This paper generalizes the classical Mycielski theorem to rectangles formed by different types of trees in measure and category settings, revealing new structural properties and nonstandard proofs via forcing extensions.
Contribution
It introduces new variants of the Mycielski theorem involving different tree structures and establishes their existence in measure and category contexts, along with nonstandard proofs using forcing.
Findings
01
Existence of perfect sets within certain trees for comeager sets
02
Construction of uniformly perfect trees avoiding specific tree bodies
03
Nonstandard proofs of Mycielski-like theorems via forcing extensions
Abstract
Two-dimensional version of the classical Mycielski theorem says that for every comeager or conull set Xβ[0,1]2 there exists a perfect set Pβ[0,1] such that PΓPβXβͺΞ. We consider generalizations of this theorem by replacing a perfect square with a rectangle AΓB, where A and B are bodies of other types of trees with AβB. In particular, we show that for every comeager GΞ΄β set GβΟΟΓΟΟ there exist a Miller tree M and a uniformly perfect tree PβM such that [P]Γ[M]βGβͺΞ and that P cannot be a Miller tree. In the case of measure we show that for every subset F of 2ΟΓ2Ο of full measure there exists a uniformly perfect tree Pβ2<Ο such that [P]Γ[P]βFβͺΞ and no side of such aβ¦
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Marcin Michalski, Robert RaΕowski, Szymon Ε»eberski, Department of Computer Science, Faculty of Fundamental Problems of Technology, WrocΕaw University of Technology, 50-370 WrocΕaw, Poland
Abstract.
Two-dimensional version of the classical Mycielski theorem says that for every comeager or conull set Xβ[0,1]2 there exists a perfect set Pβ[0,1] such that PΓPβXβͺΞ. We consider generalizations of this theorem by replacing a perfect square with a rectangle AΓB, where A and B are bodies of other types of trees with AβB. In particular, we show that for every comeager GΞ΄β set GβΟΟΓΟΟ there exist a Miller tree M and a uniformly perfect tree PβM such that [P]Γ[M]βGβͺΞ and that P cannot be a Miller tree. In the case of measure we show that for every subset F of 2ΟΓ2Ο of full measure there exists a uniformly perfect tree Pβ2<Ο such that [P]Γ[P]βFβͺΞ and no side of such a rectangle can be a body of a Silver tree or a Miller tree. We also show some properties of forcing extensions of the real line from which we derive nonstandard proofs of Mycielski-like theorems via Shoenfield Absoluteness Theorem.
The work has been partially financed by grant S50129/K1102 (0401/0052/18) from the Faculty of Fundamental Problems of Technology, WrocΕaw University of Science and Technology.
1. Introduction and notation
The motivation of this paper is the following two-dimensional version of classical Mycielski theorem (see [6]).
Theorem 1**.**
For every comeager or conull set Xβ[0,1]2 there exists a perfect set Pβ[0,1] such that PΓPβXβͺΞ, where Ξ={(x,x):xβ[0,1]}.
In the Cantor space 2Ο and the Baire space ΟΟ each perfect set has a natural combinatorial description.
Let Aβ{2,Ο} and denote A<Ο=βnβΟβAn. Let us recall that TβA<Ο is a tree on A if for each ΟβT and every nβΟ we have ΟβΎnβT.
A body of a tree TβA<Ο is the set [T]={xβAΟ:(βn)(xβΎnβT)} of infinite branches of T.
A tree TβA<Ο is called a perfect tree (or a Sacks tree), if
[TABLE]
Then PβAΟ is a perfect set if and only if P is a body of a perfect tree.
A natural question arises whether we may replace perfect trees with another types of trees.
Our general setup will be as follows. We will consider a subset X of 2ΟΓ2Ο or ΟΟΓΟΟ, of full measure or comeager, and investigate whether there exist trees T1β,T2β satisfying T2ββT2β such that [T1β]Γ[T2β]βXβͺΞ, where Ξ denotes a diagonal, i.e. Ξ={(x,x):xβS} and S is the space we work in. Natural examples of considered trees are Miller, Laver, uniformly perfect and Sliver trees.
