Recall Jensen’s diamond principle over a stationary subset $S$ of a regular uncountable cardinal $\kappa$: there exists a sequence $\langle A_\alpha\mid \alpha\in S \rangle$ such that $\{\alpha\in S\mid A\cap\alpha=A_\alpha\}$ is stationary for every $A\subseteq\kappa$. Equivalently, there exists a sequence $\langle f_\alpha:\alpha\rightarrow\alpha\mid \alpha\in S\rangle$ such that $\{\alpha\in S\mid f\restriction\alpha=f_\alpha\}$ is stationary for every function $f:\kappa\rightarrow\kappa$.

It is clear that $\diamondsuit(S)$ implies $\diamondsuit(T)$ whenever $S\subseteq T\subseteq\kappa$. Shelah proved that it is consistent that $\diamondsuit(S)$ holds, and at the same time $\diamondsuit(\omega_1\setminus S)$ fails, for some stationary co-stationary subset $S$ of $\omega_1$. Somewhat complimentary to the latter, Devlin proved that if $\diamondsuit(\kappa)$ holds, then there exists a sequence $\langle S_i\mid i<\kappa\rangle$ of pairwise disjoint subsets of $\kappa$ such that $\diamondsuit(S_i)$ holds for each $i<\kappa$.

Devlin’s gem has two surprising features. The first feature is that — for an unclear reason — many people are unaware of Devlin’s finding.

The second surprising feature is that Devlin’s proof is not entirely trivial:

Devlin defines the ideal $ND(\kappa)$ of all subsets $S$ of $\kappa$ on which $\diamondsuit(S)$ fails, proves that it is $\kappa$-complete (via Kunen’s trick for proving that $\diamondsuit^-(S)$ entails $\diamondsuit(S)$), and recalls that a $\kappa$-complete prime ideal on $\kappa$ cannot be second-order definable. Altogether, $ND(\kappa)$ is not $\kappa$-saturated.

In this short post, I’ll give a sincerely trivial proof.

**Proof of Devlin’s gem.** Let $\langle f_\alpha:\alpha\rightarrow\alpha\mid \alpha<\kappa\rangle$ be s.t. $G(f)=\{\alpha<\kappa\mid f\restriction\alpha=f_\alpha\}$ is stationary for every function $f:\kappa\rightarrow\kappa$. Write $S_i:=\{\alpha<\kappa\mid f_\alpha(0)=i\}$ for all $i<\kappa$. For every $\alpha<\kappa$, let $g_\alpha:=\bigcup\{ f_{f_\alpha(\beta)}\mid 0<\beta<\alpha\}$.

We claim that $\langle g_\alpha\mid \alpha\in S_i\rangle$ witnesses $\diamondsuit(S_i)$ for each $i<\kappa$. To see this, let $i<\kappa$ and $f:\kappa\rightarrow\kappa$ be arbitrary.

Define $g:\kappa\rightarrow\kappa$ by letting $g(0):=i$, and $g(\beta):=\min(G(f)\setminus\beta)$ whenever $0<\beta<\kappa$. As $G(g)$ is a stationary subset of $S_i$, it suffices to show that $f\restriction\alpha=g_\alpha$ for all limit $\alpha\in G(g)$. Let $\alpha\in G(g)$ be a limit ordinal. Then for all nonzero $\beta<\alpha$, $g(\beta)=f_\alpha(\beta)<\alpha$, and hence $f\restriction\beta\subseteq f_{f_\alpha(\beta)}\subseteq f\restriction\alpha$. That is, $g_\alpha=f\restriction\alpha$. $\square$

Note that the same proof shows that if $\diamondsuit(T)$ holds, then there exists a sequence $\langle S_i\mid i<\kappa\rangle$ of pairwise disjoint subsets of $T$ such that $\diamondsuit(S_i)$ holds for each $i<\kappa$.

Note also that unlike Devlin’s proof, the above proof does not build on the fact that $\kappa$ may be partitioned into $\kappa$ many pairwise disjoint stationary sets.

Gems, diamonds, it’s a good thing we don’t have a typeset of rubies, emeralds and whatnots, or else combinatorial set theory would have been filled with them!

Combinatorial candy crush. Sweet!

I dare you to write an expository paper with this title. 😀

This reminds me of the various jokes about what mathematicians call “trivial”. I stared at this on the screen for hours before I eventually figured out what’s going on here.

The problem is that $g_alpha$ is not necessarily a well-defined function for every $alpha$. Even if it is a function, its domain may not be $alpha$. Subtly included in the last part of the proof is the derivation that for sufficiently many $alpha$, the relation $g_alpha$ is in fact a function with domain $alpha$. Maybe this should be explained.

Actually the claim that $langle g_alphamidalphain S_irangle$ witnesses $diamondsuit (S_i)$ is therefore not correct, since to be diamond sequence requires that for every $alpha$, $g_alpha$ should be a function $alpha to alpha$. The definition of $g_alpha$ should be modified by appending “… if this is a function with domain $alpha$, and an arbitrary function from $alpha$ to $alpha$ otherwise.” This modification would have the dual effect of making the claim technically correct and also alerting the reader to the correct understanding of what is happening in the proof.

You are right, though the subtly is actually in what I mean by “witnesses $diamondsuit$” (as opposed to “is a $diamondsuit$-sequence”).

Generally speaking, I tend to define diamond as a sequence that guesses a particular type of object (e.g., a subset of $omega_1$), without putting additional requirements on the sequence (e.g., that the $alpha_{th}$ member of the sequence be a subset of $alpha$). See here.