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Advances in Mathematics 321 (2017) 475–485
Contents lists available at ScienceDirect
Advances in Mathematics
www.elsevier.com/locate/aim
A transversal of full outer measure
Ashutosh Kumar a,∗,1 , Saharon Shelah a,b,2
a
Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond
J Safra Campus, Givat Ram, Jerusalem 91904, Israel
b
Department of Mathematics, Rutgers, The State University of New Jersey, Hill
Center-Busch Campus, 110 Frelinghuysen Road, Piscataway, NJ 08854-8019, USA
a r t i c l e
i n f o
Article history:
Received 21 October 2015
Received in revised form 7
September 2017
Accepted 2 October 2017
Available online xxxx
Communicated by Slawomir J.
Solecki
a b s t r a c t
We show that for every partition of a set of reals into countable
sets there is a transversal of the same outer measure.
© 2017 Elsevier Inc. All rights reserved.
Keywords:
Outer measure
Forcing
1. Introduction
Our aim is to prove the following.
Theorem 1.1. Suppose Xα : α ∈ S is a partition of X ⊆ [0, 1] into countable sets. Then
there exists Y ⊆ X such that |Y ∩ Xα | = 1 for each α ∈ S and μ (Y ) = μ (X).
* Corresponding author.
E-mail addresses: akumar@math.huji.ac.il (A. Kumar), shelah@math.huji.ac.il (S. Shelah).
Supported by a Postdoctoral Fellowship at the Einstein Institute of Mathematics funded by European
Research Council grant 338821 and by NSF grant No. DMS 1101597.
2
Partially supported by European Research Council grant 338821 and by NSF grant No. DMS 1101597;
Publication No. 1068.
1
https://doi.org/10.1016/j.aim.2017.10.008
0001-8708/© 2017 Elsevier Inc. All rights reserved.
476
A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485
Here μ denotes Lebesgue outer measure on R. For partitions into finite sets, this
follows from an old result of Lusin [9] which says that any set of reals can be partitioned
into two sets of full outer measure (see Lemma 2.2). Another special case of the above
theorem was established in [8]: Every set of reals has a subset of full outer measure
that avoids rational distances. The proof given there relied on a theorem of Gitik and
Shelah [4–6] which says that forcing with a sigma ideal cannot be isomorphic to a product
of random and Cohen forcing (we give another proof of this in Theorem A.1). As a
byproduct of our proof, we get a generalization of this theorem to a larger class of
forcings (see Lemma 6.1) – For example, an ω-length finite support iteration of random
forcing. For background on forcing and generic ultrapowers, we refer the reader to [2,7].
On notation: For a set of reals X, by env(X) (envelope of X), we mean a Gδ set G
containing X such that G \ X has zero inner measure. All relations involving envelopes
are supposed to hold modulo null sets. A subset Y of X has full outer measure in X if
env(X) = env(Y ). If Y ⊆ X and env(X) = env(X \ Y ) we say that Y has positive inner
measure in X; otherwise, we say that Y has zero inner measure in X. For T ⊆ <ω 2, define
[T ] = {x ∈ 2ω : (∀n < ω)(x n ∈ T )}. For σ ∈ <ω 2, define [σ] = {x ∈ 2ω : σ x}.
In forcing, we use the convention that a larger condition is the stronger one – So p ≤ q
means q extends p. If Q, P are forcing notions, we write Q P if Q ⊆ P and every maximal
antichain in Q is also a maximal antichain in P. For an ideal I over a set X, define the
following.
• I + = P(X) \ I;
/ I;
• add(I) is the least cardinal κ satisfying: there exists F ⊆ I, |F| ≤ κ and F ∈
• For Y ∈ I + , I Y = {W ⊆ Y : W ∈ I} is the restriction of I to Y .
2. A sufficient condition
Without loss of generality, (∀α ∈ S)(|Xα | = ℵ0 ). For each α ∈ S, let Xα = {xα,n :
n < ω}. Put Yn = {xα,n : α ∈ S}. For W ⊆ S, write Yn W = {xα,n : α ∈ W }. Note
that Yn depends on the specific enumeration of Xα we fixed.
Claim 2.1. It is enough to show the following.
