Mycielski among trees

TL;DR

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

Existence of perfect sets within certain trees for comeager sets

Construction of uniformly perfect trees avoiding specific tree bodies

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 $X\subseteq [0,1]^2$ there exists a perfect set $P\subseteq [0,1]$ such that $P\times P\subseteq X\cup \Delta$. We consider generalizations of this theorem by replacing a perfect square with a rectangle $A\times B$, where $A$ and $B$ are bodies of other types of trees with $A\subseteq B$. In particular, we show that for every comeager $G_\delta$ set $G\subseteq \omega^\omega\times \omega^\omega$ there exist a Miller tree $M$ and a uniformly perfect tree $P\subseteq M$ such that $[P]\times [M]\subseteq G\cup\Delta$ and that $P$ cannot be a Miller tree. In the case of measure we show that for every subset $F$ of $2^{\omega}\times 2^\omega$ of full measure there exists a uniformly perfect tree $P\subseteq 2^{<\omega}$ such that $[P]\times[P]\subseteq F\cup\Delta$ 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.

Equations

$$(\forall \sigma \in T)(\exists \tau \in T)(\sigma \subseteq \tau \wedge (\exists i,j\in A)(i\ne j\wedge {\tau }^{⌢}i,{\tau }^{⌢}j\in T)).$$

$$Q=\{(p,q):p,q\in \mathbb{Q},\text{supp}(p)=\text{supp}(q)\text{ and }p\ne q\},$$

$$U=\bigcup _{k\in \omega }[q^k\upharpoonright (\text{supp}(q^k)+K(q^k))],$$

$$x$$

$$y$$

$$x_n$$

$$y_n$$

$$s_{n+1}=\min \{k\in {\text{succ}}_T(x_n):k>|y_n|\}$$

$$t_{n+1}=\min \{k\in {\text{succ}}_T(y_n):k>|x_{n+1}|\}$$

$$(x,y)\in ([T]\times [T])\backslash (U\cup \Delta ).$$

$$q_1\upharpoonright (\text{supp}(q)+K(q))$$

$$q_2\upharpoonright (\text{supp}(q)+K(q))$$

$$n=\max \{k\in \omega :{y_n}^{⌢}{t}_{n+1}\subseteq q_2\upharpoonright \text{supp}(q)\}.$$

$$|y_n|<|x_{n+1}|<|y_{n+1}|.$$

$$x\supseteq (x_{n+1}\upharpoonright \text{supp}(q){)}^{⌢}\underset{K(q)}{\underbrace{0^{⌢}..{.}^{⌢}0}},$$

$$y\supseteq (y_{n+1}\upharpoonright \text{supp}(q){)}^{⌢}\underset{K(q)}{\underbrace{0^{⌢}..{.}^{⌢}0}},$$

$$(\forall \sigma ,\tau \in T)(|\sigma |+1=|\tau |\wedge \sigma \subseteq \tau \wedge \sigma \in \text{split}(T)\Rightarrow \tau \notin \text{split}(T)).$$

$$n_k$$

$$s_k$$

$$B=\{x\in [T]:(\forall k\in \omega )(x(n_k)=s_k)\}$$

$$T^{\prime }=\{x\upharpoonright n:n\in \omega ,x\in B\}.$$

$$U=\bigcup_{n\in\omega}\big{[}\big{(}q^{n}_{1}\restriction\big{(}\textnormal{supp}(q^{n})\big{)}\big{)}^{\frown}(0,0)\big{]}\times\big{[}\big{(}q^{n}_{2}\restriction\big{(}\textnormal{supp}(q^{n})\big{)}\big{)}^{\frown}(1,1)\big{]}.$$

$$\displaystyle{\big{(}q_{1}\restriction\textnormal{supp}(q)\big{)}}^{\frown}(0,0)\subseteq x,$$

$$\displaystyle{\big{(}q_{2}\restriction\textnormal{supp}(q)\big{)}}^{\frown}(1,1)\subseteq y.$$

$$x(\text{supp}(q)+1)$$

$$x(\text{supp}(q)+2)$$

$${\sigma }_k={{\sigma }_k^{\prime }}^{⌢}\underset{N-|{\sigma }_k^{\prime }|}{\underbrace{0^{⌢}..{.}^{⌢}0}}$$

$$B_1=B_0\cup \{{\tau }_0,{\tau }_1\}\text{ and }{B}_2=B_1\cup \{{\tau }_{00},{\tau }_{01},{\tau }_{10},{\tau }_{11}\}.$$

$${\sigma }_1,{\sigma }_2\in \{{\sigma }^{⌢}n:\sigma \in n^{<n}\}\cup \{{\sigma }^{⌢}k:\sigma \in n^n,k<n+1\}.$$

$$M$$

$$P$$

$$(\forall n\in \omega )({\tau }_{\alpha \upharpoonright n}\subseteq y).$$

$$y\upharpoonright (|{\tau }_{a_0{a}_1...a_{n-1}}|+1)\subseteq {\tau }_{a_0{a}_1...a_{n-1}y(|{\tau }_{a_0{a}_1...a_{n-1}}|)}$$

$$\{n\in \omega :{\omega }^n\cap T\subseteq \text{split}(T)\}$$

Full text

The work has been partially financed by grant S50129/K1102 (0401/0052/18) from the Faculty of Fundamental Problems of Technology, Wrocław 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 $X\subseteq[0,1]^{2}$ there exists a perfect set $P\subseteq[0,1]$ such that $P\times P\subseteq X\cup\Delta$, where $\Delta=\{(x,x):x\in[0,1]\}.$

In the Cantor space $2^{\omega}$ and the Baire space $\omega^{\omega}$ each perfect set has a natural combinatorial description. Let $A\in\{2,\omega\}$ and denote $A^{<\omega}=\bigcup_{n\in\omega}A^{n}$. Let us recall that $T\subseteq A^{<\omega}$ is a tree on $A$ if for each $\sigma\in T$ and every $n\in\omega$ we have $\sigma\restriction n\in T$. A body of a tree $T\subseteq A^{<\omega}$ is the set ${[T]=\{x\in A^{\omega}:(\forall n)(x\restriction n\in T)\}}$ of infinite branches of $T$. A tree $T\subseteq A^{<\omega}$ is called a perfect tree (or a Sacks tree), if

