PARTITIONING THE REAL LINE INTO BOREL SETS

Abstract For which infinite cardinals 
$\kappa $
 is there a partition of the real line 
${\mathbb R}$
 into precisely 
$\kappa $
 Borel sets? Work of Lusin, Souslin, and Hausdorff shows that 
${\mathbb R}$
 can be partitioned into 
$\aleph _1$
 Borel sets. But other than this, we show that the spectrum of possible sizes of partitions of 
${\mathbb R}$
 into Borel sets can be fairly arbitrary. For example, given any 
$A \subseteq \omega $
 with 
$0,1 \in A$
 , there is a forcing extension in which 
${A = \{ n :\, \text {there is a partition of } {{\mathbb R}} \text { into }\aleph _n\text { Borel sets}\}}$
 . We also look at the corresponding question for partitions of 
${\mathbb R}$
 into closed sets. We show that, like with partitions into Borel sets, the set of all uncountable 
$\kappa $
 such that there is a partition of 
${\mathbb R}$
 into precisely 
$\kappa $
 closed sets can be fairly arbitrary.


Introduction
Hausdorff showed in [10] that the Cantor space 2 ω , and in fact any uncountable Polish space, can be expressed as an increasing union ξ<ω 1 E ξ of G δ sets.It follows that R, or any other uncountable Polish space, can be partitioned into ℵ 1 nonempty F σδ sets.This raises the question: (Q1) For which uncountable cardinals κ is there a partition of R into κ nonempty Borel sets?
Let us define the Borel partition spectrum, denoted sp(Borel), to be the answer to this question: sp(Borel) = {|P| : P is a partition of R into uncountably many Borel sets}.
The main result of this paper shows that the Borel partition spectrum can be fairly arbitrary.(A precise statement of the result can be found in Corollary 3.3 below.)For example, given any set A of positive integers, there is a forcing extension in which sp(Borel) = {ℵ n : n ∈ A} ∪ {ℵ 1 , ℵ ω , ℵ ω+1 }.
A related question can be asked for any given class of subsets of R. As R is connected, there is no partition of R into 2 or more open sets.So the descriptively simplest sets one can ask this for are the closed sets.
(Q2) For which cardinals κ is there a partition of R into κ closed sets?Making use of the Baire Category Theorem, Sierpiński proved in [22] that any partition of R into at least 2 nonempty closed sets is uncountable.(Using the modern vocabulary, his proof actually shows that any partition of R into closed sets has size ≥ cov(M).)Let us define the closed partition spectrum, denoted sp(closed), to be the answer to (Q2): sp(closed) = {|P| : P is a partition of R into at least 2 closed sets}.
We show that, like the Borel partition spectrum, sp(closed) can be fairly arbitrary -even more so, in fact, since the closed partition spectrum need not contain ℵ 1 .For example, given any set A of positive integers, there is a forcing extension in which sp(closed) = {ℵ n : n ∈ A} ∪ {ℵ ω , ℵ ω+1 }.
These results about sp(Borel) and sp(closed) are encompassed in a single theorem, Theorem 3.2 below.The proof identifies a notion of forcing that adds partitions of R into closed sets having certain prescribed sizes, while avoiding partitions of R into Borel sets with other sizes.The second of these tasks is the more difficult.It is accomplished via an isomorphism-of-names argument, similar to the folklore proof (found, e.g., in [3,Theorem 3.1]) that, after many mutually generic Cohen reals are added to a model of CH, there are no MAD families of size strictly between ℵ 1 and c.
Let us point out that analogues of (Q1) and (Q2) have been asked and answered (or partly answered) concerning other extremal families.For example, consider sp(MAD) = {κ ≥ ℵ 0 : there is a MAD family of size κ}.Hechler showed in [11] that sp(MAD) can include any prescribed set of cardinals, and Blass showed in [2] how to exclude certain cardinals from sp(MAD).Shelah and Spinas proved the strongest results in [21], showing that sp(MAD) can be rather arbitrary, especially on the regular cardinals.Similarly, the spectrum of possible sizes of maximal independent families, sp(mif), was investigated recently by Fischer and Shelah in [6,7].Like with sp(MAD), they proved that sp(mif) can be fairly arbitrary, especially on the regular cardinals.Similar work concerning maximal cofinitary groups was done by Fischer in [8].Ultimately, all these proofs share the same core idea: variations on the isomorphism-of-names argument mentioned above.In every case, the key to making this kind of argument work is to find an automorphism-rich poset that can be used to add extremal families of prescribed sizes.One notable exception to this rule is Shelah's analysis of the set of possible sizes of ultrafilter bases in [20], where he proves, from large cardinal hypotheses, that this spectrum can exhibit fairly chaotic behavior.