(): For every X ⊆ [0, 1], for every partition Xα : α ∈ S of X into ℵ0 -sized subsets, for every enumeration Xα = {xα,n : n < ω} (so we can speak of Yn ’s w.r.t. this
enumeration), there is a subset W of S such that either
(a) Y0 is null or
(b) Y0 W has positive outer measure and for all n ≥ 1, Yn W has zero inner measure
in Yn .
Proof of Claim 2.1. Assume (). It is enough to show that we can strengthen “Y0 W
has positive outer measure” to “μ (Y0 W ) ≥ 0.5(μ (Y0 ))” in () above. For then we
can inductively construct a sequence (Wi , ni ) : i < ω such that
A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485
477
•
•
•
•
the Wi ’s are pairwise disjoint subsets of S,
each n < ω equals ni for infinitely many i’s,
if m = ni = nj , i < j, then env(Ym Wi ) ∩ env(Ym Wj ) = ∅,
for each i, letting m = ni and Si = S\ {Wj : j < i}, we have μ (Ym Wi ) ≥
0.5μ [(Ym Si )\Bi ], where Bi = env( {Ym Wj : j < i, nj = m}),
• if n = ni , then μ (Yn (Si \Wi )) = μ (Yn Si ).
Claim 2.1 will immediately follow by taking Y = {Yni Wi : i < ω}.
Call a sequence Wi : i < δ (where δ ≤ ω) of pairwise disjoint subsets of S
greedy, if for each i < δ the following “greediness condition” holds: μ (Y0 Wi ) ≥
0.5(sup{μ (Y0 W ) : W ⊆ Si , env(Y0 W ) ∩ env(Y0 S\Si ) = 0, (∀n ≥ 1)(Yn W
has zero inner measure in Yn Si )}) where Si = S\ {Wj : j < i} and Yn Wi has
zero inner measure in Yn Si for every n ≥ 1. Clearly such sequences exist by () so
fix a maximal one and let W = {Wi : i < δ}. If possible, suppose Y0 W does not
have full outer measure in Y0 . So δ = ω. Using (), pick W ⊆ S\W such that Y0 W is non null, its envelope is disjoint with the envelope of Y0 W and for every n ≥ 1,
μ (Yn S\(W ∪ W )) = μ (Yn S\W ). Since μ (Y0 Wi ) goes to zero as i → ∞, we
conclude that at some stage i < ω, our sequence ceased to be greedy: A contradiction.
Hence Y0 W has full outer measure in Y0 and by taking the union of sufficiently many
Wi ’s we get the required strengthening of () above. 2
Lemma 2.2. Given n ≥ 1, A ⊆ [0, 1] and a partition {xα,k : k < n} : α ∈ T of A into
ordered sets of size n, there exists a partition Tk : k < n of T such that for every k < n,
{xα,k : α ∈ Tk } has full outer measure in {xα,k : α ∈ T }.
Proof of Lemma 2.2. Use induction on n. If n = 1, there is nothing to show. So assume
that the result holds for n ≥ 1. Arguing as in the beginning of the proof of Claim 2.1, it
is enough to show the following:
For every A ⊆ [0, 1], for every partition {xα,k : k < n + 1} : α ∈ T of A into ordered
sets of size n + 1, writing Zk = {xα,k : α ∈ T }, there exists W ⊆ T such that
• μ (Z0 W ) ≥ 0.5μ (Z0 ) (here Zk W = {xα,k : α ∈ W }),
• Zk W has zero inner measure in Zk for every 1 ≤ k ≤ n.
By inductive hypothesis, we can choose a partition Tk : 1 ≤ k ≤ n of T such that
Zk Tk has full outer measure in Zk . Using Lusin’s result [9], for each 1 ≤ k ≤ n choose
a partition Tk = Tk,0 Tk,1 such that each one of Zk Tk,0 and Zk Tk,1 has full outer
measure in Zk Tk . Define W = {Tk,i(k) : 1 ≤ k ≤ n} where i(k) ∈ {0, 1} is such that
μ (Z0 Tk,i(k) ) ≥ 0.5μ (Z0 Tk ). It is easily checked that W is as required. 2
A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485
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3. Forcing
Assume () fails. Fix a witnessing X ⊆ [0, 1], a partition Xα : α ∈ S, enumerations
Xα = {xα,n : n ∈ ω} and the corresponding Yn ’s.