[TABLE]

Then $P\subseteq A^{\omega}$ 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 $2^{\omega}\times 2^{\omega}$ or $\omega^{\omega}\times\omega^{\omega}$, of full measure or comeager, and investigate whether there exist trees $T_{1},T_{2}$ satisfying $T_{2}\subseteq T_{2}$ such that $[T_{1}]\times[T_{2}]\subseteq X\cup\Delta$, where $\Delta$ denotes a diagonal, i.e. ${\Delta=\{(x,x):x\in S\}}$ 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 $T\subseteq A^{<\omega}$ be a tree on a set $A\in\{2,\omega\}$. We will use the following notions related to trees:

• •

$\textnormal{succ}_{T}(\sigma)=\{a\in A:\sigma^{\frown}a\in T\}$;

• •

$\textnormal{split}(T)=\{\sigma\in T:|\textnormal{succ}_{T}(\sigma)|\geq 2\}$;

• •

$\textnormal{Succ}_{T}(\sigma)=\{\tau\in\textnormal{split}(T):\sigma\subseteq\tau\textnormal{ and }\neg(\exists\tau^{\prime}\in\textnormal{split}(T))(\sigma\subseteq\tau^{\prime}\subsetneq\tau)\}$;

• •

$\omega$-$\textnormal{split}(T)=\{\sigma\in T:|\textnormal{succ}_{T}(\sigma)|=\omega\}$.

A tree $T\subseteq A^{<\omega}$ is called

• •

a Miller or superperfect tree, if $(\forall\sigma\in T)(\exists\tau\in\omega$-$\textnormal{split}(T))(\sigma\subseteq\tau)$;

• •

a Laver tree, if $(\exists\sigma)(\forall\tau\in T)(\tau\subseteq\sigma\lor(\sigma\subseteq\tau\land\tau\in\omega$-$\textnormal{split}(T)))$.

We will denote the shortest splitting node of a given tree $T$ by $\textnormal{stem}(T)$. Nodes $\tau,\sigma\in A^{<\omega}$ are orthogonal (denoted by $\sigma\perp\tau$), if neither $\tau\subseteq\sigma$ nor $\sigma\subseteq\tau$. Sometimes we will be indexing nodes with nodes. In such cases for the sake of brevity we will write e.g. $\tau_{010}$ instead of $\tau_{(0,1,0)}$.

As mentioned above, we will also consider some specific types of perfect trees (see [5]). We call a perfect tree $T\subseteq A^{<\omega}$

• •

uniformly perfect, if for every $n\in\omega$ either $A^{n}\cap T\subseteq\textnormal{split}(T)$ or $A^{n}\cap\textnormal{split}(T)=\emptyset$;

• •

a Silver tree, if $(\forall\sigma,\tau\in T)(|\sigma|=|\tau|\Rightarrow(\forall a\in A)(\sigma^{\frown}a\in T\Leftrightarrow\tau^{\frown}a\in T))$.

Before we proceed let us notice that to provide an example of a comeager subset of $X^{2}$ which does not contain a rectangle $A\times B$ of sets of certain type, it is enough to show that there exists comeager set $G\subseteq X$ with $A\not\subseteq G$ or $B\not\subseteq G$. Indeed, in such a case $G\times X$ is comeager too (by Kuratowski-Ulam Theorem) and $A\times\{x\}\not\subseteq G\times X$ for every $x\in X$. 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 $T_{1}\subseteq T_{2}\subseteq\omega^{<\omega}$ of types mentioned in Introduction, satisfying $[T_{1}]\times[T_{2}]\subseteq G$ for a given comeager set $G\subseteq\omega^{\omega}\times\omega^{\omega}$. The main positive result is Theorem 8. Theorem 3 and Propositions 5 and 9 show that the main result is somehow optimal.

Let $\mathbb{Q}=\{q\in\omega^{\omega}:(\forall^{\infty}n)(q(n)=0)\}$ be a set of rationals localized in $\omega^{\omega}$. On several occasions in this section we will use some specific countable dense subset of $\omega^{\omega}\times\omega^{\omega}$. Let us define it in the following way:

[TABLE]

where $\textnormal{supp}(q)=\max\{n\in\omega:q(n)\neq 0\}+1$. Since $\textnormal{supp}(q_{1})=\textnormal{supp}(q_{2})$ for every $q=(q_{1},q_{2})\in Q$, we may naturally extend the domain of supp to $\mathbb{Q}\cup Q$ so that $\textnormal{supp}(q)=\textnormal{supp}(q_{1})$.

As a warm up let us consider a case of Laver trees.

Proposition 2**.**

There exists a dense $G_{\delta}$ set $G\subseteq\omega^{\omega}$ such that $[L]\not\subseteq G$ for every Laver tree $L.$

Proof.

Let $G=\{x\in\omega^{\omega}:(\exists^{\infty}n\in\omega)(x(n)=0)\}$. Clearly, $G$ is $G_{\delta}$ and dense. Let $L$ be a Laver tree. Let $x\in[L]$ such that $x(n)\neq 0$ for every $n\geq|\textnormal{stem}(L)|$. Then $x\in[L]\backslash G$. ∎

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 $U\subseteq\omega^{\omega}\times\omega^{\omega}$ such that $[T]\times[T]\not\subseteq U\cup\Delta$ for every Miller tree $T$.

Proof.

Let $\{q^{n}:n\in\omega\}$ be an enumeration of $Q$ and let us set

[TABLE]

where $K(q)=\max\{q_{1}(n),q_{2}(n):n\in\omega\}$ for $q=(q_{1},q_{2})\in Q$.

