# proof of $\Diamond$ is equivalent to $\clubsuit$ and continuum hypothesis

The proof that $\Diamond_{S}$ implies both $\clubsuit_{S}$ and that for every $\lambda<\kappa$, $2^{\lambda}\leq\kappa$ are given in the entries for $\Diamond_{S}$ and $\clubsuit_{S}$.

Let $A=\langle A_{\alpha}\rangle_{\alpha\in S}$ be a sequence which satisfies $\clubsuit_{S}$.

Since there are only $\kappa$ bounded subsets of $\kappa$, there is a surjective function $f:\kappa\rightarrow\operatorname{Bounded}(\kappa)\times\kappa$ where $\operatorname{Bounded}(\kappa)$ is the bounded subsets of $\kappa$. Define a sequence $B=\langle B_{\alpha}\rangle_{\alpha<\kappa}$ by $B_{\alpha}=f(\alpha)$ if $sup(B_{\alpha})<\alpha$ and $\emptyset$ otherwise. Since the set of $(B_{\alpha},\lambda)\in\operatorname{Bounded}(\kappa)\times\kappa$ 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, $D=\langle D_{\alpha}\rangle_{\alpha\in S}$ such that $x\in D_{\alpha}\leftrightarrow x\in B_{\beta}$ for some $\beta\in A_{\alpha}$. We can show that $D$ satisfies $\Diamond_{S}$.

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 $j:\kappa\rightarrow\kappa$ as follows:

• $j(0)=0$

• To find $j(\alpha)$, take $X\cap\{j(\beta)\mid\beta<\alpha\}$. 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 $\{\alpha\}\cup\{j(\beta)\mid\beta<\alpha\}$ such that $B_{\gamma}=X\cap\{j(\beta)\mid\beta<\alpha\}$.

Let $T^{\prime}=\operatorname{range}(j)$. 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 $\lambda<\kappa$ then $j(\lambda)$ includes $j(\alpha)$ for $\alpha<\lambda$, 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 $\operatorname{sup}(c\cap\alpha)=\alpha$. Then for any $\beta<\alpha$, there is some $\gamma\in c$ such that $\beta<\gamma<\alpha$ and therefore $j(\beta)<\gamma$. 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 $\alpha\in S^{\prime}$, $\beta\in\operatorname{range}(j)$, 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 $\operatorname{sup}(A_{\alpha})=\alpha$, there is some $\eta\in A_{\alpha}$ such that $\gamma<\eta$, so $j(x)\in\eta$. Since $\eta\in A_{\alpha}$, $B_{\eta}\subseteq D_{\alpha}$, and since $\eta\in\operatorname{range}(j)$, $j(\delta)\in B_{\eta}$ for any $\delta, 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 $\alpha\in C^{\prime}$.

Title proof of $\Diamond$ is equivalent to $\clubsuit$ and continuum hypothesis ProofOfDiamondIsEquivalentToclubsuitAndContinuumHypothesis 2013-03-22 12:53:57 2013-03-22 12:53:57 Henry (455) Henry (455) 6 Henry (455) Proof msc 03E45 proof that diamond is equivalent to club and continuum hypothesis Diamond Clubsuit