Jensen’s diamond principle has many equivalent forms. The translation between these forms is often straight-forward, but there is one form whose equivalence to the usual form is somewhat surprising, and Devlin’s translation from one to the other, seems a little bit of a magic. Let us provide a proof.
Lemma (Devlin, Kunen). The following are equivalent:
- There exists a sequence of countable sets $\langle \mathcal A_\alpha\mid \alpha<\omega_1\rangle$ such that for every $X\subseteq \omega_1$, $\{ \alpha<\omega_1\mid X\cap\alpha\in\mathcal A_\alpha\}$ contains an infinite ordinal.
- There exists a sequence of countable sets $\langle \mathcal B_\alpha\mid \alpha<\omega_1\rangle$ such that for every $X\subseteq\omega_1$, $\{ \alpha<\omega_1\mid X\cap\alpha\in\mathcal B_\alpha\}$ is stationary.
- $\diamondsuit_{\omega_1}$. That is, there exists a sequence $\langle A_\alpha\mid \alpha<\omega_1\rangle$ such that for every $X\subseteq\omega_1$, $\{ \alpha<\omega_1\mid X\cap\alpha=A_\alpha\}$ is stationary.
Proof. We commence with Devlin’s lemma that $(1)\Rightarrow(2)$.
Let $\overrightarrow{\mathcal A}=\langle \mathcal A_\alpha\mid \alpha<\omega_1\rangle$ be as in (1). For all $\alpha<\omega_1$, let $\mathcal B_\alpha$ be some countable elementary submodel of $(\mathcal H(\omega_1),\in)$ such that $\overrightarrow{\mathcal A},\alpha\in\mathcal B_\alpha$. We claim that $\langle\mathcal B_\alpha\mid\alpha<\omega_1\rangle$ witnesses (2).
How magical is that!
Now, let us explain..
For all infinite $\alpha<\omega_1$, we have $|\alpha|=|\omega|$, and so elementarity together with $\alpha\in\mathcal B_\alpha$ give us a bijection $\psi_\alpha:\omega\leftrightarrow\alpha$ that lies in $\mathcal B_\alpha$. For this, given a subset $A\subseteq\omega_1$, denote:
- ${A\over 2}:=\{ \delta<\omega_1\mid 2\cdot\delta\in A\}$;
- $A(\beta,\alpha)=\{\psi_\alpha(n)\mid (\beta+2n+1)\in A\}$.
The idea behind the second definition is as follows: if $\beta$ is a limit ordinal, then a subset $A$ of the interval $(\beta,\beta+\omega)$ can code subsets of $\alpha$ for arbitrarily high $\alpha<\omega_1$.
Notice that by $\alpha\in\mathcal B_\alpha$, we have $\alpha+\omega\subseteq\mathcal B_\alpha$, and moreover: $$\left\{ {A\cap\alpha\over 2},A(\beta,\alpha)\mid \beta\le\gamma<\alpha+\omega, A\in \mathcal A_\gamma\right\}\subseteq \mathcal B_\alpha.$$
Finally, suppose that we are given a subset $X\subseteq\omega_1$ along with a club $C\subseteq\omega_1$, and let us find some $\alpha\in C$ such that $X\cap\alpha\in\mathcal B_\alpha$.
By shrinking, we may assume that $C$ consists of limit infinite ordinals. For every $\beta<\omega_1$, denote $\beta^+:=\min(C\setminus\beta+1)$. Then, put $$Y:=\{ 2\cdot\delta\mid \delta\in X\}\cup\{\beta+2n+1\mid \beta<\omega_1\text{ is a limit}, n<\omega, \psi_{\beta^+}(n)\in X\}.$$
Now, by the hypothesis, we may fix an infinite $\gamma<\omega_1$ and $A\in\mathcal A_\gamma$ such that $Y\cap\gamma=A$.
Denote $\beta:=\sup((C\cup\{0\})\cap\gamma)$, and $\alpha:=\min(C\setminus\beta+1)$. Note that $\alpha\in C$. Also note that if $\gamma>\min(C)$, then $\beta\in C$, as well. We shall consider two cases.
$\blacktriangleright$ If $\gamma<\beta+\omega$, then $\beta\in C$ is a limit ordinal, and hence $X\cap\beta={Y\cap\beta\over 2}={A\cap\beta\over 2}\in\mathcal B_\beta$.
$\blacktriangleright$ If $\gamma\ge\beta+\omega$, then $\beta<\beta+\omega\le\gamma<\alpha+\omega$. Since $Y\cap(\beta+\omega)=A\cap(\beta+\omega)$, we get that $$\{ n<\omega\mid (\beta+2n+1)\in A\}=\{n<\omega\mid \psi_{\beta^+}(n)\in X\}=\psi^{-1}_{\alpha}[X\cap\alpha]$$ and hence $$X\cap\alpha=A(\beta,\alpha)\in \mathcal B_\alpha.$$
We continue with Kunen’s lemma that $(2)\Rightarrow(3)$.
