Weakening of $\diamondsuit$ implies CH

The $\diamondsuit$ principle states that there is a collection of subsets $S_\alpha \subset \alpha$ for every $\alpha < \omega_1$ such that for any $X \subset\omega_1$ the set $\{\alpha : X \cap \alpha = S_\alpha\}$ is stationary in $\omega_1$. Diamond is very easily proven to imply CH by using the fact that the above set is stationary.

However, let us consider a weakening: there is a collection of subsets $S_\alpha \subset \alpha$ for every $\alpha < \omega_1$ such that for any $X \subset\omega_1$ there will be an $ \alpha > 1$ with $X \cap \alpha = S_\alpha$.

I would like to show that this weakened version implies CH and is equivalent to Diamond itself. The first result follows from showing it is equivalent to Diamond. My main problem is the following: how can we show that given $X \subset \omega_1$ that the existence of one $\alpha$ with $X \cap \alpha = S_\alpha$ implies an unbounded many $\alpha$ that make $X \cap \alpha = S_\alpha$. Any hints would be appreciated.

Edit: Let's take the following hint. Assume our weakening of $\diamondsuit$ and let $\{S_\alpha\}_{\alpha < \omega_1}$ be our sequence, and let's construct $\{T_\alpha\}_{\alpha < \omega_1}$ by $\beta \in T_\alpha$ if and only if $\omega + \beta \in S_{\omega+\alpha}$. Fix some $X \subset \omega_1$. By assumption, we have some $\alpha > 1$ such that $X \cap \alpha = S_\alpha$. We would like to show that there is some $\beta \geq \omega$ such that $X \cap \beta = T_\beta$. This last step is all that is needed for me, as I feel comfortable with the step that shows the principle below is equivalent to $\diamondsuit$ itself.

Solution 1:

You can find a proof of this in K. Devlin, Variations on diamond, The Journal of Symbolic Logic, Volume 44, Issue 1, March 1979, pp. 51 - 58 where it is shown that the following statement is equivalent to diamond.

There exists $\langle \mathcal{A}_{\alpha}: \alpha < \omega_1 \rangle$ where each $\mathcal{A}_{\alpha}$ is a countable family of subsets of $\alpha$ such that for every $X \subseteq \omega_1$, there exists $\alpha \geq \omega$ such that $X \cap \alpha \in \mathcal{A}_{\alpha}$.