Let $T$ be a Miller tree. Without loss of generality we may assume that for every $\sigma\in T$ either $|\textnormal{succ}_{T}(\sigma)|=1$ or $|\textnormal{succ}_{T}(\sigma)|=\omega$. We will pick points

[TABLE]

from $[T]$ via induction. Let $\sigma_{0}=\tau_{0}=\textnormal{stem}(T)$. Let us assume the following notation

[TABLE]

Let us execute the step $n+1$. We set

[TABLE]

and $\sigma_{n+1}\supseteq s_{n+1}$ such that ${x_{n}}^{\frown}\sigma_{n+1}\in\textnormal{split}(T)$ and $|{x_{n}}^{\frown}\sigma_{n+1}|>|y_{n}|$. In a similar fashion we proceed with $y_{n+1}$. We set

[TABLE]

and $\tau_{n+1}\supseteq t_{n+1}$ such that ${y_{n}}^{\frown}\tau_{n+1}\in\textnormal{split}(T)$ and $|{y_{n}}^{\frown}\tau_{n+1}|>|x_{n+1}|$.

We will show that

[TABLE]

It is clear that $(x,y)\in([T]\times[T])\backslash\Delta$. Let us suppose that $(x,y)\in U$. Then there is $q\in Q$ such that $(x,y)\in[q\restriction(\textnormal{supp}(q)+K(q))]$. It follows that

[TABLE]

Let us observe that since $q_{1}\neq q_{2}$, $|\textnormal{stem}(T)|<\textnormal{supp}(q)$. Let

[TABLE]

In particular it means that $K(q)\geq t_{n+1}$ and $q_{2}\restriction\textnormal{supp}(q)\subsetneq{y_{n+1}}^{\frown}t_{n+2}$. Let us also observe that

[TABLE]

It is the case that exactly one of the following holds:

1. (1)

$q_{1}\restriction\textnormal{supp}(q)\subsetneq{x_{n+1}}^{\frown}s_{n+2}$; 2. (2)

${x_{n+1}}^{\frown}s_{n+2}\subseteq q_{1}\restriction\textnormal{supp}(q)$.

If $(1)$ is true, then

[TABLE]

which gives a contradiction, since $K(q)\geq t_{n+1}>|x_{n+1}|$ and $s_{n+2}\neq 0$.

If $(2)$ holds, then $K(q)\geq s_{n+2}$ and

[TABLE]

which is a contradiction because $K(q)\geq s_{n+2}>|y_{n+1}|$ and $t_{n+2}\neq 0$.

Therefore $(x,y)\notin U$. ∎

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 $T^{\prime}\subseteq T$ that splits and rests.

Proof.

Let $n_{0}=\min\{|\sigma|:\sigma\in\textnormal{split}(T)\}$ and $s_{0}=\min\{n\in\omega:{\sigma_{0}}^{\frown}n\in T\}$, where $\sigma_{0}\in T$ and $|\sigma_{0}|=n_{0}$. For $k>0$ let

[TABLE]

where $\sigma_{k}\in T$ satisfies $|\sigma_{k}|=n_{k}$. Now, let

[TABLE]

and set

[TABLE]

Then $T^{\prime}$ is the desired tree. ∎

Proposition 5**.**

There exists an open dense set $U\subseteq\omega^{\omega}\times\omega^{\omega}$ such that $[T]\times[T]\not\subseteq U\cup\Delta$ for any Silver tree $T$.

Proof.

Let $Q=\{q^{n}:n\in\omega\}$ and set

[TABLE]

Let $T$ be a Silver tree. Without loss of generality we may assume that $T$ splits and rests (Lemma 4). Let $(x,y)\in[T]\times[T]$, $x\neq y$, and suppose that $(x,y)\in U$. Then there is $q=(q_{1},q_{2})\in Q$ such that

[TABLE]

Clearly

[TABLE]

hence all of the nodes in $T$ of lengths supp and $\textnormal{supp}+1$ are splitting, which constitutes a contradiction with the splitting and resting property of $T$. ∎

The following lemmas are preparation to the main theorem of this section.

Lemma 6**.**

For every open dense set $U\subseteq\omega^{\omega}\times\omega^{\omega}$ and two open sets $V_{1},V_{2}\subseteq\omega^{\omega}$ there are sequences $\sigma_{1},\sigma_{2}\in\omega^{<\omega}$ satisfying $[\sigma_{1}]\subseteq V_{1}$, $[\sigma_{2}]\subseteq V_{2}$, $|\sigma_{1}|=|\sigma_{2}|$ such that $[\sigma_{1}]\times[\sigma_{2}]\subseteq U$ and $[\sigma_{2}]\times[\sigma_{1}]\subseteq U$.

Proof.

Let $U$, $V_{1}$ and $V_{2}$ be as in the formulation. $(V_{1}\times V_{2})\cap U$ is open and nonempty, therefore there are sequences $\tau_{1},\tau_{2}$ with $[\tau_{1}]\times[\tau_{2}]\subseteq(V_{1}\times V_{2})\cap U$. Repeating the argument, we find sequences $\tau_{1}^{\prime}\supseteq\tau_{1}$ and $\tau_{2}^{\prime}\supseteq\tau_{2}$ satisfying $[\tau_{2}^{\prime}]\times[\tau_{1}^{\prime}]\subseteq U$. We may assume that $|\tau_{1}^{\prime}|=|\tau_{2}^{\prime}|$ (otherwise we extend the shorter one however we like). We set $\sigma_{1}=\tau_{1}^{\prime}$ and $\sigma_{2}=\tau_{2}^{\prime}$. ∎

For $\sigma,\tau\in\omega^{<\omega}$ and $U\subseteq\omega^{\omega}\times\omega^{\omega}$ let us denote the fact that $[\sigma]\times[\tau]\subseteq U$ and $[\tau]\times[\sigma]\subseteq U$ by $\psi(\sigma,\tau,U)$. The following lemma is an extension of the previous one.