We adopt the standard set-theoretical notation (see [3]). Let TβA<Ο be a tree on a set Aβ{2,Ο}. We will use the following notions related to trees:
a Miller or superperfect tree, if (βΟβT)(βΟβΟ-split(T))(ΟβΟ);
β’
a Laver tree, if (βΟ)(βΟβT)(ΟβΟβ¨(ΟβΟβ§ΟβΟ-split(T))).
We will denote the shortest splitting node of a given tree T by stem(T). Nodes Ο,ΟβA<Ο are orthogonal (denoted by Οβ₯Ο), if neither ΟβΟ nor ΟβΟ. Sometimes we will be indexing nodes with nodes. In such cases for the sake of brevity we will write e.g. Ο010β instead of Ο(0,1,0)β.
As mentioned above, we will also consider some specific types of perfect trees (see [5]). We call a perfect tree TβA<Ο
a Silver tree, if (βΟ,ΟβT)(β£Οβ£=β£Οβ£β(βaβA)(Οβ’aβTβΟβ’aβT)).
Before we proceed let us notice that to provide an example of a comeager subset of X2 which does not contain a rectangle AΓB of sets of certain type, it is enough to show that there exists comeager set GβX with Aξ βG or Bξ βG. Indeed, in such a case GΓX is comeager too (by Kuratowski-Ulam Theorem) and AΓ{x}ξ βGΓX for every xβX. The same is true for the measure case thanks to Fubini Theorem. This observation gives weight to Propositions 2 and 10.
2. Category Case
In this section we will focus on finding trees T1ββT2ββΟ<Ο of types mentioned in Introduction, satisfying [T1β]Γ[T2β]βG for a given comeager set GβΟΟΓΟΟ. The main positive result is Theorem 8. Theorem 3 and Propositions 5 and 9 show that the main result is somehow optimal.
Let Q={qβΟΟ:(ββn)(q(n)=0)} be a set of rationals localized in ΟΟ. On several occasions in this section we will use some specific countable dense subset of ΟΟΓΟΟ. Let us define it in the following way:
[TABLE]
where supp(q)=max{nβΟ:q(n)ξ =0}+1. Since supp(q1β)=supp(q2β) for every q=(q1β,q2β)βQ, we may naturally extend the domain of supp to QβͺQ so that supp(q)=supp(q1β).
As a warm up let us consider a case of Laver trees.
Proposition 2**.**
There exists a dense GΞ΄β set GβΟΟ such that [L]ξ βG for every Laver tree L.
Proof.
Let G={xβΟΟ:(ββnβΟ)(x(n)=0)}. Clearly, G is GΞ΄β and dense. Let L be a Laver tree. Let xβ[L] such that x(n)ξ =0 for every nβ₯β£stem(L)β£. Then xβ[L]\G.
β
Let us notice that every nonempty open set is a body of a Laver tree.
The following theorem shows that the perfect set in Mycielski Theorem cannot be replaced with a body of Miller tree.
Theorem 3**.**
There exists an open dense set UβΟΟΓΟΟ such that [T]Γ[T]ξ βUβͺΞ for every Miller tree T.
Proof.
Let {qn:nβΟ} be an enumeration of Q and let us set
[TABLE]
where K(q)=max{q1β(n),q2β(n):nβΟ} for q=(q1β,q2β)βQ.
Let T be a Miller tree. Without loss of generality we may assume that for every ΟβT either β£succTβ(Ο)β£=1 or β£succTβ(Ο)β£=Ο. We will pick points
[TABLE]
from [T] via induction. Let Ο0β=Ο0β=stem(T). Let us assume the following notation
[TABLE]
Let us execute the step n+1. We set
[TABLE]
and Οn+1ββsn+1β such that xnββ’Οn+1ββsplit(T) and β£xnββ’Οn+1ββ£>β£ynββ£. In a similar fashion we proceed with yn+1β. We set
[TABLE]
and Οn+1ββtn+1β such that ynββ’Οn+1ββsplit(T) and β£ynββ’Οn+1ββ£>β£xn+1ββ£.
