Show that the following are equivalent:
(1) $\diamondsuit$
(2) The existence of $B_\alpha\subseteq\alpha\times\alpha\ (\alpha<\omega_1)$ so that the set $\{\alpha<\omega_1\mid B\cap (\alpha\times\alpha)=B_\alpha\}$ is stationary for any $B\subseteq\omega_1\times\omega_1$
My question is actually pretty simple. I think I've figured out the harder direction [$(1)\Rightarrow(2)$] by slightly modifying the first part of the argument in Theorem 7.14.
I take the other direction to be pretty obvious, but I'm actually having a hard time articulating why, exactly. Any help would be a appreciated.
HINT: Let $h:\omega_1\times\omega_1\to\omega_1$ be a bijection. Let $C=\{\alpha\in\omega_1:h[\alpha\times\alpha]=\alpha\}$, and show that $C$ is a club set in $\omega_1$. Then consider the sets $h[B_\alpha]$ for $\alpha\in C$.
Added: That was the first thing that occurred to me, but there is an easier argument. For $\alpha\in\omega_1$ let $C_\alpha=\{\xi\in\omega_1:\langle\xi,\xi\rangle\in B_\alpha\}$. If $A\subseteq\omega_1$, let $A'=\{\langle\xi,\xi\rangle:\xi\in A\}$; then
$$\{\alpha\in\omega_1:A\cap\alpha=C_\alpha\}\supseteq\{\alpha\in\omega_1:A'\cap(\alpha\times\alpha)=B_\alpha\}$$
is stationary.