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A set $S \subset \omega_1 \times \omega$ is a large rectangle if $S = A \times B$ where $A$ is uncountable, and $B$ is infinite.

Assuming the continuum hypothesis, is there necessarily a set $T \subset \omega_1 \times \omega$ such that every large rectangle intersects both $T$ and its complement?

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Yes this is true. Let $T_{\alpha}\subseteq \omega$ be a family of sets and define $T=\bigcup\limits_{\alpha<\omega_1}\{\alpha\}\times T_{\alpha}$. We want the sets $T_{\alpha}$ to have the property that for each infinite $B\subseteq\omega$, $$B\cap T_{\alpha}\neq \emptyset$$ $$B\cap (\omega - T_{\alpha})\neq \emptyset$$ both hold for all but countably many $\alpha<\omega_1$.

Such a family can be constructed using CH, as follows, let $B_{\alpha}$ be an enumeration in order type $\omega_1$ of all infinite subseteq of $\omega$. To define $T_{\beta}$ we need only a set such that

$$B_{\alpha}\cap T_{\beta}\neq \emptyset$$ $$B_{\alpha}\cap (\omega -T_{\beta})\neq \emptyset$$ For all $\beta<\alpha$.

This is easy to accomplish let $B_i:i<\omega$ be an enumeration of the sets $B_{\alpha}$ for $\alpha<\beta$ and for each $i$ chose $x_i,y_i\in B_i$ and place $x_i\in T_{\beta}$ and $y_i\in \omega-T_{\beta}$. Thus at any stage there are a finite number of forbidden $x_j$ and $y_j$'s but as $B_i$ is infinite a choice always exists. This constructs $T_{\beta}$ and thus $T$.

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Here is my write up of the same proof... what am I getting wrong?

Given that $CH \rightarrow \omega_1$ can be well ordered, let $\langle X_{\alpha} \ | \ \alpha < \omega_1 \rangle$ denote a well ordering of all subsets $X \subseteq \omega$. We will create sets $T_{\beta} \subseteq \omega$ by recursion as follows:

Step 0: $T_0 = \emptyset$

...

Step $\beta$: For all $X_{\alpha}$ with $\alpha < \beta$ find some $x_{\alpha,\beta}, y_{\alpha,\beta} \in X_{\alpha}$ in such a way that we can place $x_{\alpha,\beta} \in T_{\beta}$ and claim $y_{\alpha,\beta} \notin T_{\beta}$ for all $\alpha < \beta$.

Now we can define the desired $T$ as follows:

$$ T = \bigcup_{\beta < \omega_1}\{\beta\} \times T_{\beta}$$

Now consider an arbitrary large rectangle $S \subseteq \omega_1 \times \omega$. Looking at only the $\omega$ components of $S$, call it $B$, we can say $B = S_{\alpha}$ for some $\alpha < \omega_1$. For all $\beta > \alpha$ we have some $x_{\alpha,\beta} \in S_{\alpha}$ and $x_{\alpha,\beta} \in T_{\beta}$. Because $A$ is uncountable and $B$ is countably infinite, we can find a $(\tau, x_{\alpha,\beta})$... and I'm stuck...

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