Lemma 7**.**

For every open dense set $U\subseteq\omega^{\omega}\times\omega^{\omega}$, a finite sequence of open sets $(V_{k}:0\leq k<n)$ in $\omega^{\omega}$ there is a sequence of sequences $(\sigma_{k}:0\leq k<n)$ such that:

1. (1)

$[\sigma_{k}]\subseteq V_{k}$* for all $0\leq k<n$,* 2. (2)

$|\sigma_{k}|=|\sigma_{l}|$* for all $0\leq k,l<n$,* 3. (3)

$\psi(\sigma^{k}_{l},\sigma^{l}_{k},U)$* for all distinct $0\leq k,l<n$.*

Proof.

Let $U$ and $(V_{k}:k<n)$ be as in the formulation. Applying Lemma 6 multiple times we will construct inductively a sequence $(\sigma_{l}^{k}:k,l<n;\,k\neq l)$ of sequences satisfying:

1. (1)

$(\forall k,l<n)(l\neq k\Rightarrow[\sigma^{k}_{l}]\subseteq V_{k})$; 2. (2)

$(\forall k)([\sigma^{k}_{0}]\supseteq[\sigma^{k}_{1}]\supseteq...\supseteq[\sigma^{k}_{k-1}]\supseteq[\sigma^{k}_{k+1}]\supseteq...\supseteq[\sigma^{k}_{n-1}])$; 3. (3)

$\psi(\sigma^{k}_{l},\sigma^{l}_{k},U)$ holds for all distinct $k,l<n$.

At the step [math] first we find $\sigma^{0}_{1}$ and $\sigma^{1}_{0}$ such that $[\sigma^{0}_{1}]\subseteq V_{0}$, $[\sigma^{1}_{0}]\subseteq V_{1}$ and $\psi(\sigma^{0}_{1},\sigma^{1}_{0},U)$. Then for every $k<n,k\neq 0,1,$ we choose $\sigma^{0}_{k}$ and $\sigma^{k}_{0}$ satisfying $[\sigma^{0}_{k}]\subseteq[\sigma^{0}_{k-1}],[\sigma^{k}_{0}]\subseteq V_{k}$, and $\psi(\sigma^{0}_{k},\sigma^{k}_{0},U)$.

Let us execute the step $k$, $0<k<n$. We pick $\sigma^{k}_{k+1}$ and $\sigma^{k+1}_{k}$ satisfying $[\sigma^{k}_{k+1}]\subseteq[\sigma^{k}_{k-1}]$, $[\sigma^{k+1}_{k}]\subseteq[\sigma^{k+1}_{k-1}]$ and $\psi(\sigma^{k}_{k+1},\sigma^{k+1}_{k},U)$. For $l>k+1$ we find $\sigma^{k}_{l}$ and $\sigma^{l}_{k}$ such that $[\sigma^{k}_{l}]\subseteq[\sigma^{k}_{l-1}]$, $[\sigma^{l}_{k}]\subseteq[\sigma^{l}_{k-1}]$ and $\psi(\sigma^{k}_{l}\sigma^{l}_{k},U)$. The construction is completed.

Let us set $\sigma_{k}^{\prime}=\sigma^{k}_{n-1}$ for every $k<n$. If lengths of $\sigma_{k}^{\prime},0\leq k<n$ are the same, then we set $\sigma_{k}=\sigma_{k}^{\prime}$ for each $0\leq k<n$. If not, let $N=\max\{|\sigma_{k}^{\prime}|:0\leq k<n\}$, and let us set

[TABLE]

for each $0\leq k<n$. Then lengths of these sequences match and properties established during the construction are not compromised. ∎

Theorem 8**.**

For every comeager set $G$ of $\omega^{\omega}\times\omega^{\omega}$ there exists a Miller tree $M\subseteq\omega^{<\omega}$ and a uniformly perfect tree $P\subseteq M$ such that $[P]\times[M]\subseteq G\cup\Delta.$

Proof.

Let us assume that $G=\bigcap_{n\in\omega}U_{n}$ where $(U_{n})_{n\in\omega}$ is a descending sequence of open dense subsets of $\omega^{\omega}\times\omega^{\omega}$. We will construct recursively a sequence $(B_{n}:n\in\omega)$ of sets. $B_{n}=\{\tau_{\sigma}:\sigma\in n^{\leq n}\}$ should consist of nodes satisfying:

1. (1)

$\tau_{\emptyset}=\emptyset$, $\tau_{\sigma_{1}}\subseteq\tau_{\sigma_{2}}$ for $\sigma_{1}\subseteq\sigma_{2}$ and ${\tau_{\sigma}}^{\frown}k\subseteq\tau_{\sigma^{\frown}k}$; 2. (2)

$\tau_{\sigma^{\frown}k}\cap\tau_{\sigma^{\frown}j}=\tau_{\sigma}$ for $\sigma\in n^{<n}$ and distinct $k,j<n$; 3. (3)

for $n>0$ and all $\tau,\tau^{\prime}\in B_{n}\backslash B_{n-1}$ $\psi(\tau,\tau^{\prime},U_{n})$ holds; 4. (4)

If $\sigma_{1},\sigma_{2}\in\{0,1\}^{\leq n}$ then $|\tau_{\sigma_{1}}|=|\tau_{\sigma_{2}}|$.

At the step [math] we set $\tau_{\emptyset}=\emptyset$ and $B_{0}=\{\tau_{\emptyset}\}$. Next, we set $\tau_{0},\tau_{1}\supseteq\tau_{\emptyset}$ so that $\psi(\tau_{0},\tau_{1},U_{2})$ (Lemma 6), and $\tau_{00},\tau_{01}\supseteq\tau_{0},\tau_{10},\tau_{11}\supseteq\tau_{1}$ with accordance to Lemma 7. We set

[TABLE]

Now, let us assume that we already have a set $B_{n}$ with the above properties and let us execute the step $n+1$, $n>1$. First we set $\tau_{\sigma^{\frown}n}$ for $\sigma\in n^{<n}$ and $\tau_{\sigma^{\frown}k}$, $\sigma\in n^{n}$, $k<n+1$, in such a way that they have the same lengths, propagate the condition $(1)$ and $(2)$, and $\psi(\tau_{\sigma_{1}},\tau_{\sigma_{2}},U_{n+1})$ for all distinct

[TABLE]

Next, we set $\tau_{\sigma^{\frown}k}$ for $\sigma\in(n+1)^{<n+1}\backslash n^{\leq n}$ and $k<n+1$ in a similar fashion.