We will show that
[TABLE]
It is clear that (x,y)β([T]Γ[T])\Ξ. Let us suppose that (x,y)βU. Then there is qβQ such that (x,y)β[qβΎ(supp(q)+K(q))]. It follows that
[TABLE]
Let us observe that since q1βξ =q2β, β£stem(T)β£<supp(q). Let
[TABLE]
In particular it means that K(q)β₯tn+1β and q2ββΎsupp(q)βyn+1ββ’tn+2β. Let us also observe that
[TABLE]
It is the case that exactly one of the following holds:
(1)
q1ββΎsupp(q)βxn+1ββ’sn+2β;
2. (2)
xn+1ββ’sn+2ββq1ββΎsupp(q).
If (1) is true, then
[TABLE]
which gives a contradiction, since K(q)β₯tn+1β>β£xn+1ββ£ and sn+2βξ =0.
If (2) holds, then K(q)β₯sn+2β and
[TABLE]
which is a contradiction because K(q)β₯sn+2β>β£yn+1ββ£ and tn+2βξ =0.
Therefore (x,y)β/U.
β
Next result is concerned with replacing a perfect tree with a Silver tree. First, let us define some useful property of perfect trees. We will say that a perfect tree T splits and rests, if
[TABLE]
Lemma 4**.**
For every Silver tree T there exists a Silver tree Tβ²βT that splits and rests.
Proof.
Let n0β=min{β£Οβ£:Οβsplit(T)} and s0β=min{nβΟ:Ο0ββ’nβT}, where Ο0ββT and β£Ο0ββ£=n0β. For k>0 let
[TABLE]
where ΟkββT satisfies β£Οkββ£=nkβ. Now, let
[TABLE]
and set
[TABLE]
Then Tβ² is the desired tree.
β
Proposition 5**.**
There exists an open dense set UβΟΟΓΟΟ such that [T]Γ[T]ξ βUβͺΞ for any Silver tree T.
Proof.
Let Q={qn:nβΟ} and set
[TABLE]
Let T be a Silver tree. Without loss of generality we may assume that T splits and rests (Lemma 4). Let (x,y)β[T]Γ[T], xξ =y, and suppose that (x,y)βU. Then there is q=(q1β,q2β)βQ such that
[TABLE]
Clearly
[TABLE]
hence all of the nodes in T of lengths supp and supp+1 are splitting, which constitutes a contradiction with the splitting and resting property of T.
β
The following lemmas are preparation to the main theorem of this section.
Lemma 6**.**
For every open dense set UβΟΟΓΟΟ and two open sets V1β,V2ββΟΟ there are sequences Ο1β,Ο2ββΟ<Ο satisfying [Ο1β]βV1β, [Ο2β]βV2β, β£Ο1ββ£=β£Ο2ββ£ such that [Ο1β]Γ[Ο2β]βU and [Ο2β]Γ[Ο1β]βU.
For Ο,ΟβΟ<Ο and UβΟΟΓΟΟ let us denote the fact that [Ο]Γ[Ο]βU and [Ο]Γ[Ο]βU by Ο(Ο,Ο,U). The following lemma is an extension of the previous one.
Lemma 7**.**
For every open dense set UβΟΟΓΟΟ, a finite sequence of open sets (Vkβ:0β€k<n) in ΟΟ there is a sequence of sequences (Οkβ:0β€k<n) such that:
(1)
[Οkβ]βVkβ* for all 0β€k<n,*
2. (2)
β£Οkββ£=β£Οlββ£* for all 0β€k,l<n,*
3. (3)
Ο(Οlkβ,Οklβ,U)* for all distinct 0β€k,l<n.*
Proof.
Let U and (Vkβ:k<n) be as in the formulation. Applying Lemma 6 multiple times we will construct inductively a sequence (Οlkβ:k,l<n;kξ =l) of sequences satisfying:
Ο(Οlkβ,Οklβ,U) holds for all distinct k,l<n.
At the step [math] first we find Ο10β and Ο01β such that [Ο10β]βV0β, [Ο01β]βV1β and Ο(Ο10β,Ο01β,U).
Then for every k<n,kξ =0,1, we choose Οk0β and Ο0kβ satisfying [Οk0β]β[Οkβ10β],[Ο0kβ]βVkβ, and Ο(Οk0β,Ο0kβ,U).
