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Taken out of ch II, Kunen. Need to show the following two versions of $\Diamond$ are equivalent:

$\Diamond_\kappa$: There are $A_\alpha \subset \alpha$ for $\alpha < \kappa$ such that for each $A \subset \kappa$ the set $\{\alpha < \kappa : A \cap \alpha = A_\alpha\} $ is stationary.

$\Diamond_\kappa'$: There are $A_\alpha \subset \mathcal P(\alpha)$ for $\alpha < \kappa$ such that each $|A_\alpha| \leq \alpha$ and for each $A \subset \kappa$, $\{\alpha < \kappa : A \cap \alpha \in A_\alpha\}$ is stationary.

$\Diamond_\kappa \Leftarrow \Diamond_\kappa'$ was shown on one of the lectures (for $\omega_1$ instead of $\kappa$) and it involved using $f: \omega_1 \leftrightarrow \omega \times \omega_1$ as a set-building function for the transition. Is the other direction more straightforward? Seems like it's possible to build sets of $\diamond_\kappa'$ from sets of $\diamond_\kappa$ by applying an easy transformation to them, but the condition $|A_\alpha| \leq \alpha$ seems tricky here.

Any help would be appreciated.

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up vote 4 down vote accepted

Hint: Consider $\mathcal A_\alpha=\{A_\alpha\}$ where $\langle A_\alpha: \alpha<\omega_1\rangle $ is your $\diamondsuit_\kappa $ sequence.

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Thanks! Don't know how I missed such an easy one. – Pavel Jul 2 '12 at 17:46

To prove $\diamondsuit_\kappa\to\diamondsuit_\kappa'$, let $\langle A_\xi:\xi<\kappa\rangle$ be a $\diamondsuit$-sequence for $\kappa$. Now for each $\xi<\kappa$ let $\mathscr{A}_\xi=\{A_\xi\}$. Clearly $\mathscr{A}_\xi\subseteq\wp(\xi)$, and $|\mathscr{A}_\xi|\le|\xi|$.

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Thanks! An elegant one. – Pavel Jul 2 '12 at 17:49

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