Variations on diamond

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:

1. There exists a sequence  of countable sets $⟨{\mathcal{A}}_{\alpha }\mid \alpha <{\omega }_1⟩$ such that for every $X\subseteq {\omega }_1$,  $\{\alpha <{\omega }_1\mid X\cap \alpha \in {\mathcal{A}}_{\alpha }\}$ contains an infinite ordinal.
2. There exists a sequence of countable sets $⟨{\mathcal{B}}_{\alpha }\mid \alpha <{\omega }_1⟩$ such that for every $X\subseteq {\omega }_1$,  $\{\alpha <{\omega }_1\mid X\cap \alpha \in {\mathcal{B}}_{\alpha }\}$ is stationary.
3. ${\mathrm{♢}}_{{\omega }_1}$. That is, there exists a sequence $⟨A_{\alpha }\mid \alpha <{\omega }_1⟩$ 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 $\overrightarrow{\mathcal{A}}=⟨{\mathcal{A}}_{\alpha }\mid \alpha <{\omega }_1⟩$ 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 $⟨{\mathcal{B}}_{\alpha }\mid \alpha <{\omega }_1⟩$ 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:

• $\frac{A}{2}:=\{\delta <{\omega }_1\mid 2\cdot \delta \in A\}$;
• $A(\beta ,\alpha )=\{{\psi }_{\alpha }(n)\mid (\beta +2n+1)\in A\}$.

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: $$\{\frac{A\cap \alpha }{2},A(\beta ,\alpha )\mid \beta \le \gamma <\alpha +\omega ,A\in {\mathcal{A}}_{\gamma }\}\subseteq {\mathcal{B}}_{\alpha }.$$

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.

$▸$ If $\gamma <\beta +\omega$, then $\beta \in C$ is a limit ordinal, and hence $X\cap \beta =\frac{Y\cap \beta }{2}=\frac{A\cap \beta }{2}\in {\mathcal{B}}_{\beta }$.

$▸$ 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 $$\{n<\omega \mid (\beta +2n+1)\in A\}=\{n<\omega \mid {\psi }_{{\beta }^{+}}(n)\in X\}={\psi }_{\alpha }^{-1}[X\cap \alpha ]$$ and hence $$X\cap \alpha =A(\beta ,\alpha )\in {\mathcal{B}}_{\alpha }.$$

We continue with Kunen’s lemma that $(2)\Rightarrow (3)$.
Let $⟨{\mathcal{B}}_{\alpha }\mid \alpha <{\omega }_1⟩$ be as in (2). Fix a bijection $\pi :{\omega }_1\leftrightarrow \omega \times {\omega }_1$. For every $\alpha <{\omega }_1$, fix an enumeration ${\mathcal{B}}_{\alpha }=\{B_{\alpha }^n\mid n<\omega \}$, and denote $A_{\alpha }^n:=\{\delta <{\omega }_1\mid (n,\delta )\in \pi [B_{\alpha }^n]\}$. We claim that there exists some $n<\omega$ such that $⟨A_{\alpha }^n\mid \alpha <{\omega }_1⟩$ witnesses ${\mathrm{♢}}_{{\omega }_1}$.
Suppose not, then for every $n<\omega$, there exists $X_n\subseteq {\omega }_1$ and a club $D_n\subseteq {\omega }_1$ such that $X_n\cap \alpha \ne A_{\alpha }^n$ 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 $\alpha \in D_n$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

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:

1. There exists a sequence  of sets $⟨{\mathcal{A}}_{\alpha }\mid \alpha <\kappa ⟩$ 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 $\ge \log (\kappa )$.
2. ${\mathrm{♢}}_{\kappa }$. That is, there exists a sequence $⟨A_{\alpha }\mid \alpha <\kappa ⟩$ 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 $⟨A_{\alpha }\mid \alpha <{\omega }_1⟩$ 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 $X\cap \omega =A_{\omega }$;
• $n=min\{m<\omega \mid m\notin X\}$ is well defined. Then $X\cap (n+1)=A_{n+1}$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

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 $⟨{\mathcal{B}}_{\alpha }\mid \alpha <{\omega }_1⟩$ such that for every $X\subseteq {\omega }_1$,  $\{\alpha <{\omega }_1\mid X\cap \alpha \in {\mathcal{B}}_{\alpha }\}$ is stationary, but no choice $⟨A_{\alpha }\mid \alpha <{\omega }_1⟩\in \prod _{\alpha <{\omega }_1}{\mathcal{B}}_{\alpha }$ witnesses ${\mathrm{♢}}_{{\omega }_1}$.

We also have the following:

Proposition. Suppose that $\kappa$ is a regular uncountable cardinal, $S\subseteq E_{\ge \log (\kappa )}^{\kappa }$ is stationary, and there exists a sequence  of sets $⟨{\mathcal{A}}_{\alpha }\mid \alpha <\kappa ⟩$ 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 ${\mathrm{♢}}_S$ holds.

Corollary. For every infinite regular cardinal $\lambda$, and every stationary $S\subseteq {\lambda }^{+}$, ${\mathrm{♢}}_S$ holds iff here exists a sequence  of sets $⟨{\mathcal{A}}_{\alpha }\mid \alpha <{\lambda }^{+}⟩$ such that for every $X\subseteq {\lambda }^{+}$, there exists $\alpha \in S$ with $X\cap \alpha \in {\mathcal{A}}_{\alpha }$ and $|{\mathcal{A}}_{\alpha }|\le \lambda$.