2. sp(Borel) and sp(closed) do not depend on R We begin this section by observing that the definition of sp(Borel) does not depend on R, and remains unchanged when R is replaced by any other uncountable Polish space: Theorem 2.1.If X is any uncountable Polish space, then sp(Borel) = {κ > ℵ 0 : there is a partition of X into κ Borel sets} .
Proof.By a theorem of Kuratowski (see [14,Theorem 15.6]), any two uncountable Polish spaces are Borel isomorphic: in other words, there is a bijection It turns out that the same is true for sp(closed): if X is any uncountable Polish space, then sp(closed) = {κ > ℵ 0 : there is a partition of X into κ closed sets} .A related theorem is proved by Miller in [16,Theorem 3]: R can be partitioned into ℵ 1 closed sets if and only if some uncountable Polish space can be, if and only if every uncountable Polish space can be.We wish to prove the same, but with any uncountable κ in place of ℵ 1 .Miller's proof does not readily adapt to this task, because it uses in an essential way the fact that ℵ 1 is the smallest uncountable cardinal.So we take a different approach.First we need a few lemmas.

Lemma 2.2. For any uncountable cardinal κ, the following are equivalent:
(1) There is a partition of 2 ω into κ closed sets.
(2) There is a partition of ω ω into κ compact sets.
(3) For every uncountable Polish space X, there is a partition of X into κ compact sets.(4) For some uncountable compact Polish space X, there is a partition of X into κ F σ sets.
(1) ⇒ (2): Suppose P is a partition of 2 ω into κ closed sets.We claim first that there is a closed X ⊆ 2 ω such that P ↾ X is a partition of X into κ nowhere dense closed sets (nowhere dense in X, that is).
To see this, we define a descending transfinite sequence of closed subsets of 2 ω .Let X 0 = 2 ω , and if α is a limit ordinal, take X α = ξ<α X ξ .At stage α, given X α , form X α+1 by removing any open subset of X α contained in a single member of P: Clearly, P ↾ X is a partition of X into compact nowhere dense sets.Furthermore, at any stage α < β of our recursion, Now, we claim there is a subspace Y of X such that Y ≈ ω ω and P ↾ Y is a partition of Y into κ compact sets.Fix a countable basis B for X, and for every U ∈ B fix some follows from the Alexander-Urysohn characterization of ω ω as the unique nowhere compact, zero-dimensional Polish space without isolated points (see [14,Theorem 7.7]).
(2) ⇒ (3): Suppose P is a partition of ω ω into κ compact sets, and let X be any uncountable Polish space.Decompose X into its scattered part and perfect part: i.e., let X = Y ∪ Z, where Y is countable and Z is closed in X (hence still Polish) and Z has no isolated points.By [14,Exercise 7.15], there is a continuous bijection f : (3) ⇒ (4): This implication is obvious, since "every uncountable Polish space" includes some compact spaces, and compact sets are F σ .
(4) ⇒ (1): Suppose X is an uncountable compact Polish space.By [14, Theorem 7.4], there is a continuous surjection f : But every F σ subset of 2 ω can be partitioned into countably many closed sets.(This observation is attributed to Luzin in [16,Theorem 2].)Thus, by breaking up any non-closed members of Q into countably many closed pieces, we can refine Q to obtain a partition of 2 ω into κ closed sets.
Define the Sierpiński cardinal ń to be the minimum size of a cardinal satisfying the equivalent statements in Lemma 2.2: that is, ń = min {|P| : P is a partition of 2 ω into uncountably many closed sets} = min {|P| : P is a partition of ω ω into compact sets} .
Recall that the dominating number d is equal to the smallest covering of ω ω by compact sets.Hence d ≤ ń.And clearly ń ≤ c, because ω ω can be partitioned into singletons.Thus we may consider ń to be a cardinal characteristic of the continuum.Quite a bit is known already about this cardinal.The main results are due to Stern [24], Miller [16], Newelski [18], and Spinas [23], who studied this cardinal implicitly without giving it a name, and Hrušak [12,13], who denotes it a T .The name "Sierpiński cardinal" and the notation ń were suggested by Banakh in [1].
A set S of cardinals is closed under singular limits if for every singular cardinal κ, if S ∩ κ is unbounded in κ then κ ∈ S.
Lemma 2.3.The set S = {κ : there is a partition of 2 ω into κ closed sets} is closed under singular limits.
Theorem 2.4.Let κ be an uncountable cardinal.Then all six statements of the following form are equivalent: Some/every uncountable Polish space can be partitioned into κ compact/closed/F σ sets.