For each n ≥ 1, let An ⊆ S be such that Y0 An is null and for every A ⊆ S\An ,
if Y0 A is null, then Yn A ⊆ env(Yn An ). Let W0 = {An : n ≥ 1}. Let Cn =
env(Yn W0 ), Bn = env(Yn ) \ Cn . We can assume that μ(Bn ) > 0 for infinitely many n –
Otherwise we can use Lemma 2.2 to get a contradiction to the failure of (). By ignoring
the Yn ’s for which Bn is null, we can also assume that μ(Bn ) > 0 for every n ≥ 1.
Replace S by S\W0 . By modifying the Yn ’s, we can assume that env(Yn ) = Bn . This
ensures the following: If W ⊆ S is such that Y0 W is null, then for every n ≥ 1, Yn W
has zero inner measure in Yn .
For n ≥ 1, let Wn ⊆ S be such that Yn Wn is null and if W ⊆ S\Wn is such that
Yn W is null, then for all m ≥ n + 1, Ym W has zero inner measure in Ym .
Let I = {W ⊆ S : μ(Y0 W ) = 0}. Then I is a sigma ideal on S.
Let P = P(S)/I be forcing with I. Observe that Z ∈ P iff (∃n ≥ 1)(Yn Z has positive inner measure in Yn ). In fact, by Lemma 2.2, we can assume that Z ∈ P
iff (∃∞ n)(Yn Z has positive inner measure in Yn ). For n ≥ 1, let Qn = {Z ⊆ S : (∃B ⊆
Bn ) (B is Borel positive and Z = {α ∈ S \ Wn : xα,n ∈ B ∩ Yn })}. Note that if Z ∈ Qn ,
then Yn Z has positive inner measure in Yn and hence Y0 Z is non null. So Qn ⊆ P
for every n ≥ 1. Let Pn = {Z ∈ P : Z ∩ Wn = 0}.
Lemma 3.1. The following hold:
(1)
(2)
(3)
(4)
For each n ≥ 1, Qn Pn ;
{Qn : n ≥ 1} is dense in P;
Each Qn is isomorphic to random forcing;
P adds a Cohen real.
Proof of Lemma 3.1. (1) Suppose {Zk : k < θ ≤ ω} is a maximal antichain in Qn but
not in Pn . Let Z ∈ Pn be incompatible with Zk for every k < θ. So Z ⊆ S \ Wn , Yn Z is
null and Y0 Z is non null. Choose m > n such that Ym Z has positive inner measure
in Ym . But now Z ⊆ S \ Wn , Yn Z is null and Ym Z has positive inner measure in
Ym which is impossible by our choice of Wn .
(2) Let Z ∈ P. Choose n ≥ 1 such that Yn Z has positive inner measure in Yn . Let
B ⊆ Bn be Borel positive such that B ∩ (Yn (S \ Z)) = ∅. Let Z = {α ∈ (Z \ Wn ) :
xα,n ∈ B}. Then Z ∈ Qn and Z ⊆ Z.
(3) Should be clear.
(4) Construct a tree Aσ : σ ∈ <ω 2 of subsets of S such that the following hold.
(a) A∅ = S, and for every σ ∈ <ω 2,
(b) Aσ is a disjoint union of Aσ0 and Aσ1 ,
(c) for every 0 ≤ k ≤ |σ| + 1, μ (Yk Aσ0 ) = μ (Yk Aσ1 ) = μ (Yk Aσ ).
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For Clause (c), we make use of Lemma 2.2 and Lusin’s result [9]. We claim that if
G is P-generic over V , then in V [G], the real x = {σ ∈ <ω 2 : Aσ ∈ G} is Cohen
over V . To see this, suppose that D ⊆ <ω 2 is dense and x ∈
/ {[σ] : σ ∈ D}. Then
S\ {Aσ : σ ∈ D} = Z ∈ G so that Y0 Z is non null. But then, for some n ≥ 1,
Yn Z has positive inner measure in Yn . So Z meets Aσ for every σ extending some σ0 ,
|σ0 | = n, and hence also meets some condition in {Aσ : σ ∈ D}: A contradiction. 2
Note that Lemma 3.1 implies that P satisfies ccc.
4. A dichotomy
Claim 4.1. Let Q P be atomless. Then forcing with Q adds a new real.
