would imply that h Ah∩V ∩V0 is piecewise continuous as well, and thusV∩V0 would contradict the fact thatxish-irreducible outside A.
Finally observe that iff:X → Y is the pointwise limit of a sequence of functionshfm:X→Y |m∈ωi, then we have the following property: ifx∈X
andU0, U1, . . . are pairwise disjoint open sets such that for infinitely manyn’s there is anm for which fm(x) ∈Un, then f(x)∈/ Un for each n (otherwise,
fm(x)∈Un for all but finitely manym’s contradicting our hypothesis).
Now we are ready to prove Theorem 2.1. The proof essentially uses recur-sively Lemma 2.3 applied to smaller and smaller subspaces ofX to construct some sequences, and Lemma 2.4 will guarantee that at each stage the con-struction can be carried out. This is the reason for which we have proved both the Lemmas for arbitrary functionshwith domain a generic subsetX0 of the Polish spaceX: in fact we will generally apply them to the restriction of the original functionf to some subset ofX, that is withh=f X0.
Proof of Theorem 2.1. Assume thatf:X →Y is of Baire class 1 (hence also
Σ0
2-measurable) but not piecewise continuous, and let hfn | n ∈ ωi be a
sequence of continuous functions which converges pointwise to f. We will construct an open set ˆU ⊆Y such thatf−1( ˆU) is a completeΣ0
2-set, and this will imply thatf is not a∆0
2-function. To be more specific, we will construct (together with ˆU) a continuous reduction from theΣ0
2-complete set S={z∈ω2| ∃i∀j ≥i(z(j) = 0)}
tof−1( ˆU), i.e. a functiong:ω2→X such that
z∈S ⇐⇒ g(z)∈f−1( ˆU).
The functiongwill be defined using aweakCantor schemehVs|s∈<ω2i(that
is a classical Cantor scheme in which we drop the conditionVsa0∩Vsa1=∅) such that for everys, t∈<ω2 we have:
1) Vsis an open subset ofX;
2) ifs(tthen Cl(Vt)⊆Vs;
3) diam(Vs)≤2−length(s).
It is straightforward to check that, given such a scheme, the functiong:ω2→
X:z7→T
n∈ωVzn is well-defined (by the completeness ofX) and continuous
The construction will be carried out by recursion on the rank ofs∈<ω2
with respect to the orderdefined by
st ⇐⇒ length(s)<length(t)∨(length(s) = length(t)∧s≤lext), where≤lexis the usual lexicographical order on<ω2 (the strict part ofwill be denoted by≺). In fact we will define, together with a schemehVs|s∈<ω2i
with the properties above, a sequence hxs | s ∈ <ω2i of points of X and a
sequencehUs|s∈<ω2iof subsets ofY such that for everys∈<ω2:
i) xs∈Vs;
ii) f(xs)∈Us;
iii) Usis basic open and for everyt∈<ω2 we have thatUsandUtare either
equal or strongly disjoint;
iv) there is somem∈ω such thatfm“Vs⊆Us;
v) xtisf-irreducible outsideAfor every ts, whereA=St0sUt0;
vi) ifs=s0a1 thenUs=6 Utfor every2ts0a0 (and therefore, in particular,
for everyt⊆s0).
As already noted, to construct these sequences we will recursively apply Lemma 2.3 to the restriction off to smaller and smaller pieces.
At the first stage, letxandU be given as in Lemma 2.3 applied to the whole f, and letV =f−1
m (U) wherem∈ωis such thatfm(x)∈U (such anmmust
exists by the fact thatf is the limit of the fn’s). Then put V∅ =V, x∅ =x
andU∅ =U. Now lets 6=∅ and suppose we have defined Vt, xt and Ut for
t≺s. If the last digit ofsis a 0, that iss=s0a0, then simply putVs=W,
xs = xs0 and Us = Us0, where W is any open set such that Cl(W) ⊆ Vs0,
xs ∈ W and diam(W) ≤ 2−length(s). Otherwise s =s0a1: by the inductive
hypothesis, condition v) implies thath0=f Af∩Vs0, whereA=S
ts0a0Ut,
is not piecewise continuous (otherwise, sincexs0a0∈Vs0a0⊆Vs0,xs0a0should bef-reducible outsideA).
