Many diamonds from just one

Recall Jensen’s diamond principle over a stationary subset $S$ of a regular uncountable cardinal $κ$ : there exists a sequence $⟨ A_{α} ∣ α \in S ⟩$ such that ${α \in S ∣ A \cap α = A_{α}}$ is stationary for every $A \subseteq κ$ . Equivalently, there exists a sequence $⟨ f_{α} : α \to α ∣ α \in S ⟩$ such that ${α \in S ∣ f ↾ α = f_{α}}$ is stationary for every function $f : κ \to κ$ .

It is clear that $♢ (S)$ implies $♢ (T)$ whenever $S \subseteq T \subseteq κ$ . Shelah proved that it is consistent that $♢ (S)$ holds, and at the same time $♢ (ω_{1} ∖ S)$ fails, for some stationary co-stationary subset $S$ of $ω_{1}$ . Somewhat complimentary to the latter, Devlin proved that if $♢ (κ)$ holds, then there exists a sequence $⟨ S_{i} ∣ i < κ ⟩$ of pairwise disjoint subsets of $κ$ such that $♢ (S_{i})$ holds for each $i < κ$ .

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 $N D (κ)$ of all subsets $S$ of $κ$ on which $♢ (S)$ fails, proves that it is $κ$ -complete (via Kunen’s trick for proving that $♢^{-} (S)$ entails $♢ (S)$ ), and recalls that a $κ$ -complete prime ideal on $κ$ cannot be second-order definable. Altogether, $N D (κ)$ is not $κ$ -saturated.

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

Proof of Devlin’s gem. Let $⟨ f_{α} : α \to α ∣ α < κ ⟩$ be s.t. $G (f) = {α < κ ∣ f ↾ α = f_{α}}$ is stationary for every function $f : κ \to κ$ . Write $S_{i} := {α < κ ∣ f_{α} (0) = i}$ for all $i < κ$ . For every $α < κ$ , let $g_{α} := ⋃ {f_{f_{α} (β)} ∣ 0 < β < α}$ .

We claim that $⟨ g_{α} ∣ α \in S_{i} ⟩$ witnesses $♢ (S_{i})$ for each $i < κ$ . To see this, let $i < κ$ and $f : κ \to κ$ be arbitrary.
Define $g : κ \to κ$ by letting $g (0) := i$ , and $g (β) := min (G (f) ∖ β)$ whenever $0 < β < κ$ . As $G (g)$ is a stationary subset of $S_{i}$ , it suffices to show that $f ↾ α = g_{α}$ for all limit $α \in G (g)$ . Let $α \in G (g)$ be a limit ordinal. Then for all nonzero $β < α$ , $g (β) = f_{α} (β) < α$ , and hence $f ↾ β \subseteq f_{f_{α} (β)} \subseteq f ↾ α$ . That is, $g_{α} = f ↾ α$ .

Note that the same proof shows that if $♢ (T)$ holds, then there exists a sequence $⟨ S_{i} ∣ i < κ ⟩$ of pairwise disjoint subsets of $T$ such that $♢ (S_{i})$ holds for each $i < κ$ .

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

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6 Responses to Many diamonds from just one

Asaf Karagila says:

January 6, 2015 at 6:21 pm

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!

- saf says:
  
  January 7, 2015 at 3:38 am
  
  Combinatorial candy crush. Sweet!
  
  - Asaf Karagila says:
    
    January 15, 2015 at 7:06 pm
    
    I dare you to write an expository paper with this title.
    
Ari B. says:

January 8, 2015 at 8:22 am

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

Actually the claim that $⟨ g_{α} ∣ α \in S_{i} ⟩$ witnesses $♢ (S_{i})$ is therefore not correct, since to be diamond sequence requires that for every $α$ , $g_{α}$ should be a function $α \to α$ . The definition of $g_{α}$ should be modified by appending “… if this is a function with domain $α$ , and an arbitrary function from $α$ to $α$ 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.

- saf says:
  
  January 8, 2015 at 9:15 am
  
  You are right, though the subtly is actually in what I mean by “witnesses $♢$ ” (as opposed to “is a $♢$ -sequence”).
  
  Generally speaking, I tend to define diamond as a sequence that guesses a particular type of object (e.g., a subset of $ω_{1}$ ), without putting additional requirements on the sequence (e.g., that the $α^{t h}$ member of the sequence be a subset of $α$ ). See here.
  
'saf says:

October 21, 2020 at 3:45 pm

For yet another proof, see Theorem 3.7 of this paper.

Many diamonds from just one

6 Responses to Many diamonds from just one

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