Proof.It is clear that (every-compact) ⇒ (every-closed) ⇒ (every-F σ ), and that (some-compact) ⇒ (some-closed) ⇒ (some-F σ ).Also, Lemma 2.2 implies that (every-F σ ) ⇒ (some-compact): because if every uncountable Polish space can be partitioned into κ F σ sets, then in particular some uncountable compact Polish space can be, and by Lemma 2.2 this implies ω ω can be partitioned into κ compact sets.Thus, to prove the theorem, we need to show (some-F σ ) ⇒ (every-compact).So let X be some uncountable Polish space, and suppose P is a partition of X into κ F σ sets.If κ = ń, then by definition, there is a partition of ω ω into ń compact sets, and by Lemma 2.2 every uncountable Polish space can be partitioned into κ compact sets, and we are done.So let us suppose κ > ń.We consider two cases.
For the first case, suppose κ is regular.By the definition of ń and Lemma 2.2, there is a partition Q of X into ń compact sets.Because |P| = κ > ń and κ is regular, there is some K ∈ Q such that P ↾ K = κ.Thus P ↾ K is a partition of K into κ compact subsets.But K is a compact Polish space, and uncountable because |K| ≥ P ↾ K = κ.Thus some compact uncountable Polish space can be partitioned into κ F σ sets.Invoking Lemma 2.2 again, this implies every uncountable Polish space can be partitioned into κ compact sets.
For the second case, suppose κ is singular.As in the first case, there is a partition Q of X into ń compact sets.If P ↾ K = κ for some K ∈ Q, then we may argue as in the first case and conclude that every uncountable Polish space can be partitioned into κ compact sets, and we are done.So let us suppose instead that P ↾ K < κ for every K ∈ Q.Let λ be any cardinal with ń < λ < κ.Because |P| = κ > λ > ń = |Q|, there is some K ∈ Q such that P ↾ K ≥ λ.Let µ = P ↾ K , and note that µ < κ (by the third sentence of this paragraph).Now, as in the previous paragraph, K is a compact uncountable Polish space that can be partitioned into µ F σ sets.By Lemma 2.2, this implies 2 ω can be partitioned into µ closed sets.Because λ was an arbitrary cardinal below κ and λ ≤ µ < κ, this shows that S = {µ : there is a partition of 2 ω into µ closed sets} is unbounded below κ.By Lemma 2.3, κ ∈ S. By Lemma 2.2, this implies every uncountable Polish space can be partitioned into κ compact sets.
Corollary 2.5.For any uncountable Polish space X, We note that the inequality ń ≥ d has been observed before, and can be considered folklore.It is (arguably) implicit in Miller [16], and was observed explicitly by Hrušak in [13] and later by Banakh in [1].Anticipating the main theorem in Section 3, note that this inequality gives us an easy way of excluding an initial segment of the uncountable cardinals from sp(closed): simply make d big.
Proof.This follows from Corollary 2.5 and Lemma 2.3.
It is worth pointing out that the same result holds for sp(Borel), by a very similar argument.The analogous result also holds for the set sp(MAD) mentioned in the introduction [11,Theorem 3.1], again by a similar argument.
Let {B ξ : ξ < λ} be a partition of 2 ω into λ Borel sets.Observe that Thus we may (and do) assume that B ξ is uncountable for each ξ < λ.Every uncountable Borel set contains a closed subspace homeomorphic to 2 ω .For each ξ < ν, fix C ξ ⊆ B ξ with C ξ ≈ 2 ω , and let P ξ be a partition of We end this section with some terminology regarding trees, and two open questions regarding the Sierpiński cardinal ń.
Recall that a subtree of 2 <ω is a subset of 2 <ω that is closed under taking initial segments.A subtree T of 2 <ω is pruned if every node of T has a successor in for any subtree T of 2 <ω ; and conversely, for every closed C ⊆ 2 ω there is a subtree T of 2 <ω with C = [[T ]] (for details, see [14,Chapter 2]).Similar terminology is used for subtrees of ω <ω or of 2 <k .
Representing closed sets with trees in this way, Theorem 2.4 states that κ ∈ sp(closed) if and only if κ > ℵ 0 and there is a MAD family of κ subtrees of 2 <ω .Similarly, κ ∈ sp(closed) if and only if there is a MAD family of κ finitely branching, pruned subtrees of ω <ω .These characterizations of sp(closed) explain Hrušák's notation, writing a T for min sp(closed) .This relationship between ń and a raises the following questions.Question 2.9.Is it consistent that ń has countable cofinality?