Let us execute the step k, 0<k<n. We pick Οk+1kβ and Οkk+1β satisfying [Οk+1kβ]β[Οkβ1kβ], [Οkk+1β]β[Οkβ1k+1β] and Ο(Οk+1kβ,Οkk+1β,U). For l>k+1 we find Οlkβ and Οklβ such that [Οlkβ]β[Οlβ1kβ], [Οklβ]β[Οkβ1lβ] and Ο(ΟlkβΟklβ,U). The construction is completed.
Let us set Οkβ²β=Οnβ1kβ for every k<n. If lengths of Οkβ²β,0β€k<n are the same, then we set Οkβ=Οkβ²β for each 0β€k<n. If not, let N=max{β£Οkβ²ββ£:0β€k<n}, and let us set
[TABLE]
for each 0β€k<n. Then lengths of these sequences match and properties established during the construction are not compromised.
β
Theorem 8**.**
For every comeager set G of ΟΟΓΟΟ there exists a Miller tree MβΟ<Ο and a uniformly perfect tree PβM such that [P]Γ[M]βGβͺΞ.
Proof.
Let us assume that G=βnβΟβUnβ where (Unβ)nβΟβ is a descending sequence of open dense subsets of ΟΟΓΟΟ. We will construct recursively a sequence (Bnβ:nβΟ) of sets. Bnβ={ΟΟβ:Οβnβ€n} should consist of nodes satisfying:
(1)
Οβ β=β , ΟΟ1βββΟΟ2ββ for Ο1ββΟ2β and ΟΟββ’kβΟΟβ’kβ;
2. (2)
for n>0 and all Ο,Οβ²βBnβ\Bnβ1βΟ(Ο,Οβ²,Unβ) holds;
4. (4)
If Ο1β,Ο2ββ{0,1}β€n then β£ΟΟ1βββ£=β£ΟΟ2βββ£.
At the step [math] we set Οβ β=β and B0β={Οβ β}. Next, we set Ο0β,Ο1ββΟβ β so that Ο(Ο0β,Ο1β,U2β) (Lemma 6), and Ο00β,Ο01ββΟ0β,Ο10β,Ο11ββΟ1β with accordance to Lemma 7. We set
[TABLE]
Now, let us assume that we already have a set Bnβ with the above properties and let us execute the step n+1, n>1. First we set ΟΟβ’nβ for Οβn<n and ΟΟβ’kβ, Οβnn, k<n+1, in such a way that they have the same lengths, propagate the condition (1) and (2), and Ο(ΟΟ1ββ,ΟΟ2ββ,Un+1β) for all distinct
[TABLE]
Next, we set ΟΟβ’kβ for Οβ(n+1)<n+1\nβ€n and k<n+1 in a similar fashion.
This completes the construction. Let us set B=βnβΟβBnβ and
[TABLE]
Clearly, M is a Miller tree. Furthermore, PβM is a uniformly perfect tree thanks to the condition (4). We will show that [P]Γ[M]βGβͺΞ. Let (x,y)β[P]Γ[T], xξ =y. We claim that there exists Ξ±βΟΟ such that
[TABLE]
We will define Ξ±=(a0β,a1β,...) via induction. Let us observe that yβΎ1βΟy(0)β and Οy(0)β is the shortest sequence from B possessing such a property. Therefore, yβΎβ£Οy(0)ββ£=Οy(0)β, otherwise there would be ΟβB such that Οy(0)ββΟ and yβΎβ£Οy(0)ββ£βΟ, which is a contradiction. We set a0β=y(0).
Next, let us assume that we already have a strictly ascending sequence (akβ:k<n) of natural numbers with a property Οa0βa1β...anβ1βββy for every k<n. As previously, we see that
[TABLE]
and that Οa0βa1β...anβ1βy(β£Οa0βa1β...anβ1βββ£)β is the shortest sequence from B with such a property. Hence Οa0βa1β...anβ1βy(β£Οa0βa1β...anβ1βββ£)ββy, so we set anβ=y(β£Οa0βa1β...anβ1βββ£). This completes the definition of Ξ±.