Proof of Claim 4.1. Since Q satisfies ccc (as Q P and P satisfies ccc), it is enough to
show that every generic extension contains a new ω-sequence of members of V . Towards
a contradiction, suppose p ∈ Q forces that no such sequence appears in the extension.
Let α be the least ordinal such that for some ν ∈ V Q and q ∈ Q, q ≥ p and q Q
ν : α → V ∧ν ∈
/ V . It is clear that α is regular uncountable. Choose q ≥ p, ν ∈ V Q
such that q Q ν : α → V ∧ ν ∈
/ V . For each β < α, choose (qβ,n , νβ,n ) : n < ω
such that Aβ = {qβ,n : n < ω} is a maximal antichain in Q above q, νβ,n ∈ V and
qβ,n Q ν β = νβ,n . Choose N < ω such that W = {β < α : Aβ ∩ QN = ∅} has
size α. For each β ∈ W , let nβ < ω be such that qβ,nβ ∈ QN . Since QN is isomorphic to
random forcing we can find W ∈ [W ]α such that {qβ,nβ : β ∈ W } consists of pairwise
compatible conditions. Put ν = {νβ,nβ : β ∈ W }. Since Q satisfies ccc, there exists
q ≥ q such that q Q |GQ ∩ {qβ,nβ : β ∈ W }| = α. But now q ν = ν ∈ V :
A contradiction. 2
Lemma 4.2. Suppose ν is a P-name such that P ν ∈ 2ω \ V . Then for every p ∈ P there
exist q ≥ p and a Borel function B : 2ω → ω ω such that either q P B(ν) is an infinitely
often equal real over V or q P V [ν] is a random real extension of V .
Proof of Lemma 4.2. For simplicity, assume that p is the empty condition. For each
n ≥ 1, let T̊n ∈ V Qn be such that for every Qn -generic Gn , T̊n [Gn ] = {σ ∈ <ω 2 : (∀p ∈
Gn )(∃q ∈ P)(q ≥ p and q P σ ≺ ν)}. Clearly, Qn T̊n is a leafless subtree of <ω 2. We
break our proof into two cases.
Claim 4.3. Suppose for some n ≥ 1 and q ∈ Qn , q Qn T̊n is not a perfect tree. Then
there exists p ∈ P, p ≥ q such that p P V [ν] is a random real extension of V .
Proof of Claim 4.3. Choose q1 ∈ Qn , q1 ≥ q and σ ∈ <ω 2 such that q1 Qn σ ∈ T̊n and
T̊n has a unique branch above σ. Choose p ∈ P, p ≥ q1 such that p P σ ≺ ν. Let GP
be P-generic over V with p ∈ GP . Put GQn = GP ∩ Qn . Then, ν[GP ] ∈ V [GQn ] since it
is the unique branch through T̊n [GQn ] above σ. Since intermediate models in a random
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A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485
real extension are also random real extensions (as a complete subalgebra of a measure
algebra is also a measure algebra), the claim follows. 2
Claim 4.4. Suppose for every n ≥ 1, Qn T̊n is a perfect tree. Then for some Borel
function B, P (∀x ∈ ω ω ∩ V )(∃∞ l)(B(ν)(l) = x(l)).
Proof of Claim 4.4. We will think of conditions in Qn as the set of branches through a
perfect subtree S ⊆ <ω 2 such that for each σ ∈ S, μ([S] ∩ [σ]) > 0 – This is a dense set
of conditions in Qn . Let us call such trees fat. Let (pn,k , Sn,k , gn,k , Gn,k , fn,k ) : k < ω
satisfy the following.
•
•
•
•
•
•
•
•
•
{pn,k : k < ω} is a maximal antichain in Qn ;
Sn,k is a fat perfect subtree of <ω 2 and pn,k = [Sn,k ];
gn,k is a function with domain Sn,k ;
For each σ ∈ Sn,k , gn,k (σ) is a finite subtree of ≤h 2 for some h < ω such that each
terminal node of gn,k (σ) has length h = hn,k (|σ|) – So hn,k depends only on |σ|;
Whenever σ τ are from Sn,k , gn,k (σ) is a subtree of gn,k (τ );
Gn,k is a function from pn,k to perfect subtrees of <ω 2;
For every x ∈ pn,k , the height of gn,k (x m) goes to infinity as m goes to infinity
and the limiting tree is Gn,k (x);
pn,k Qn “Gn,k (r̊n ) = T̊n is a perfect tree” where r̊n is the Qn -name for the random
real added by Qn ;
fn,k : ω → ω is increasing such that if σ τ are from Sn,k and |τ | ≥ fn,k (|σ|), then
for every maximal node ν1 ∈ gn,k (σ) there are ≥ 2|σ|+3 + hn,k (|σ|) maximal nodes
ν2 ν1 in gn,k (τ ) – fn,k is well defined because for every x ∈ [Sn,k ], Gn,k (x) is a
perfect tree.