The corresponding question for a was solved by Brendle in [4], where he used Shelah's template forcing technique to obtain a model of a = ℵ ω .It is relatively easy to make ń singular of uncountable cofinality: e.g., by adding ℵ ω 1 Cohen reals to a model of CH, we get a model where It is also simple to prove the consistency of a < ń.This holds, for example, in the Cohen model, where We note that if ń < a is consistent, then proving it is likely very difficult.This is because ń < a implies d < a.The consistency of d < a was an open question for a long time, solved by Shelah in [20].Shelah's technique will not work, however, for obtaining a model of ń < a: his technique makes ń large for the same reasons it makes a large.(Roughly, by an "averaging of names" argument, an ultrapower P κ /U forces that if ń ≥ κ then ń = c.)

Forcing an (almost) arbitrary spectrum
Every κ ∈ sp(Borel) is an example what Blass calls a "∆ 1 1 characteristic" in [2].Blass proves [2, Theorems 8 and 9] that there can be many cardinals between ℵ 1 and c that are not ∆ 1 1 characteristics, and therefore are not in sp(Borel).For example, it is consistent to have the set of ∆ 1  1 characteristics be equal to precisely {ℵ n : n is a power of 17} ∪ {ℵ ω , ℵ ω+1 }.Blass' method does not guarantee that any of these ∆ 1 1 characteristics will be in sp(Borel), except of course for ℵ 1 and c.But his results do show at least that sp(Borel) can contain large gaps, and that it is possible to surgically exclude specific cardinals from sp(Borel).(We should mention that the consistency of having ℵ 2 / ∈ sp(Borel) predates Blass' work, and is due to Miller [17].) In the other direction, extending an earlier result of his [16, Theorem 4], Arnie Miller and the author showed in [5, Theorem 3.12] that: In particular, for any set C of cardinals, there is a ccc forcing extension in which C ⊆ sp(closed).We now sketch a proof of this theorem, in order to introduce some of the ideas used in the proof of the main theorem below, but in a simpler context.This sketch can be skipped by the reader requiring no such introduction.
Proof sketch of Theorem 3.1.Given X ⊆ 2 ω , let T X be the poset whose conditions are pairs (t, B), where • There is some k ∈ ω such that t is a subtree of 2 <k , and t is pruned: i.e., if σ ∈ t with |σ| < k, then σ has a proper extension in t. • B is a finite subset of 2 ω \ X, and b ↾ k is a branch of t for every b ∈ B. The ordering on T X is defined by: (t ′ , B ′ ) extends (t, B) if and only if B ′ ⊇ B and t ′ is an end extension of t (meaning that t = t ′ ∩ 2 <k for some k).
For any X ⊆ 2 ω , T X is σ-centered.The poset T X generically adds an infinite pruned subtree T of 2 <ω , defined from a generic filter G on T X as T = {t : (t, B) ∈ G for some B}.(Equivalently, T is the evaluation in V [G] of the name Ṫ = { σ, q : q = (t, B) for some B, and σ ∈ t} .)In the extension, [[T ]] is a closed subset of 2 ω disjoint from X.
Let T ω X denote the finite support product of countably many copies of T X .For any X ⊆ 2 ω , T ω X is σ-centered.The poset T ω X generically adds countably many infinite pruned subtrees T 0 , T 1 , T 2 , . . . of 2 <ω , and in the extension, In other words, T ω X generically adds an F σ subset of 2 ω that intersects the ground model reals in precisely the complement of X. (Note: this poset may be familiar: it is the one usually used for showing that Martin's Axiom implies every < c-sized subset of 2 ω is a Q-set.) Let C denote the set of cardinals in [ℵ 1 , κ].We now define a finite support iteration of length ω 1 as follows.At stage 0, force with the poset Q 0 = C κ of finite partial functions κ → 2, in order to add a set of κ mutually generic Cohen reals.Let {c ξ : ξ < κ} be an enumeration of these Cohen reals in V Q 0 , and for each µ ∈ C, let P µ 0 = {{c ξ } : ξ < µ}.Note that P µ 0 is a µ-sized collection of disjoint subsets of R: we think of P µ 0 as a first approximation to a µ-sized partition we are trying to build.At a later stage α of the iteration, suppose we have already obtained, for each µ ∈ C, a µ-sized collection P µ α of disjoint subsets of R. In V Qα , define X µ α = P µ α for each µ ∈ C, and then obtain V Q α+1 from V Qα by forcing with µ∈C T ω X µ α .This adds countably many generic trees T µ α,0 , T µ α,1 , T µ α,2 , . . .for each µ ∈ C, and in V Q α+1 we define then there is some α < ω 1 with x ∈ V Qα .At that stage of the iteration, either Either way, x ∈ P µ α+1 ⊆ P µ .Thus P µ is a partition of 2 ω into F σ sets in V Qω 1 , and this means µ ∈ sp(closed) by Corollary 2.5.