Now, let us fix NβΟ. There exists Nβ²β₯N such that ΟΞ±βΎNββBNβ²β\BNβ²β1β. Furthermore, there exists Οβ2Nβ² such that ΟΟββx. Then [ΟΟβ]Γ[ΟΞ±βΎNβ]βUNβ²ββUNβ, hence also (x,y)βUNβ. N was chosen arbitrarily, thus (x,y)βG.
β
Let us make some remarks. The Miller tree T in the above theorem has a nice property. For each ΟβT the set succTβ(Ο)=Ο or β£succTβ(Ο)β£=1. Let us also observe that one cannot make this Miller tree uniformly perfect.
Proposition 9**.**
There exists a GΞ΄β set G such that [T]ξ βG for every uniformly perfect Miller tree.
Proof.
For every nβΟ let Gnβ=βqβQβ[qβΎ(supp(q)+K(q)+n))]. Let T be a uniformly perfect Miller tree. Without loss of generality we may assume that for every ΟβT we have β£succTβ(Ο)β£β{1,Ο}. Let {nkβ:kβΟ} be an enumeration of
[TABLE]
in an ascending order. We find xβ[T] such that x(nkβ)>nk+1β for each kβ₯0. Let N>n0β and let us suppose that xβGNβ. Then there exists qβQ such that qβΎ(supp(q+K(q)+N))βx. If supp(q)<n0β, then x(n0β)=0, a contradiction. Let us assume that supp(q)β₯n0β then, and let
[TABLE]
Let us notice that m>0. nmβ1ββ€supp(q), hence K(q)β₯x(nmβ1β)>nmβ, which implies that x(nmβ)=0. A contradiction, thus the proof is complete.
β
Let us observe that each nonempty open set contains a body of uniformly perfect Miller tree, e.g. a basic clopen set.
3. Measure Case
This section is devoted to possible enhancements of two-dimensional Mycieski theorem for the measure case. Proposition 10 mirrors Proposition 2. It shows that we may exclude Miller trees from further considerations. Hence, in consecutive results we work in the Cantor space. The main theorem of this section (Theorem 11) shows that we can inscribe the square of a body of uniformly perfect tree into a set of measure one (modulo diagonal). Proposition 13 shows that it is not true in the case of Silver trees and Proposition 14 shows that no one-dimensional counterexample is feasible.
Proposition 10**.**
Let ΞΌ be a strictly positive probabilistic measure on ΟΟ. Then there exists an FΟβ set F of measure 1 such that [T]ξ βF for every Miller tree T.
Proof.
Let Ξ΅nβ=2n1β for every n>0. We will construct inductively a sequence (Fnβ:nβΟ) of closed subsets of ΟΟ. For every nβΟ let us set
[TABLE]
and for k>1 let
[TABLE]
Then we set Tnβ=βiβΟβTinβ and Fnβ=[Tnβ]. Finally, let F=βnβΟβFnβ. To see that F is the desired set, let us approximate its measure. For each nβΟ we have
[TABLE]
β
From now on we will work in 2Ο exclusively. By Ξ» we will denote standard product measure on 2Ο. We will use the same notation for standard product measure on 2ΟΓ2Ο.
Let F be a subset of 2ΟΓ2Ο of full measure. Then there exists a uniformly perfect tree Tβ2<Ο satisfying [T]Γ[T]βFβͺΞ.
Proof.
Let Fβ2ΟΓ2Ο be a set of full measure and let us assume that F=βnβΟβFnβ, where (Fnβ:nβΟ) is an ascending sequence of closed sets. Let us fix a sequence Ξ΅nβ=22n+31β, nβΟ.
