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?


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 $T$.

| cite | improve this answer | |

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...

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.