This completes the construction. Let us set $B=\bigcup_{n\in\omega}B_{n}$ and

[TABLE]

Clearly, $M$ is a Miller tree. Furthermore, $P\subseteq M$ is a uniformly perfect tree thanks to the condition $(4)$. We will show that $[P]\times[M]\subseteq G\cup\Delta$. Let $(x,y)\in[P]\times[T]$, $x\neq y$. We claim that there exists $\alpha\in\omega^{\omega}$ such that

[TABLE]

We will define $\alpha=(a_{0},a_{1},...)$ via induction. Let us observe that $y\restriction 1\subseteq\tau_{y(0)}$ and $\tau_{y(0)}$ is the shortest sequence from $B$ possessing such a property. Therefore, $y\restriction|\tau_{y(0)}|=\tau_{y(0)}$, otherwise there would be $\tau\in B$ such that $\tau_{y(0)}\subsetneq\tau$ and $y\restriction|\tau_{y(0)}|\subseteq\tau$, which is a contradiction. We set $a_{0}=y(0)$.

Next, let us assume that we already have a strictly ascending sequence $(a_{k}:k<n)$ of natural numbers with a property $\tau_{a_{0}a_{1}...a_{n-1}}\subseteq y$ for every $k<n$. As previously, we see that

[TABLE]

and that $\tau_{a_{0}a_{1}...a_{n-1}y(|\tau_{a_{0}a_{1}...a_{n-1}}|)}$ is the shortest sequence from $B$ with such a property. Hence $\tau_{a_{0}a_{1}...a_{n-1}y(|\tau_{a_{0}a_{1}...a_{n-1}}|)}\subseteq y$, so we set $a_{n}=y(|\tau_{a_{0}a_{1}...a_{n-1}}|)$. This completes the definition of $\alpha$.

Now, let us fix $N\in\omega$. There exists $N^{\prime}\geq N$ such that $\tau_{\alpha\restriction N}\in B_{N^{\prime}}\backslash B_{N^{\prime}-1}$. Furthermore, there exists $\sigma\in 2^{N^{\prime}}$ such that $\tau_{\sigma}\subseteq x$. Then $[\tau_{\sigma}]\times[\tau_{\alpha\restriction N}]\subseteq U_{N^{\prime}}\subseteq U_{N}$, hence also $(x,y)\in U_{N}$. $N$ was chosen arbitrarily, thus $(x,y)\in G$. ∎

Let us make some remarks. The Miller tree $T$ in the above theorem has a nice property. For each $\tau\in T$ the set $\textnormal{succ}_{T}(\sigma)=\omega$ or $|\textnormal{succ}_{T}(\sigma)|=1$. Let us also observe that one cannot make this Miller tree uniformly perfect.

Proposition 9**.**

There exists a $G_{\delta}$ set $G$ such that $[T]\not\subseteq G$ for every uniformly perfect Miller tree.

Proof.

For every $n\in\omega$ let $G_{n}=\bigcup_{q\in\mathbb{Q}}[q\restriction(\textnormal{supp}(q)+K(q)+n))]$. Let $T$ be a uniformly perfect Miller tree. Without loss of generality we may assume that for every $\sigma\in T$ we have $|\textnormal{succ}_{T}(\sigma)|\in\{1,\omega\}$. Let $\{n_{k}:k\in\omega\}$ be an enumeration of

[TABLE]

in an ascending order. We find $x\in[T]$ such that $x(n_{k})>n_{k+1}$ for each $k\geq 0$. Let $N>n_{0}$ and let us suppose that $x\in G_{N}$. Then there exists $q\in\mathbb{Q}$ such that $q\restriction(\textnormal{supp}(q+K(q)+N))\subseteq x$. If $\textnormal{supp}(q)<n_{0}$, then $x(n_{0})=0$, a contradiction. Let us assume that $\textnormal{supp}(q)\geq n_{0}$ then, and let

[TABLE]

Let us notice that $m>0$. $n_{m-1}\leq\textnormal{supp}(q)$, hence $K(q)\geq x(n_{m-1})>n_{m}$, which implies that $x(n_{m})=0$. A contradiction, thus the proof is complete. ∎

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 $\mu$ be a strictly positive probabilistic measure on $\omega^{\omega}$. Then there exists an $F_{\sigma}$ set $F$ of measure $1$ such that $[T]\not\subseteq F$ for every Miller tree $T$.

Proof.

Let $\varepsilon_{n}=\frac{1}{2^{n}}$ for every $n>0$. We will construct inductively a sequence $(F_{n}:n\in\omega)$ of closed subsets of $\omega^{\omega}$. For every $n\in\omega$ let us set

[TABLE]

and for $k>1$ let

[TABLE]

Then we set $T_{n}=\bigcup_{i\in\omega}T^{n}_{i}$ and $F_{n}=[T_{n}]$. Finally, let $F=\bigcup_{n\in\omega}F_{n}$. To see that $F$ is the desired set, let us approximate its measure. For each $n\in\omega$ we have

[TABLE]

∎

From now on we will work in $2^{\omega}$ exclusively. By $\lambda$ we will denote standard product measure on $2^{\omega}$. We will use the same notation for standard product measure on $2^{\omega}\times 2^{\omega}$.

Let $\sigma\in 2^{<\omega}$. For a given set $A\subseteq[\sigma]^{2}$ let us denote $A^{s}=A\cap A^{-1}$, where $A^{-1}=\{(x,y):(y,x)\in A\}$. Let us observe that if $\lambda(A)=(1-\varepsilon)\lambda([\sigma])^{2}$, then $\lambda(A^{s})=(1-2\varepsilon)\lambda([\sigma])^{2}$. For every set $B$ we will denote a set of its density points by $B^{*}$.