The existence of such objects follows from the following well known fact about random
forcing.
Fact 4.5. Let B denote the random forcing whose conditions are viewed as compact positive
measure subsets of 2ω . Let τ be a B-name for a real in 2ω . Then for every p ∈ B, there
exist a fat tree S ⊆ 2<ω and a (uniformly) continuous function G : [S] → 2ω such that
letting q = [S], q ≥ p and q B G(r˚B ) = τ where r̊B is the canonical name for the random
real added by B.
Claim 4.6. Given k, m < ω and n ≥ 1 there exist m , l = lH and H where m > m,
l < ω and H : l 2 → {0, 1} such that () holds: For every σ ∈ Sn,k ∩ m 2, the set {τ ∈
Sn,k ∩ m 2 : τ σ and (∀η ∈ gn,k (σ))(range(H {η η : η ∈ l 2 ∩ gn,k (τ )} = {0, 1})}
m
has size at least |{τ ∈ Sn,k ∩ m 2 : τ σ}|(1 − 2−2 ).
Proof of Claim 4.6. Fix n ≥ 1 and k, m < ω. Let N0 = max{|gn,k (σ) ∩ hn,k (m) 2| : σ ∈
Sn,k ∩ m 2}. Then N0 ≤ 2hn,k (m) . Put m = fn,k (m). Let l = hn,k (m ) be the common
A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485
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height of gn,k (τ ) for τ ∈ Sn,k ∩ m 2. Let p = min{|{η η : η ∈ l 2 ∩ gn,k (τ )}| : τ ∈
Sn,k ∩ m 2 ∧ η ∈ gn,k (τ m) ∩ hn,k (m) 2}. Note that p ≥ 2m+3 + hn,k (m).
For each ρ ∈ l 2, we choose H(ρ) ∈ {0, 1} with equal probability – So for
l
A ⊆ {H : H : l 2 → {0, 1}}, the probability P r(H ∈ A) = |A|2−2 . For
each σ ∈ Sn,k ∩ m 2, τ ∈ Sn,k ∩ m 2 with τ σ, say that H satisfies (σ, τ )
if for every η ∈ gn,k (σ), the range of H {η η : η ∈ l 2 ∩ gn,k (τ )} is
−p+1
{0, 1}. Note that P r(H satisfies (σ, τ ))
. It follows that for each
≥ 1 − N0 2
m
m
σ ∈ Sn,k ∩ 2, the probability that τ σ : τ ∈ Sn,k ∩ 2 ∧ H satisfies (σ, τ ) ≥
m m
1 − 2−2 τ σ : τ ∈ Sn,k ∩ m 2 is at least 1 − N0 2−p+1 22 . So for an appropriate H to exist, it suffices to have 2m N0 2−p+1 22
p ≥ hn,k (m) + 2m+3 > log(N0 ) + 2m + m + 1. 2
m
< 1. But this is clear since
For each n ≥ 1, k < ω, choose (mn,k,j , Hn,k,j , sn,k,j ) = (mj , Hj , sj ) : j < ω satisfying
the following.
• The mj ’s are increasing with j;
• For each j, () of Claim 4.6 holds with m = mj , m = mj+1 and H = Hj+1 ;
• sj : j < ω lists ω infinitely often.
Define a Borel function B : 2ω → ω ω as follows: For each η ∈ 2ω , l < ω, letting
(n, k, j) = l (so l codes a triplet via some bijection between 3 ω and ω) if there exists
some j1 > j such that Hn,k,j1 (η lHn,k,j1 ) = 1, then pick the least such j1 and define
B(η)(l) = sn,k,j1 . If there is no such j1 , put B(η)(l) = 0.
Claim 4.7. For every p ∈ P, x ∈ ω ω , l < ω, there exist q ≥ p and l1 > l such that
q P B(ν)(l1 ) = x(l1 ).