We shall construct inductively:
β’
a collection of clopen sets {[ΟΟβ]:Οβ2<Ο};
β’
two sequences of natural numbers (knβ:nβΟ) and (Nnβ:nβΟ\{0});
β’
a sequence of pairs ((xnβ,ynβ):nβΟ\{0}) from 2ΟΓ2Ο;
β’
a collection of points {tΟβ:Οβ2<Ο} from 2Ο;
β’
a sequence (Bnβ:nβΟ) of subsets of 2ΟΓ2Ο;
satisfying the following conditions for all Ο,Ξ·β2<Ο and nβΟ:
Next, let (x1β,y1β)βB0ββ, with a requirement x1βξ =y1β, and set
[TABLE]
Then set
[TABLE]
and let tiββ2Ο be such that tiββΎN1β=Οiβ and tiβ(n)=0 for nβ₯N1β, iβ{0,1}. Also, set
[TABLE]
and
[TABLE]
Let us execute the step n+1, n>0. We pick (xn+1β,yn+1β)βBnββ, xn+1βξ =yn+1β, and set
[TABLE]
Then for every Οβ{0,1}n let
[TABLE]
and for iβ{0,1} let
[TABLE]
Let us set
[TABLE]
Finally, let us set
[TABLE]
Since
[TABLE]
we may carry on with the construction, thus it is complete. A set
[TABLE]
is the uniformly perfect tree we were looking for.
β
Now, let us recall the notion of small sets (see [1]) connected to null subsets of 2Ο.
Definition 12**.**
Aβ2Ο* is a small set if there is a partition A of Ο into finite sets and a collection (Jaβ)aβAβ such that Jaββ2a, βaβAβ2β£aβ£β£Jaββ£β<β and*
[TABLE]
Let us remark that each small set is a null set. Moreover, every null set is a union of two small sets (see [1]).
The space 2ΟΓ2Ο is canonically homeomorphic to 2Ο, so it is natural to consider a notion of small set in 2ΟΓ2Ο.
Let us assume that at the step n+1 we have a sequence (Οkβ:kβ€n) of finite [math]-1 sequences. Let 0kβ=(k0,0,...,0ββ) and for every sβ2n+1 let us denote
[TABLE]
and assume that a set
[TABLE]
has a positive measure. Let xn+1ββBnββ. Then
[TABLE]
for every sβ2n+1. Let us observe that for a given sequence Οβ2<Ο satisfying Ο0n+1βββ’Οβxn+1β it is also true that Οsββ’Οβxn+1β+tsβ for every sβ2n+1. Hence, we may pick Οn+1β such that for every sβ2n+1
[TABLE]
Similarly to the first step, we see that for every sβ2n+1
[TABLE]
and eventually
[TABLE]
This allows us to carry on with the construction, thus it is complete. Then
[TABLE]
is the desired Silver tree.
β
4. Nonstandard proofs
In this section we prove a result concerned with implications of adding a Cohen real. As a consequence we obtain a nonstandard proof of strengthened two-dimensional version of Mycielski Theorem (see [6]). We use Shoenfield Absoluteness Theorem. Using similar methods we prove a strengthened Egglestone Theorem (see [2]).
By canonical Polish spaces we understand countable products of ΟΟ,2Ο,[0,1],R and Perf(R) - a space of perfect subsets of R. We say that Ο is Ξ£21β-sentence if for some canonical Polish spaces X,Y and Borel set BβXΓYΟ is of the form:
[TABLE]
The Borel set B has its so called Borel code bβΟΟ (see [4]). The triple (X,Y,b) is a parameter of our Ξ£21β-sentence Ο.
Now, let us recall Schoenfield Absoluteness Theorem.
Theorem 15** (Schoenfield).**
Let MβN be standard transitive models of ZFC and Ο1NββM. Let Ο be a Ξ£21β-sentence with a parameter from model M. Then
[TABLE]
Let us observe that if N is a generic extension of standard transitive model M of ZFC then OrdM=OrdN and Ο1NββM.
A method of providing nonstandard proofs of mentioned theorems will be as follows. We start with a standard transitive model M of ZFC and find a generic extension N of M in which the theorem can be easily proved. Then we verify that the theorem forms a Ξ£21β-sentence. We apply Schoenfield Absoluteness Theorem to deduce that it is true in the ground universe M.
Before we proceed let us introduce some additional notation. For a tree TβΟ<Ο we define
[TABLE]
Let us recall that for a tree TβΟΟ and a node ΟβT we define
[TABLE]
We will denote a height of a given tree T by ht(T)=rankTβ(β ).