Theorem 11**.**

Let $F$ be a subset of $2^{\omega}\times 2^{\omega}$ of full measure. Then there exists a uniformly perfect tree $T\subseteq 2^{<\omega}$ satisfying $[T]\times[T]\subseteq F\cup\Delta$.

Proof.

Let $F\subseteq 2^{\omega}\times 2^{\omega}$ be a set of full measure and let us assume that $F=\bigcup_{n\in\omega}F_{n}$, where $(F_{n}:n\in\omega)$ is an ascending sequence of closed sets. Let us fix a sequence $\varepsilon_{n}=\frac{1}{2^{2n+3}}$, $n\in\omega$. We shall construct inductively:

• •

a collection of clopen sets $\{[\tau_{\sigma}]:\sigma\in 2^{<\omega}\}$;

• •

two sequences of natural numbers $(k_{n}:n\in\omega)$ and $(N_{n}:n\in\omega\backslash\{0\})$;

• •

a sequence of pairs $((x_{n},y_{n}):n\in\omega\backslash\{0\})$ from $2^{\omega}\times 2^{\omega}$;

• •

a collection of points $\{t_{\sigma}:\sigma\in 2^{<\omega}\}$ from $2^{\omega}$;

• •

a sequence $(B_{n}:n\in\omega)$ of subsets of $2^{\omega}\times 2^{\omega}$;

satisfying the following conditions for all $\sigma,\eta\in 2^{<\omega}$ and $n\in\omega$:

1. (1)

$\tau_{\sigma}\subseteq\tau_{\eta}\Leftrightarrow\sigma\subseteq\eta$; 2. (2)

$|\sigma|=|\eta|\Rightarrow[\tau_{\sigma}]\times[\tau_{\eta}]\cap F_{k_{|\sigma\cap\eta|}}\neq\emptyset$; 3. (3)

$|\sigma|=|\eta|\Rightarrow|\tau_{\sigma}|=|\tau_{\eta}|;$ 4. (4)

The set

[TABLE]

has a positive measure.

Let $\tau_{\emptyset}=\emptyset,t_{\emptyset}=0^{\frown}0^{\frown}...,$ and set

[TABLE]

Next, let $(x_{1},y_{1})\in B_{0}^{*}$, with a requirement $x_{1}\neq y_{1}$, and set

[TABLE]

Then set

[TABLE]

and let $t_{i}\in 2^{\omega}$ be such that $t_{i}\restriction N_{1}=\tau_{i}$ and $t_{i}(n)=0$ for $n\geq N_{1}$, $i\in\{0,1\}$. Also, set

[TABLE]

and

[TABLE]

Let us execute the step $n+1$, $n>0$. We pick $(x_{n+1},y_{n+1})\in B_{n}^{*}$, $x_{n+1}\neq y_{n+1}$, and set

[TABLE]

Then for every $\sigma\in\{0,1\}^{n}$ let

[TABLE]

and for $i\in\{0,1\}$ 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. ∎

Now, let us recall the notion of small sets (see [1]) connected to null subsets of $2^{\omega}$.

Definition 12**.**

$A\subseteq 2^{\omega}$* is a small set if there is a partition $\mathcal{A}$ of $\omega$ into finite sets and a collection $(J_{a})_{a\in\mathcal{A}}$ such that $J_{a}\subseteq 2^{a}$, $\sum_{a\in\mathcal{A}}\frac{|J_{a}|}{2^{|a|}}<\infty$ 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 $2^{\omega}\times 2^{\omega}$ is canonically homeomorphic to $2^{\omega}$, so it is natural to consider a notion of small set in $2^{\omega}\times 2^{\omega}$.

Proposition 13**.**

There exist a small set $A\subseteq 2^{\omega}\times 2^{\omega}$ such that $(A\cap[S]\times[S])\setminus\Delta\neq\emptyset$ for any Silver tree $S\subseteq 2^{<\omega}.$

Proof.

Let $\{I_{n}\}_{n\in\omega}$ be a partition of $\omega$ into finite segments such that $|I_{n}|\geq n$.

Clearly, $\{I_{n}\times I_{m}\}_{n,m\in\omega}$ forms a partition of $\omega\times\omega$. Define

[TABLE]

Notice that $\displaystyle\frac{|J_{n,n}|}{2^{|I_{n}\times I_{n}|}}=\frac{1}{2^{|I_{n}|}}\leq\frac{1}{2^{n}}.$ So

[TABLE]

is a small set. Let $S$ be a Silver tree. Let $x,y\in[S]$ be such that $(\forall^{\infty}k)(x(k)=y(k))$, but $x\neq y$. Clearly, $(x,y)\in A\setminus\Delta$. ∎

Proposition 14**.**

Every closed subset of $2^{\omega}$ of positive Lebesgue measure contains a Silver tree.

Proof.

Let $F\subseteq 2^{\omega}$ be a closed set of positive measure. Let $\varepsilon_{n}=\frac{1}{2^{n+3}}$ for every $n\in\omega$. Let $x_{0}$ be a density point of $F$ and let $\sigma_{0}\subseteq x_{0}$ such that $\lambda([\sigma_{0}]\cap F)>(1-\varepsilon_{0})\lambda([\sigma_{0}])$. Since

[TABLE]

we have

[TABLE]

Let $t_{1}=(\underbrace{0,0,...,0}_{|\sigma_{0}|},1,0,...)$ and let us observe that $([{\sigma_{0}}^{\frown}1]\cap F)-t_{1}\subseteq[{\sigma_{0}}^{\frown}0]$. Since

[TABLE]

for $i\in\{0,1\}$, we have

[TABLE]

Let us assume that at the step $n+1$ we have a sequence $(\sigma_{k}:k\leq n)$ of finite [math]-$1$ sequences. Let $0_{k}=(\underbrace{0,0,...,0}_{k})$ and for every $s\in 2^{n+1}$ let us denote

