proof of $\mathrm{◇}$ is equivalent to $\mathrm{♣}$ and continuum hypothesis

The proof that ${\mathrm{◇}}_S$ implies both ${\mathrm{♣}}_S$ and that for every , $2^{\lambda }\le \kappa$ are given in the entries for ${\mathrm{◇}}_S$ and ${\mathrm{♣}}_S$.

Let $A={⟨A_{\alpha }⟩}_{\alpha \in S}$ be a sequence which satisfies ${\mathrm{♣}}_S$.

Since there are only $\kappa$ bounded subsets of $\kappa$, there is a surjective function $f:\kappa \to \mathrm{Bounded}(\kappa )\times \kappa$ where $\mathrm{Bounded}(\kappa )$ is the bounded subsets of $\kappa$. Define a sequence by $B_{\alpha }=f(\alpha )$ if and $\mathrm{\varnothing }$ otherwise. Since the set of $(B_{\alpha },\lambda )\in \mathrm{Bounded}(\kappa )\times \kappa$ such that $B_{\alpha }=T$ is unbounded for any bounded subset $T$, it follow that every bounded subset of $\kappa$ occurs $\kappa$ times in $B$.

We can define a new sequence, $D={⟨D_{\alpha }⟩}_{\alpha \in S}$ such that $x\in D_{\alpha }\leftrightarrow x\in B_{\beta }$ for some $\beta \in A_{\alpha }$. We can show that $D$ satisfies ${\mathrm{◇}}_S$.

First, for any $\alpha$, $x\in D_{\alpha }$ means that $x\in B_{\beta }$ for some $\beta \in A_{\alpha }$, and since $B_{\beta }\subseteq \beta \in A_{\alpha }\subseteq \alpha$, we have $D_{\alpha }\subseteq \alpha$.

Next take any $T\subseteq \kappa$. We consider two cases:

$T$ is bounded

The set of $\alpha$ such that $T=B_{\alpha }$ forms an unbounded sequence $T^{\prime }$, so there is a stationary $S^{\prime }\subseteq S$ such that $\alpha \in S^{\prime }\leftrightarrow A_{\alpha }\subset T^{\prime }$. For each such $\alpha$, $x\in D_{\alpha }\leftrightarrow x\in B_i$ for some $i\in A_{\alpha }\subset T^{\prime }$. But each such $B_i$ is equal to $T$, so $D_{\alpha }=T$.

$T$ is unbounded

We define a function $j:\kappa \to \kappa$ as follows:

• $j(0)=0$

• To find $j(\alpha )$, take . This is a bounded subset of $\kappa$, so is equal to an unbounded series of elements of $B$. Take $j(\alpha )=\gamma$, where $\gamma$ is the least number greater than any element of such that .

Let $T^{\prime }=\mathrm{range}(j)$. This is obviously unbounded, and so there is a stationary $S^{\prime }\subseteq S$ such that $\alpha \in S^{\prime }\leftrightarrow A_{\alpha }\subseteq T^{\prime }$.

Next, consider $C$, the set of ordinals less than $\kappa$ closed under $j$. Clearly it is unbounded, since if then $j(\lambda )$ includes $j(\alpha )$ for , and so induction gives an ordinal greater than $\lambda$ closed under $j$ (essentially the result of applying $j$ an infinite number of times). Also, $C$ is closed: take any $c\subseteq C$ and suppose $\sup (c\cap \alpha )=\alpha$. Then for any , there is some $\gamma \in c$ such that and therefore . So $\alpha$ is closed under $j$, and therefore contained in $C$.

Since $C$ is a club, $C^{\prime }=C\cap S^{\prime }$ is stationary. Suppose $\alpha \in C^{\prime }$. Then $x\in D_{\alpha }\leftrightarrow x\in B_{\beta }$ where $\beta \in A_{\alpha }$. Since $\alpha \in S^{\prime }$, $\beta \in \mathrm{range}(j)$, and therefore $B_{\beta }\subseteq T$. Next take any $x\in T\cap \alpha$. Since $\alpha \in C$, it is closed under $j$, hence there is some $\gamma \in \alpha$ such that $j(x)\in \gamma$. Since $\sup (A_{\alpha })=\alpha$, there is some $\eta \in A_{\alpha }$ such that , so $j(x)\in \eta$. Since $\eta \in A_{\alpha }$, $B_{\eta }\subseteq D_{\alpha }$, and since $\eta \in \mathrm{range}(j)$, $j(\delta )\in B_{\eta }$ for any , and in particular $x\in B_{\eta }$. Since we showed above that $D_{\alpha }\subseteq \alpha$, we have $D_{\alpha }=T\cap \alpha$ for any $\alpha \in C^{\prime }$.