The main theorem is proved with a modification of this poset, with two major changes.First, the set C will not necessarily be an interval of cardinals, but will consist only of those cardinals we wish to add to sp(closed).Second, instead of a true iteration, we use a streamlined modification.This modified iteration has actual finite sequences (rather than names for them) for the working parts of the T X µ α , which seems to be necessary for proving that certain permutations of the Cohen reals extend to automorphisms of the entire poset.(And these automorphisms are essential for excluding cardinals / ∈ C from sp(Borel).)The definition of this modified iteration is reminiscent of the template forcing notions in [3], but with a well ordered template, so that it is essentially an iteration.

Theorem 3.2. Let C be a set of uncountable cardinals such that
• min(C) is regular, • C has a maximum with cf(max(C)) > ω, • C is closed under singular limits, and Assuming GCH holds up to max(C), there is a ccc forcing extension in which sp(closed) = C, and furthermore, if min(C) < λ / ∈ C, then λ / ∈ sp(Borel).
Before proving this theorem, let us deduce some relatively easy corollaries from it, including the results mentioned in the introduction.
• C is closed under singular limits, and Assuming GCH holds up to max(C), there is a ccc forcing extension in which Proof.This follows immediately from the previous theorem.Proof.This is proved in exactly the same way as the previous corollary, but without ℵ ω and ℵ ω+1 put into C.
Proof of Theorem 3.2.Let C be a set of uncountable cardinals satisfying the hypotheses listed in the statement of the theorem, and assume GCH holds up to max(C).Let κ = min(C) and let θ = max(C).
For each µ ∈ C, let I µ = µ × {µ, θ + }. (These are merely indexing sets, and for all practical purposes one may think of each I µ as a set of atoms, or urlements, in the set-theoretic universe.The relevant properties of these I µ 's are that they are pairwise disjoint, each I µ has size µ, and the I µ do not "interact" in any accidental way with any other sets in the proof.)Let I = µ∈C I µ , and let P 0 denote the poset of finite partial functions from I × ω to 2. Note that this is equivalent to the standard poset C θ for adding θ mutually generic Cohen reals.For each i ∈ I, let c i denote in V P 0 the Cohen real added by P 0 in coordinate i.More formally, c i denotes in V P 0 the evaluation of the name ċi = { (n, j), p : p(i, n) = j} .
Next we define, recursively, a poset P α , ≤ Pα for each α ≤ κ with α > 0. At every stage of the recursion, P α is defined so that P β is a sub-poset of P α .Conditions in P α are finite partial functions on (I × ω) ∪ (α × C × ω), and the P α are defined so that for any β < α, the restriction of any condition in At limit stages, we take P α = ξ<α P ξ and ≤ Pα = ξ<α ≤ P ξ .In other words, P α , ≤ Pα is the direct limit of P ξ , ≤ P ξ : ξ < α for limit α.
This completes the recursive definition of the P α and ≤ Pα .It is easy to see that ≤ Pα is a partial order on P α for all α ≤ κ, and that P β is a sub-poset of P α whenever β ≤ α ≤ κ.Lemma 3.6.P β is a complete sub-poset of P α for all β ≤ α ≤ κ.
Proof.Fix β ≤ α ≤ κ.It is clear that if p, q ∈ P α and p and q are incompatible in P α , then p ↾ β and q ↾ β are incompatible in P β .Conversely, if two conditions are incompatible in P β , they remain incompatible in P α .So to prove the lemma, it suffices to show that every maximal antichain in P β is also a maximal antichain in P α .Suppose A is a maximal antichain in P β , and let p ∈ P α .By the maximality of A, there is some q ∈ A such that q is compatible with p ↾ β in P β .Let r ∈ P β be a common extension of q and p ↾ β.Define a function s on (I × ω) ∪ (α × C × ω) by setting s = r on (I × ω) ∪ (β × C × ω) and setting s = p on (I × ω) ∪ ((α \ β) × C × ω).Then s ∈ P α , and s is a common extension of p and q in P α .As p was arbitrary, this shows A is a maximal antichain in P α .Lemma 3.7.P κ has the ccc.