Proof of Claim 4.7. Choose n ≥ 1, k < ω, q ≥ p such that q ∈ Qn and q, pn,k are
compatible. Let S ⊆ <ω 2 be a fat perfect subtree such that [S] ⊆ q
∩ pn,k . Choose j < ω
m+i
(1 −2−2 ) > 0.5,
sufficiently large and σ ∈ S such that, letting m = mn,k,j , we have
i≥0
(n, k, j) = l1 > l, σ ∈ m 2 and [S] has more than 99 percent measure in [σ]. Choose j1 > j
such that sn,k,j1 = x(l1 ). Now using (), we can find τ ∈ S, τ σ, τ : mn,k,j1 → 2,
l
η ∈ gn,k (τ ) ∩ 2 Hn,k,j1 such that for each j < j < j1 , Hn,k,j (η lHn,k,j ) = 0 and
Hn,k,j1 (η) = 1. Choose q1 ∈ P such that q1 ≥ [S] ∩ [τ ] and q1 P η ν. It follows that
for each j < j < j1 , q1 P Hn,k,j (ν lHn,k,j ) = 0 and q1 P Hn,k,j1 (ν lHn,k,j1 ) = 1.
Hence q1 P B(ν)(l1 ) = sn,k,j1 = x(l1 ). 2
This finishes the proof of Claim 4.4 and Lemma 4.2.
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5. A non meager set in V P
Recall that non(Meager) denotes the least cardinality of a non meager set of reals.
Lemma 5.1. Suppose V |= non(Meager) ≤ κ. Then V P |= non(Meager) ≤ κ.
Proof of Lemma 5.1. Let A be the collection of all quadruples s = (m, n, k, h) where
m = mi : i < ω, n = ni : i < ω and k = ki : i < ω are sequences in ω, m0 = 0,
the mi ’s are strictly increasing and h = hi : i < ω where hi : ki 2 → [mi ,mi+1 ) 2.
Let {xα : α < κ} be non meager in A where the topology over A is generated by
declaring finite restrictions of members of A clopen. For α < κ, define ẙα ∈ 2ω ∩ V P by
ẙα [mα,i , mα,i+1 ) = hα,i (r̊nα,i kα,i ) where mα,i , nα,i , kα,i , hα,i correspond to xα and
r̊j is the random real added by Qj . It suffices to show the following.
Claim 5.2. P Y̊ = {ẙα : α < κ} is non meager.
Proof of Claim 5.2. Suppose not. Then we can find p ∈ P, T̊ ∈ V P , W ⊆ κ, n ≥ 1
such that, p P T̊ is a nowhere dense leafless subtree of <ω 2, {xα : α ∈ W } is
non meager and for every α ∈ W , ∃qα ≥ p such that qα ∈ Qn and qα P ẙα ∈
[T̊ ]. Let T̊n ∈ V Qn be such that T̊n [GQn ] = {σ ∈ <ω 2 : (p ∈ P)(p ≥ p ,
p is compatible with every condition in GQn and p P σ ∈
/ T̊ )}. Note that for every
q ∈ Qn , if q ≥ p , then q Qn [T̊n ] is nowhere dense. We need the following
Claim 5.3. There exist q ∈ Qn , S ⊆ 2<ω , W ⊆ W , H such that the following hold.
• S is a fat perfect tree and [S ] = q ≥ p ;
• H is a function with domain S ;
• For each σ ∈ S , H(σ) = (H0 (σ), H1 (σ)) = (m, t) where m < ω and t ⊆ <m 2 is a
subtree;
• If σ τ are from S , then H0 (σ) ≤ H0 (τ ) and H1 (σ) = H1 (τ ) ∩ <H0 (σ) 2;
• For every k < ω, there exists k < ω such that for every σ ∈ S ∩ k 2, H0 (σ) ≥ k;
• For every r ∈ [S ], {H1 (r k) : k < ω} is nowhere dense in <ω 2;
• q Qn T̊n = {H1 (r̊n k) : k < ω};
• {xα : α ∈ W } is non meager;
• For every α ∈ W , there exists qα ∈ Qn such that qα ≥ q and qα P ẙα ∈ [T̊ ].