Let $\langle \mathcal B_\alpha\mid \alpha<\omega_1\rangle$ be as in (2). Fix a bijection $\pi:\omega_1\leftrightarrow\omega\times\omega_1$. For every $\alpha<\omega_1$, fix an enumeration $\mathcal B_\alpha=\{ B^n_\alpha\mid n<\omega\}$, and denote $A^n_\alpha:=\{ \delta< \omega_1\mid (n,\delta)\in \pi[B^n_\alpha]\}$. We claim that there exists some $n<\omega$ such that $\langle A^n_\alpha\mid \alpha<\omega_1\rangle$ witnesses $\diamondsuit_{\omega_1}$.
Suppose not, then for every $n<\omega$, there exists $X_n\subseteq\omega_1$ and a club $D_n\subseteq\omega_1$ such that $X_n\cap\alpha\neq A^n_\alpha$ for every $\alpha\in D_n$. Let
- $X:=\{ \pi^{-1}(n,\delta)\mid n<\omega,\delta\in X_n\}$;
- $D:=\{\alpha\in\bigcap_{n<\omega}D_n\mid \pi^{-1}[\omega\times\alpha]\subseteq\alpha\}$.
Then $X\subseteq\omega_1$, and $D$ is a club in $\omega_1$. Fix $\alpha\in D$ such that $X\cap\alpha\in\mathcal B_\alpha$. Fix $n<\omega$ such that $X\cap\alpha=B_\alpha^n$. Then $A_\alpha^n=X_n\cap\alpha$, contradicting the fact that $\alpha\in D_n$. $\square$
The preceding generalizes as follows. Recall that $\log(\kappa)$ stands for the least cardinal $\mu\le\kappa$ for which $2^\mu\ge\kappa$. Then:
Lemma (Matet). For every uncountable cardinal $\kappa$ such that $\log(\kappa)<\kappa$, the following are equivalent:
- There exists a sequence of sets $\langle \mathcal A_\alpha\mid \alpha<\kappa\rangle$ such that $|\mathcal A_\alpha|\le|\alpha|$ for all $\alpha<\kappa$, and for every $X\subseteq\kappa$, $\{ \alpha<\kappa\mid X\cap\alpha\in\mathcal A_\alpha\}$ contains an ordinal $\ge\log(\kappa)$.
- $\diamondsuit_\kappa$. That is, there exists a sequence $\langle A_\alpha\mid \alpha<\kappa\rangle$ such that for every $X\subseteq\kappa$, $\{ \alpha<\kappa\mid X\cap\alpha = A_\alpha\}$ is stationary.
(Indeed, $\log(\kappa)$ will be used to code subsets of arbitrarily high $\alpha<\kappa$, as subsets of intervals of the form $(\beta,\beta+\log(\kappa))$.)
In contrast, let us point the following ZFC fact.
Observation. There exists a sequence $\langle A_\alpha\mid\alpha<\omega_1\rangle$ such that for every $X\subseteq\omega_1$, the set $\{ \alpha<\omega_1\mid X\cap\alpha=A_\alpha\}$ contains a non-zero ordinal $\le\omega$.
Proof. Let $A_{n+1}:=n$ for all $n<\omega$, and $A_\omega=\omega$.
Now given, $X\subseteq\omega_1$ one of the following holds:
- $X\cap\omega=\omega$. Then $X\cap\omega=A_\omega$;
- $n=\min\{ m<\omega\mid m\not\in X\}$ is well defined. Then $X\cap(n+1)=A_{n+1}$. $\square$
For completeness, we also mention the following theorem of Shelah.
Theorem (Shelah). It is consistent that GCH holds, and there exists a sequence of countable sets $\langle \mathcal B_\alpha\mid \alpha<\omega_1\rangle$ such that for every $X\subseteq\omega_1$, $\{ \alpha<\omega_1\mid X\cap\alpha\in\mathcal B_\alpha\}$ is stationary, but no choice $\langle A_\alpha\mid\alpha<\omega_1\rangle\in\prod_{\alpha<\omega_1}\mathcal B_\alpha$ witnesses $\diamondsuit_{\omega_1}$.
We also have the following:
Proposition. Suppose that $\kappa$ is a regular uncountable cardinal, $S\subseteq E^\kappa_{\ge\log(\kappa)}$ is stationary, and there exists a sequence of sets $\langle \mathcal A_\alpha\mid \alpha<\kappa\rangle$ such that $|\mathcal A_\alpha|\le|\alpha|$ for all $\alpha<\kappa$, and for every $X\subseteq\kappa$, $\{ \alpha\in S\mid X\cap\alpha\in\mathcal A_\alpha\}$ is nonempty. Then $\diamondsuit_S$ holds.
Corollary. For every infinite regular cardinal $\lambda$, and every stationary $S\subseteq\lambda^+$, $\diamondsuit_S$ holds iff here exists a sequence of sets $\langle \mathcal A_\alpha\mid \alpha<\lambda^+\rangle$ such that for every $X\subseteq\lambda^+$, there exists $\alpha\in S$ with $X\cap\alpha\in\mathcal A_\alpha$ and $|\mathcal A_\alpha|\le\lambda$.