[TABLE]

and assume that a set

[TABLE]

has a positive measure. Let $x_{n+1}\in B_{n}^{*}$. Then

[TABLE]

for every $s\in 2^{n+1}$. Let us observe that for a given sequence $\sigma\in 2^{<\omega}$ satisfying ${\tau_{0_{n+1}}}^{\frown}\sigma\subseteq x_{n+1}$ it is also true that ${\tau_{s}}^{\frown}\sigma\subseteq x_{n+1}+t_{s}$ for every $s\in 2^{n+1}$. Hence, we may pick $\sigma_{n+1}$ such that for every $s\in 2^{n+1}$

[TABLE]

Similarly to the first step, we see that for every $s\in 2^{n+1}$

[TABLE]

and eventually

[TABLE]

This allows us to carry on with the construction, thus it is complete. Then

[TABLE]

is the desired Silver tree. ∎

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 $\omega^{\omega},2^{\omega},[0,1],\mathbb{R}$ and $\textnormal{Perf}(\mathbb{R})$ - a space of perfect subsets of $\mathbb{R}$. We say that $\varphi$ is $\Sigma_{2}^{1}$-sentence if for some canonical Polish spaces $X,Y$ and Borel set $B\subseteq X\times Y$ $\varphi$ is of the form:

[TABLE]

The Borel set $B$ has its so called Borel code $b\in\omega^{\omega}$ (see [4]). The triple $(X,Y,b)$ is a parameter of our $\Sigma^{1}_{2}$-sentence $\varphi$. Now, let us recall Schoenfield Absoluteness Theorem.

Theorem 15** (Schoenfield).**

Let $M\subseteq N$ be standard transitive models of ZFC and ${\omega_{1}^{N}\subseteq M.}$ Let $\varphi$ be a $\Sigma_{2}^{1}$-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 $Ord^{M}=Ord^{N}$ and $\omega_{1}^{N}\subseteq M.$

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 $\Sigma_{2}^{1}$-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 $T\subseteq\omega^{<\omega}$ we define

[TABLE]

Let us recall that for a tree $T\subseteq\omega^{\omega}$ and a node $\sigma\in T$ we define

[TABLE]

We will denote a height of a given tree $T$ by $\textnormal{ht}(T)=\textnormal{rank}_{T}(\emptyset)$.

We say that a tree $T\subseteq\omega^{<\omega}$ is

• •

evenly cut if and there is $n\in\omega$ such that $\textnormal{tips}(q)\subseteq\omega^{n}$ and $\textnormal{ht}(q)=n$;

• •

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 $P\subseteq\omega^{\omega}$ 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 $\sigma\in\omega^{<\omega}$ there is an interval $I\subseteq\omega$ such that for every $x\in P$ we have $x\upharpoonright I=\sigma.$

Theorem 16**.**

After adding one Cohen real there is a perfect slalom tree $T$ such that ${[T]\times[T]\subseteq W\cup\Delta}$ for every dense $G_{\delta}$ set $W\subseteq\omega^{\omega}\times\omega^{\omega}$ 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 ${[T]\times[T]\subseteq U\cup\Delta}$ for every open dense set $U\subseteq\omega^{\omega}\times\omega^{w}$ coded in $V$. Let us define a poset $(\mathcal{C},\leq)$ as follows:

[TABLE]

and for every $p,q\in\mathcal{C}$

[TABLE]

Clearly, $(\mathcal{C},\leq)$ is a forcing adding one Cohen real. Let $G\subseteq\mathcal{C}$ be any $\mathcal{C}$-generic filter over $V.$ In $V[G]$ let us define a generic set ${T_{G}=\bigcup G}$. We have the following

Claim**.**

The following statements are true:

1. (1)

$T_{G}$* is a slalom perfect tree.* 2. (2)

For any open dense set $U\subseteq\omega^{\omega}$ in $V$ and natural $n$ a set

[TABLE]

is dense in $(\mathcal{C},\leq)$. 3. (3)

Fix a name $\dot{x}\in V^{\mathcal{C}}$ and $p,q\in G$. Assume that

[TABLE]

Then there exists $r\in G$ and $m\geq n$ such that $r\leq p,q$ and $r\Vdash\dot{x}\upharpoonright m\in\textnormal{tips}(q).$

Proof of the Claim.

(1) follows from the density argument. That is, to see that $T_{G}$ is a perfect tree let us observe that for every $p\in\mathcal{C}$ and every $t\in p$ the set

[TABLE]

is defined in $V$ and it is dense below $p.$ To prove that $T_{G}$ is a slalom tree it is enough to observe that for every $s\in\omega^{<\omega}$ the following ground model set

[TABLE]

is dense in $\mathcal{C}$.

(2) follows directly from Lemma 7.

To show (3) let $n\in\omega$, $p,q\in G$ and $\dot{x}\in V^{\mathcal{C}}$ such that ${p\Vdash\dot{x}\in\dot{[T_{G}]}}$ and ${p\Vdash\dot{x}\upharpoonright n\in q.}$ Let $m_{0}=\textnormal{ht}(q)$. Then there exists $r^{\prime},q^{\prime}\in G$ such that ${r^{\prime}\Vdash\dot{x}\upharpoonright m_{0}\in q^{\prime}.}$ $G$ is a filter, hence there exists $r\in G$ such that $r\leq q,q^{\prime},r^{\prime},p$ and $r\Vdash\dot{x}\upharpoonright{m_{0}}\in q^{\prime}$, so ${r\Vdash\dot{x}\upharpoonright{m_{0}}\in r}$. Let us observe that $r\leq q$ and $\dot{x}_{G}\upharpoonright m_{0}\notin q.$ Then there is $t\in\textnormal{tips}(q)$ such that ${r\Vdash t\subseteq\dot{x}_{G}\upharpoonright m_{0}\subseteq\dot{x}_{G}.}$ ∎