We say that a tree TβΟ<Ο is
β’
evenly cut if and there is nβΟ such that tips(q)βΟn and ht(q)=n;
β’
a slalom tree if
[TABLE]
Let observe that the definition of slalom trees is arithmetic and so it is absolute between transitive models of ZFC. We will say that a set PβΟΟ is slalom perfect if it is a body of a perfect slalom tree. Let us notice that for every slalom perfect set P and every ΟβΟ<Ο there is an interval IβΟ such that for every xβP we have xβΎI=Ο.
Theorem 16**.**
After adding one Cohen real
there is a perfect slalom tree T such that [T]Γ[T]βWβͺΞ for every dense GΞ΄β set WβΟΟΓΟΟ from the ground model.
Proof.
Let V be a ground model of ZFC. We will show that after adding one Cohen real to V there is a perfect tree T such that [T]Γ[T]βUβͺΞ for every open dense set UβΟΟΓΟw coded in V. Let us define a poset (C,β€) as follows:
[TABLE]
and for every p,qβC
[TABLE]
Clearly, (C,β€) is a forcing adding one Cohen real. Let GβC be any C-generic filter over V. In V[G] let us define a generic set
TGβ=βG. We have the following
Claim**.**
The following statements are true:
(1)
TGβ* is a slalom perfect tree.*
2. (2)
For any open dense set UβΟΟ in V and natural n a set
(1) follows from the density argument. That is, to see that TGβ is a perfect tree let us observe that for every pβC and every tβp the set
[TABLE]
is defined in V and it is dense below p. To prove that TGβ is a slalom tree it is enough to observe that for every sβΟ<Ο the following ground model set
The dense open set U from the ground model was chosen arbitrarily, hence [TGβ]Γ[TGβ]βWβͺΞ for any dense GΞ΄β planar set W from V[G].
β
Theorem 17**.**
For every GβGΞ΄β dense set in ΟΟΓΟΟ there exists a slalom perfect set PβΟΟ satisfying PΓPβGβͺΞ.
Proof.
Now let V be a transitive model of ZFC and WβV be a GΞ΄β dense set in ΟΟΓΟΟ. Let GβC-generic filter over V. Then by Theorem 16 there is a perfect tree TGβ in V[G] such that [TGβ]Γ[TGβ]βWβͺΞ. Here WβV, hence the formula
[TABLE]
is Ξ£21β-sentence with a parameter from V. By Shoenfield Absoluteness Theorem the above formula holds in V.
β
Our next result is concerned with a generalization of Egglestone Theorem. In [7] such a generalization was proved using Shoenfield Absoluteness Theorem.
We will give yet another proof of this result.
In [7] the author worked with a generic extension in which cof(I)=Ο1β<c, Iβ{M,N} (M and N denote ideals of meager and null sets respectively). Our proof is based on a generic extension in which Ο2β<add(I)β€c.
Let us recall that for ideals IβP(X),JβP(Y) we define a Fubini product IβJ of these ideals in the following way
[TABLE]
where Bor(XΓY) is a family of Borel subsets of XΓY and Bxβ={yβY:(x,y)βB} is a vertical section of the set B (similarly we define a horizontal section By).
We say that the pair (I,J) has a Fubini Property, if for every Borel set BβXΓY
[TABLE]
If (I,I) has a Fubini Property, then we will simply say that I has it. Let us notice that Kuratowski-Ulam Theorem and Fubini Theorem imply that M and N respectively possess the Fubini Property.
We work in Vβ² universe. Let Z={xβR:GxβcββI}. By the Fubini Property ZcβI. Then β£Zβ£=cβ₯Ο3β. Let us choose any set TβZ of cardinality Ο2β. Since Ο2β<add(I), the complement of a set βtβTβGtββ is in I. Let BβBor(R) such that BcβI, BββtβTβGtββ and consider a set
A={xβR:BβGxββ}.
Clearly, A is coanalytic. Since T has a size Ο2β and TβA, A contains a perfect subset P. It implies that Vβ² is a model for the following formula
[TABLE]
It is Ξ£21β, hence by Shoenfield Absoluteness Theorem it also holds in V.
β
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