Proof of Claim 5.3. First using Fact 4.5, choose a maximal antichain {qi, : i < ω} of
conditions qi, ∈ Qn such that for each i < ω, qi, ≥ p and there are Si, and H for
this qi, satisfying all but the last two clauses about W . Then note that W = i<ω Wi,
where Wi, = {α ∈ W : (∃qα ∈ Qn )(qα ≥ qi, ∧ qα P ẙα ∈ [T̊ ])}. Choose i < ω,
such that {xα : α ∈ Wi , } is non meager and take q = qi , , W = Wi , and the
corresponding S and H. 2
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Claim 5.4. C = {x ∈ A : (∃i < ω)(∀σ ∈ S ∩ 2kx,i )(∀ρ ∈ mx,i 2)(nx,i = n ∧ H0 (σ) ≥
mx,i+1 ∧ ρ ∪ hx,i (σ) ∈
/ H1 (σ))} is open dense in A. Here mx,i , nx,i , kx,i , hx,i correspond
to x ∈ A.
Proof of Claim 5.4. Let (m, n, l, h) be the restriction of some member of A to some i < ω.
Extend (m, n, l, h) to a member x ∈ A as follows. Put nx,i = n . Choose mx,i > mi−1 .
For each r ∈ [S ], let τr ∈ 2<ω , kr < ω be such that (∀ρ ∈ mx,i 2), [ρ τr ] ∩ [H1 (r)] = ∅
and H0 (r kr ) > mx,i + |τr | where H1 (r) = {H1 (r k) : k < ω}. Choose F ⊆ [S ]
finite such that {[r kr ] : r ∈ F } covers [S ]. Choose mx,i+1 > max{H0 (r kr ) : r ∈ F }.
Choose kx,i such that for every r ∈ F , kx,i > kr and for every σ ∈ S ∩ kx,i 2, H0 (σ) ≥
mx,i+1 . For σ ∈ S ∩ kx,i 2, choose hx,i (σ) ∈ [mx,i ,mx,i+1 ) 2 such that for some r ∈ F ,
r kr = σ kr and for all t < |τr |, hx,i (σ)(mx,i + t) = τr (t). Extend x arbitrarily to a
member of A. It is easily checked that x ∈ C. 2
Now let α ∈ W be such that xα ∈ C as witnessed by i < ω. Choose qα ∈ Qn such
that qα ≥ q and qα P ẙα ∈ [T̊ ]. Let σ ∈ kα,i 2 be such that q = qα ∩ [σ] has positive
measure. Notice that q P ẙα mα,i+1 ∈
/ T̊n hence for some q1 ∈ P, q1 ≥ q ≥ qα and
q1 P ẙα ∈
/ [T̊ ]: A contradiction. 2
6. Contradiction
The following lemma contradicts Lemma 3.1 and thus finishes the proof of Theorem 1.1.
Lemma 6.1. The following is impossible.
(1) I is an ideal over S, add(I A) = κ > ℵ0 for every A ∈ I + ;
(2) (Qn , pn ) : n < ω satisfies the following: For every n < ω, pn ∈ P, Qn P≥pn
where P≥pn = {p ∈ P : p ≥ pn }, Qn is forcing isomorphic to random forcing and
{Qn : n < ω} is dense in P;
(3) P is forcing isomorphic to P(S)/I;
(4) Forcing with P adds a Cohen real.
Proof of Lemma 6.1. Let G be P(S)/I-generic over V and let j : V → N ⊆ V [G] be
the generic ultrapower embedding with critical point κ. Let f : S → κ represent κ.
Define J = {A ⊆ κ : f −1 [A] ∈ I}. Let Q = P(κ)/J . Then Q P – So Q adds a
new real by Claim 4.1. Let H be Q-generic over V and let k : V → M ⊆ V [H] be the
generic ultrapower embedding. Note that κ N ∩ V [G] ⊆ N and κ M ∩ V [H] ⊆ M (see
Proposition 2.14, [2]) and we’ll use this freely below. We can divide our proof into two
cases.
Case 1: There is a real r ∈ ω ω ∩ M such that r is infinitely often equal to every real
in V . Let rα : α < κ represent r. Then for every x ∈ ω ω ∩ V , there exists α < κ such
484
A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485
that rα and x agree infinitely often. It follows that there is a non meager set of size κ
in V . Since P adds a random real, V ∩ 2ω is meager in V [G]. Let B be a meager Fσ -set
coded in V [G] that contains V ∩2ω . Then B is also coded in N . So by elementarity of j, it
follows that every set of reals in V of size < κ is meager in V . So V |= non(Meager) = κ.