Proof.Let A 0 be an uncountable subset of P κ .By the ∆-system lemma, there is an uncountable As there are only finitely many functions R ∩ (I × ω) → 2, this implies there is some uncountable A 2 ⊆ A 1 such that p ↾ (R ∩ (I × ω)) = q ↾ (R ∩ (I × ω)) for any p, q ∈ A 2 .For every p ∈ A 2 , if j ∈ R \ (I × ω), then p(j) = (t, B) for some finite subtree t of 2 <ω .As there are only countably many finite subtrees of 2 <ω , there is some uncountable A 3 ⊆ A 2 such that for each j ∈ R \ (I × ω), there is some fixed t j such that for any p ∈ A 3 , p(j) = (t j , B p j ) for some B p j .But any two members of A 3 are compatible: if p, q ∈ A 3 then R is a common extension of p and q.Thus P κ has no uncountable antichains, and is ccc.(In fact we have shown a bit more: that P κ has property K.) If β ≤ α, then because P β ⊆ P α , we may (and do) consider every P β -name to be a P α -name as well.Let us also set the convention that in V Pα , the evaluation of a P α -name is indicated by removing its dot.So, for example, c i denotes in V Pα (for any α ≤ κ) the Cohen real added by P 0 in coordinate i, and T ξ µ,n denotes in V Pα , for any α ≥ ξ, the evaluation of the name Ṫ ξ µ,n .Lemma 3.8.
is a partition of 2 ω .Observe that every member of P is an F σ subset of 2 ω , and |P| = µ.(It is clear that each T α µ,n is a subtree of 2 <ω , which means the sets of the form n∈ω [[T α µ,n ]] are all F σ .)Thus if P is a partition, then µ ∈ sp(closed) by Theorem 2.4.
The members of P come in two types, so in order to show they are pairwise disjoint, we have three things to prove.First, if i, i ′ ∈ I µ and i = i ′ , then clearly Second, fix i ∈ I µ , α < κ, and n ∈ ω.We claim To see this, fix some p ∈ P κ : we will find an extension r of p forcing , if necessary, we may (and do) assume (α, µ, n) ∈ dom(p).Let (t, B) = p(α, µ, n), with t a pruned subtree of 2 <k .From the definition of P κ , we know p ↾ α Pα ḃ = ċi for every ḃ ∈ B. Using this, and the fact that B is finite, there is an extension q of p in P α , and some finite sequence σ ∈ 2 ℓ for some ℓ > k, such that q Pα " ċi ↾ ℓ = σ but ḃ ↾ ℓ = σ for all ḃ ∈ B." Let t ′ be the largest subtree of 2 <ℓ+1 that is an end extension of t and that does not contain σ.(In other words, τ ∈ t ′ if and only if τ ∈ 2 <ℓ+1 \ {σ} and τ ↾ k ∈ t.This is a subtree of 2 <ℓ+1 , and it is an end extension of t because ℓ > k and σ ∈ 2 ℓ .)Define a function r on (I × ω) ∪ (κ × C × ω) by setting Then r ∈ P κ , and r σ / ∈ Ṫ α µ,n and ċi ↾ ℓ = σ, which means r ċi / ∈ [[ Ṫ α µ,n ]].As p was arbitrary, this shows that To see this, fix p ∈ P κ : we will find an extension of p forcing [[T β µ,m ]]∩ [[T α µ,n ]] = ∅.Extending p to p∪{ (β, µ, m), (∅, ∅) , (α, µ, n), (∅, ∅) }, if necessary, we may (and do) assume (β, µ, m), (α, µ, n) ∈ dom(p).Let (t β , B β ) = p(β, µ, m), where t β is a pruned subtree of 2 <k β for some k β ∈ ω, and let (t α , B α ) = p(α, µ, n), where t α is a pruned subtree of 2 <kα for some k α ∈ ω.From the definition of P κ , we know that p ↾ ((I × ω) ∪ (α × C × ω)) Pα ḃ = ȧ for every ḃ ∈ B β and ȧ ∈ B α .Using this, and the fact that B β and B α are both finite, there is an extension q of p in P α , and some ℓ > k β , k α , such that q "decides" all the ḃ and ȧ up to ℓ, and in such a way that witnesses ḃ = ȧ for all ḃ ∈ B β and ȧ ∈ B α .More precisely: for each ḃ ∈ B β , there is a particular branch c( ḃ) of 2 <ℓ+1 such that q Pα " ḃ ↾ ℓ = c( ḃ)"; and similarly, for each ȧ ∈ B α , there is a particular branch c( ȧ) of 2 <ℓ+1 such that q Pα " ȧ ↾ ℓ = c( ȧ)"; and furthermore, c( ḃ) ↾ ℓ = c( ȧ) ↾ ℓ for all ḃ ∈ B β and ȧ ∈ B α .Let t ′ β be the largest end extension of t β that is a pruned subtree of 2 <ℓ , and let t ′′ β be the end extension of t ′ β to a pruned subtree of 2 <ℓ+1 containing on level ℓ the nodes: β and σ = c( ȧ) for any ȧ ∈ B α .Similarly, let t ′ α be the largest end extension of t α that is a pruned subtree of 2 <ℓ , and let t ′′ α be the end extension of t ′ α to a pruned subtree of 2 <ℓ+1 containing on level ℓ the nodes: α and σ = c( ḃ) for any ḃ ∈ B β .Observe that t ′′ β and t ′′ α have no common nodes on level ℓ.Define a function r on (I × ω) ∪ (κ × C × ω) by setting Then r ∈ P κ , and r t ′′ β is a subtree of Ṫ β µ,m and r t ′′ α is a subtree of Ṫ α µ,n .Because t ′′ β and t ′′ α share no nodes on level ℓ, r Thus P is a pairwise disjoint collection of F σ subsets of 2 ω in V Pκ .To finish the proof, we must show also that P = 2 ω .Fix x ∈ 2 ω in V Pκ .Because P κ has the ccc, and each P α is a complete sub-poset of P κ , there is some ] for some ξ < α, then we're done.If not, let ẋ be a nice P α -name for x, and fix some p ∈ P κ such that p " ẋ = ċi for all i ∈ I µ , and ẋ / ∈ [[ Ṫ ξ µ,n ]] for all ξ < α and n ∈ ω."Because p has finite support, there is some Proof.A straightforward transfinite induction on α shows that |P α | = θ for all α ≤ κ. (For the base case, clearly |P 0 | = θ, and the limit case is also clear.For the successor case, suppose |P α | = θ.Because P α has the ccc, there are θ ℵ 0 = θ nice P α -names for reals, and it follows that |P α+1 | = θ.) In particular, |P κ | = θ, and because P κ has the ccc, it follows that there are θ ℵ 0 = θ nice P κ -names for reals.Hence c ≤ θ in V Pκ .
On the other hand, it is obvious that c ≥ θ in V Pκ , because of the Cohen reals added by P 0 .
Proof.By Lemma 3.8 and Corollary 2.6, κ ≥ ń ≥ d ≥ cov(M) in V Pκ .So to prove the lemma, it suffices to show that 2 ω cannot be covered with < κ meager sets in V Pκ .Suppose λ < κ and {C ξ : ξ < λ} is a collection of closed nowhere dense subsets of 2 ω in V Pκ .For each C ξ , let A ξ be a Borel code that evaluates to C ξ .Because P κ has the ccc and each P α is a complete sub-poset of P κ , there is for each ξ some α < κ such that A ξ ∈ V Pα .Because κ is regular and λ < κ, there is some particular α < κ such that A ξ ∈ V Pα for all ξ < λ.So to prove the lemma, it suffices to show that for each α < κ there is some c ∈ V Pκ not contained in any closed nowhere dense subset of 2 ω coded in V Pα (i.e., we want c to be Cohen-generic over V Pα ).
Then r ∈ P κ , and r forces c ∈ U .As U was an arbitrary open dense subset of 2 ω coded in V Pα , this shows c is Cohen-generic over V Pα .
From the last two lemmas, it follows that sp(closed) ⊆ [κ, θ] in V Pκ , and if λ > θ then λ / ∈ sp(Borel).To finish proving the theorem, it remains only to show that if λ ∈ [κ, θ] and λ / ∈ C, then λ / ∈ sp(Borel).This is where the isomorphism-of-names argument comes in.This argument requires a rich supply of automorphisms of P κ , which we describe next.
If φ : I → I is a permutation, then for every set x we define φ(x) to be the set obtained from x by replacing every i ∈ I in the transitive closure of x with φ(i).This is well defined, because if i, j ∈ I then i is not in the transitive closure of j. (One may think of I as a set of urelements, φ as a permutation of them, and φ as the automorphism of the universe induced by φ.)Alternatively, φ is described via well-founded recursion by the relation φ(x) = φ(y) : y ∈ x .
In particular, if p ∈ P κ , then and this expression can be taken as a recursive definition of φ on P κ , and the natural extension of φ to the class of P κ -names.Lemma 3.11.Let φ be a permutation of I such that φ ↾ I µ is a permutation of I µ for all µ ∈ C. Then φ ↾ P α is an automorphism of P α for all α ≤ κ.