Now let $\dot{x},\dot{y}\in V^{\mathcal{C}}$ and $p\in G$, $k\in\omega$ be such that ${p\Vdash\dot{x},\dot{y}\in\dot{[T_{G}]}}$ and ${\dot{x}\upharpoonright{k}\neq\dot{y}\upharpoonright{k}.}$ Then there are $p_{1},q_{x},q_{y}\in G$ such that $p_{1}\leq p$ and for some $n_{x},n_{y}>k$ we have ${p_{1}\Vdash\dot{x}\restriction n_{x}\in q_{x}\land\dot{y}\restriction n_{y}\in q_{x}}$. Since $G$ is a filter in $(\mathcal{C},\leq)$, there exist $p^{\prime},q\in G$ such that $p^{\prime}\leq p_{1},$ $q\leq q_{x},q_{y}$ and $p^{\prime}\Vdash\dot{x}\restriction n_{x},\dot{y}\restriction n_{y}\in q.$ By (1) of Claim there is a condition $q^{\prime}\in G$ such that $q^{\prime}\leq q$ and for every $t,s\in\textnormal{tips}(q^{\prime})$ if $t\neq s$ then $[t]\times[s]\subseteq U.$ By (2) of Claim we can find a generic condition $r\in G$ such that $r\leq p^{\prime}$ and there are $s,t\in\textnormal{tips}(q^{\prime})$ such that $r\Vdash\dot{x}\restriction m_{x},\dot{y}\restriction=s\land m_{y}=t$ for some $m_{x}\geq n_{x}\geq k$ and $m_{y}\geq n_{y}\geq k.$ Then for the $r\in G$ we have

[TABLE]

The dense open set $U$ from the ground model was chosen arbitrarily, hence $[T_{G}]\times[T_{G}]\subseteq W\cup\Delta$ for any dense $G_{\delta}$ planar set $W$ from $V[G]$. ∎

Theorem 17**.**

For every $G\in G_{\delta}$ dense set in $\omega^{\omega}\times\omega^{\omega}$ there exists a slalom perfect set $P\subseteq\omega^{\omega}$ satisfying $P\times P\subseteq G\cup\Delta.$

Proof.

Now let $V$ be a transitive model of ZFC and $W\in V$ be a $G_{\delta}$ dense set in $\omega^{\omega}\times\omega^{\omega}.$ Let $G\subseteq\mathcal{C}$-generic filter over $V.$ Then by Theorem 16 there is a perfect tree $T_{G}$ in $V[G]$ such that $[T_{G}]\times[T_{G}]\subseteq W\cup\Delta.$ Here $W\in V$, hence the formula

[TABLE]

is $\Sigma^{1}_{2}$-sentence with a parameter from $V$. By Shoenfield Absoluteness Theorem the above formula holds in $V.$ ∎

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 $\rm cof(\mathcal{I})=\omega_{1}<\mathfrak{c}$, $\mathcal{I}\in\{\mathcal{M},\mathcal{N}\}$ ($\mathcal{M}$ and $\mathcal{N}$ denote ideals of meager and null sets respectively). Our proof is based on a generic extension in which $\omega_{2}<\textnormal{add}(I)\leq\mathfrak{c}.$

Let us recall that for ideals $\mathcal{I}\subseteq P(X),\mathcal{J}\subseteq P(Y)$ we define a Fubini product $\mathcal{I}\otimes\mathcal{J}$ of these ideals in the following way

[TABLE]

where $\textnormal{Bor}(X\times Y)$ is a family of Borel subsets of $X\times Y$ and $B_{x}=\{y\in Y:(x,y)\in B\}$ is a vertical section of the set $B$ (similarly we define a horizontal section $B^{y}$). We say that the pair $(\mathcal{I},\mathcal{J})$ has a Fubini Property, if for every Borel set $B\subseteq X\times Y$

[TABLE]

If $(\mathcal{I},\mathcal{I})$ has a Fubini Property, then we will simply say that $\mathcal{I}$ has it. Let us notice that Kuratowski-Ulam Theorem and Fubini Theorem imply that $\mathcal{M}$ and $\mathcal{N}$ respectively possess the Fubini Property.

Theorem 18** ([7], Thm 4 and Thm 5).**

Let $\mathbb{R}\supseteq\mathcal{I}\in\{\mathcal{M},\mathcal{N}\}$ and $G\subseteq\mathbb{R}^{2}$ be a Borel set such that $G^{c}\in\mathcal{I}\otimes\mathcal{I}.$ Then there are two ets $B,P\subseteq\mathbb{R}$ such that $B\times P\subseteq G$, $B^{c}\in\mathcal{I}$ and $P\in\textnormal{Perf}(\mathbb{R})$.

Proof.

Let $V$ be a universe of ZFC such that $G\in V$ and let $V^{\prime}$ be its extension such that $\omega_{2}<add(\mathcal{I}).$ Let $b\in\omega^{\omega}\cap V$ be a Borel code for $G.$ Let $G^{\star}$ be a Borel subset of $\mathbb{R}^{2}$ decoded by $b$ in $V^{\prime}.$ By the absolutness of Borel codes of sets from $\mathcal{I}$ it is the case that ${B^{\star}}^{c}$ is in $\mathcal{I}$ in $V^{\prime}.$

We work in $V^{\prime}$ universe. Let $Z=\{x\in\mathbb{R}:\;G_{x}^{\star c}\in\mathcal{I}\}$. By the Fubini Property $Z^{c}\in\mathcal{I}$. Then $|Z|=\mathfrak{c}\geq\omega_{3}.$ Let us choose any set $T\subseteq Z$ of cardinality $\omega_{2}$. Since $\omega_{2}<\textnormal{add}(\mathcal{I})$, the complement of a set $\bigcap_{t\in T}G^{\star}_{t}$ is in $\mathcal{I}.$ Let $B\in\textnormal{Bor}(\mathbb{R})$ such that $B^{c}\in\mathcal{I}$, $B\subseteq\bigcap_{t\in T}G^{\star}_{t}$ and consider a set $A=\{x\in\mathbb{R}:\;B\subseteq G^{\star}_{x}\}.$ Clearly, $A$ is coanalytic. Since $T$ has a size $\omega_{2}$ and $T\subseteq A$, $A$ contains a perfect subset $P.$ It implies that $V^{\prime}$ is a model for the following formula

[TABLE]

It is $\Sigma^{1}_{2}$, hence by Shoenfield Absoluteness Theorem it also holds in $V$. ∎