Claim. There are xs ∈ Vs0 and Us ⊆ Y such that f(xs) ∈ Us, Us is basic
open and strongly disjoint from A (which in particular implies Us 6= Ut for
everyts0a0), andx
tisf-irreducible outsideA∪Us for everyts.
Proof of the Claim. By Lemma 2.3 applied toh0 there must be anx0 ∈Vs0
andx0 ish0-irreducible outsideU0 (hence alsoh0-irreducible outsideA∪U0, since range(h0)∩Cl(A) =∅, and thusf-irreducible outsideA∪U0). If there is somets0a0 such thatxtisf-reducible outsideA∪U0, we can again apply
Lemma 2.3 toh1=h0(A∪U0)f (h1 is not piecewise continuous becausex0 is h0-irreducible outsideA∪U0) to find x1 ∈Vs0 and a basic open U1 such
thatf(x1)∈U1, U1 is strongly disjoint fromA∪U0, and x1 ish1-irreducible outsideA∪U0∪U1 (hence also f-irreducible outsideA∪U1). Moreover, by Lemma 2.4 it must be the case that also xt is f-irreducible outside A∪U1 (in factxt must be f-irreducible outsideA∪U for every open set U which
is strongly disjoint from A∪U0). Now, it could be the case that there is anothert0 s0a0 such that xt0 is f-reducible outsideA∪U1: if this is the
case, apply Lemma 2.3 toh2=h1(A∪U0∪U1)f to getx2andU2such that
f(x2)∈U2,U2is basic open and strongly disjoint fromA∪U0∪U1, andx2is h2-irreducible outsideA∪U0∪U1∪U2(hence, in particular,x2isf-irreducible outsideA∪U2). By Lemma 2.4 again, we must have that both xtandxt0 are
f-irreducible outside A∪U2. Arguing inductively in this way, after (at most) k=|{t∈<ω2|ts0a0}|+ 1-stages we will have found some x
k =xs ∈Vs0
andUk =Ussuch thatf(xs)∈Us,Usis basic open and strongly disjoint from
A, xs is f-irreducible outside A∪Us, and xt isf-irreducible outside A∪Us
for everyts0a0 as well. Claim
LetW ⊆Xbe an open neighborhood ofxssuch that diam(W)≤2−length(s),
Cl(W)⊆Vs0andfm“W ⊆Usfor somem, and defineVs=W. This completes
the recursive definition of the sequences required.
It is easy to check that the scheme hVs | s ∈ <ω2i and the sequences
hxs | s ∈ <ω2i and hUs | s ∈ <ω2i constructed in this way are as required,
i.e. that they satisfy 1)–3) and i)–vi). Now put ˆU = S
s∈<ω2Us, and let
g:ω2 → X be obtained from hV
s | s ∈ <ω2i as described above. We have
only to check that g is a reduction of S to f−1( ˆU). LethU
k | k ∈ωi be an
enumeration without repetitions of hUs | s ∈<ω2i, so that by condition iii)
theUk’s are pairwise disjoint and ˆU =Sk∈ωUk. Ifz∈S, then for some ¯n∈ω
we will have thatxzm=xzn¯ = ¯xfor every m≥n, therefore¯ g(z) = ¯xand f(g(z)) =f(¯x)∈Uzn¯ ⊆ U. Assume nowˆ z /∈S: by conditions vi) and iv), for infinitely many k’s there is somem ∈ ω such that fm(g(z)) ∈ Uk (since
g(z)∈ Vzn for every n∈ ω), and thereforef(g(z))∈/ Uˆ by the observation
preceding this proof.
References
[2] J. E. Jayne and C. A. Rogers,First level Borel functions and isomorphisms, J. Math. pures et appl.61(1982), 177–205.
[3] S. Solecki,Decomposing Borel sets and functions and the structure of Baire
class 1 functions, J. Amer. Math. Soc.11 (1998), no. 3, 521–550.
[4] S. Solecki,Covering analytic sets by families of closed sets, J. Symb. Log.