Together with the fact that φ is invertible (with inverse φ−1 = φ −1 ), this shows that the image of P α under φ is a poset, and is naturally isomorphic to P α as witnessed by φ.But it is not immediately clear that the image of P α under φ is equal to P α , which is of course required for our claim that φ ↾ P α is an automorphism of P α .(In fact, one may show that this would not be true for α ≥ 1 if φ were not required to fix all the I µ .)This is proved for φ and its inverse, simultaneously, by induction on α.
The base case α = 0 is clear.And for limit α, if φ ↾ P ξ is an automorphism of P ξ , for all ξ < α, then φ ↾ P α is an automorphism of P α because P α , ≤ Pα is the direct limit of P ξ , ≤ P ξ : ξ < α .The same applies to φ−1 .
Because p ∈ P α , t is a pruned subtree of 2 <k for some k ∈ ω, and (by the inductive hypothesis, that φ ↾ P γ is an automorphism of P γ ), { φ( ḃ) : ḃ ∈ B} is a finite set of nice P γ -names for members of 2 ω .Furthermore, by applying the automorphism φ of P γ to the displayed statement in the last bullet point in the definition of the conditions in P α , we get that for every ḃ ∈ B, φ(p) ↾ ((I × ω) ∪ (γ × C × ω)) Pγ φ( ḃ) ↾ j ∈ t for every j < k, φ( ḃ) = φ( ċi ) for every i ∈ I µ , and φ( ḃ) is not a branch of φ( Ṫ ξ µ,m ) for any ξ < γ and m ∈ ω.
In other words, φ(p) satisfies the final bullet point in the definition of the P α conditions.Hence φ(p) ∈ P α , as claimed.
This shows that (the restriction of) φ is an injective morphism from P α to P α .To get surjectivity, simply note that the same argument applies to φ−1 = φ −1 as well.
From now on, we work in the ground model.Let λ be a cardinal such that λ ∈ [κ, θ] and λ / ∈ C. Suppose Ḃα : α < λ is a set of nice P κ -names for Borel codes for subsets of R. (Any standard method of constructing Borel codes can be used for the proof, provided only that the codes are hereditarily countable sets.But for concreteness, let us take a Borel code to be a subset of ω.)We let B α denote the evaluation of the name Ḃα in V Pκ , and we let Bα ⊆ R denote the interpretation of B α .
We aim to show that Bα : α < λ is not a partition of R in V Pκ .To this end, suppose q ∈ P κ and q forces each of the Bα is nonempty, and Bα ∩ Bβ = ∅ whenever α = β.We will show that q also forces α<λ Bα = R.
Fix a cardinal ν ≤ λ such that α ℵ 0 < ν for all cardinals α < ν, and such that C contains no cardinals in the interval [ν, λ].If λ is neither singular nor the successor of a singular cardinal, then we may simply take λ = ν.Otherwise, using the last two bullet points in our description of C, there is an infinite interval of cardinals below λ and disjoint from C, and we may take ν to be any successor-of-a-successor cardinal in this interval.In either case, α ℵ 0 < ν for all cardinals α < ν by the GCH.
For the base case, P 0 ↾ D 0 is just the set of finite partial functions D 0 → 2, so P α ↾ D 0 = ℵ 0 ≤ κ because D 0 is countable.Because P 0 ↾ D 0 has the ccc, there are ℵ 0 nice P 0 ↾ D 0 -names for subsets of ω.
For the other cases, fix α ≤ κ, and suppose P ξ ↾ D 0 ≤ κ and there are ≤ κ nice P ξ ↾ D 0 -names for subsets of ω, for all ξ < α.A condition in P α ↾ D 0 is defined in the same way as a condition in P α , except that we restrict the ḃ in that definition to members of P γ ↾ D 0 .Using the inductive hypothesis on P γ ↾ D 0 -names, it follows that P α ↾ D 0 ≤ κ.Each nice P α ↾ D 0 -name ẋ for a subset of ω has the form ẋ = { n, p : p ∈ A n }, where each A n is an antichain in P α ↾ D 0 .Because P α has the ccc, each A n is countable.Thus there are κ ℵ 0 = κ names ẋ with this form.
We now proceed to define a new P κ -name Ḃλ for a subset of ω.
Applying standard facts about automorphisms of forcing posets, φ0,λ (q) the evaluation of φ0,λ ( Ḃλ ) codes a nonempty Borel subset of R.
Let B λ denote the evaluation of Ḃλ in V Pκ , and let Bλ be the Borel set that it codes.So, in particular, the displayed statement above implies that q = φ0,λ (q) Bλ = ∅.To show q α<λ Bα = R, it now suffices to show q Bλ ∩ Bβ = ∅ for all β < λ.