Suppose that the continuum hypothesis holds. I'm trying to prove that there is a set $T\subseteq\omega_1\times\omega$ such that every set $S$ with $S=A\times B\subseteq\omega_1\times\omega$ with $A$ being uncountable and $B$ being infinite meets both $T$ and $T$'s complement in $\omega_1\times\omega$. Not really sure how to begin: I don't see how the continuum hypothesis could be of any help here.

  • $\begingroup$ $CH$ might come in handy since we would have that $|\mathcal{P}(\omega)| = \omega_1$. $\endgroup$ – Kyle May 26 '16 at 20:27
  • $\begingroup$ Since $\omega_1$ is countabkle and $\omega$ is infinite, it appears that $\omega_1\times \omega$ itself is a possible $S$ -- and if that is the case, then it will be impossible for any $T$ to meet its complement ... On the other hand, if the complement of $S=\omega_1\times\omega$ is not to be met, then just take $T=\omega_1\times\omega$. Are you sure you have stated the problem accurately? $\endgroup$ – Henning Makholm May 26 '16 at 20:33
  • $\begingroup$ You're right! I've edited to be cleaerer $\endgroup$ – R. Thomas May 26 '16 at 21:11

Let $\langle X_i : i < \omega_1 \rangle$ list all infinite subsets of $\omega$. For each $i < \omega_1$, choose $Y_i \subseteq \omega$ such that for every $j < i$, both $Y_i \cap X_j$ and $(\omega \setminus Y_i) \cap X_j$ are infinite. Put $T = \{(i, n) : i < \omega_1, n \in Y_i\}$. Now check.


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