Hence N |= non(Meager) = j(κ) > κ. By Lemma 5.1, V [G] and hence N has a non
meager set of size κ: A contradiction.
Case 2: No real in M is infinitely often equal to every real in V . By Lemma 4.2, for
every new real r ∈ M , V [r] is a random real extension of V . Using Claim 4.1, choose
r ∈ V [H] random over V . Then r ∈ M . Let rα : α < κ represent r. Then {rα : α < κ}
is a non null set in V . Since P adds a Cohen real, V ∩ 2ω is null in V [G]. Let B be
a null Gδ -set coded in V [G] that contains V ∩ 2ω . Then B is also coded in N . So by
elementarity of j, it follows that every set of reals in V of size < κ is null in V . In
particular, for every γ < κ, {rα : α < γ} is null in V . By considering k(rα : α < κ), it
follows that {rα : α < κ} is null in M . Let r be a null Borel set coded in M witnessing
this. It follows that V [r ] is not a random real extension of V : A contradiction. 2
Appendix A
Let Randomθ (resp. Cohenθ ) denote the forcing for adding θ random (resp. Cohen) reals. Randomθ is the measure algebra on the product measure space [0, 1]θ (with Lebesgue
measure on [0, 1]) and Cohenθ is the forcing for adding a function in ω×θ 2 using finite
approximations. Gitik and Shelah [4–6] proved that forcing with a sigma ideal cannot
be isomorphic to Random1 × Cohen1 . In [3] (Section 546), Fremlin asked if the result
generalizes to a product that adds more of these reals. We show that this is the case.
Theorem A.1. Let θ1 , θ2 be positive cardinals. Let I be a sigma ideal over X. Then forcing
with P(X)/I is not isomorphic to P = Randomθ1 × Cohenθ2 .
Proof of Theorem A.1. Suppose not. By restricting I, we can assume add(I A) = κ
for every A ∈ I + . Note that since P is ccc, κ > ℵ1 . Let G be P-generic over V and let
j : V → M ⊆ V [G] be the generic ultrapower embedding with critical point κ.
The following is well known [1].
Fact A.2. V Cohen1 |= cov(Null) = ℵ1 . Here, cov(Null) is the least cardinality of a family
of null sets whose union covers R.
So V [G] and therefore M (as κ M ∩ V [G] ⊆ M ) thinks that cov(Null) = ℵ1 . Hence also
V |= cov(Null) = ℵ1 . Let Ni : i < ω1 be a sequence of null Gδ -sets witnessing this in V .
By elementarity of j, M |= {Ni : i < ω1 } = R. But this is impossible as V [G] and
therefore M contains a random real over V which cannot belong to any of the Ni ’s. 2
A. Kumar, S. Shelah / Advances in Mathematics 321 (2017) 475–485
485
References
[1] T. Bartoszynski, H. Judah, Set Theory: On the Structure of the Real Line, A K Peters, Wellesley
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[2] M. Foreman, Ideals and generic elementary embeddings, in: M. Foreman, A. Kanamori (Eds.), Handbook of Set Theory, vol. 2, Springer, 2010, pp. 885–1147.
[3] D.H. Fremlin, Measure Theory, vol. 5: Set-Theoretic Measure Theory, Part II, 2009.
[4] M. Gitik, S. Shelah, Forcing with ideals and simple forcing notions, Israel J. Math. 68 (1989) 129–160.
[5] M. Gitik, S. Shelah, More on simple forcing notions and forcings with ideals, Ann. Pure Appl. Logic
59 (1993) 219–238.
[6] M. Gitik, S. Shelah, More on real-valued measurable cardinals and forcing with ideals, Israel J. Math.
124 (2001) 221–242.
[7] A. Kanamori, The Higher Infinite, 2nd edition, Springer Monogr. Math., Springer, Berlin, 2003.
[8] A. Kumar, Avoiding rational distances, Real Anal. Exchange 38 (2) (2012/2013) 493–498.
[9] N. Lusin, Sur la decomposition des ensembles, C. R. Acad. Sci. Paris 198 (1934) 1671–1674.
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