6
$\begingroup$

In the Jensen-Kunen manuscript on combinatorial principles, they define the notion of a subtle cardinal:

Definition. A regular cardinal $\kappa$ is subtle if for all $C$ club, $(A_\alpha)_{\alpha\in C}$ a sequence such that $A_\alpha\subseteq\alpha$ for all $\alpha$, there exists $\alpha,\beta\in C$ such that $\alpha<\beta$ and $A_\alpha=\alpha\cap A_\beta$.

They then proceed to show that $\lozenge(\kappa)$ holds for $\kappa$ subtle with the following argument:

By induction on limit $\alpha<\kappa$ define $(S_\alpha,C_\alpha)$ such that $S_\alpha\subseteq \alpha$, $C_\alpha$ club in $\alpha$ and $\forall\gamma\in C_\alpha\;S_\gamma\ne\gamma\cap S_\alpha$ if possible.

Claim: Let $S\subseteq\kappa$, then $\{\alpha<\kappa\;\vert\;S\cap\alpha=S_\alpha\}$ is stationary.

Proof: Suppose not. Then there is a pair $(S,C)$, $C$ club in $\kappa$ and $\forall\alpha\in C\;S_\alpha\ne\alpha\cap S$.

Let $C^*$ be the set of limit points in $C$. Then $a\in C^*\to\forall\gamma\in C_\alpha\;S_\gamma\ne\gamma\cap S_\alpha$.

By the subtlety of $\kappa$, however, there are $\alpha,\beta\in C^*$ such that $\alpha<\beta$, $C_\alpha=\alpha\cap C_\beta$, $S_\alpha=\alpha\cap S_\beta$. Hence $\alpha\in C_\beta$. Contradiction!

I'm trying to understand this proof but have a few questions. First of all, how do we know that $C_\alpha$ and $C_\beta$ are defined? The definition of $(S_\alpha,C_\alpha)$ says "if possible" so what if at $\alpha,\beta$ it is not possible to find these sets?

Secondly, the definition gives us a way to find $\alpha,\beta$ such that $A_\alpha=\alpha\cap A_\beta$, but how does one do this for $(S_\alpha,C_\alpha)$ simultaneously as in the proof? Would it be necessary to code $\kappa\times\kappa$ with $\kappa$ by some bijection?

$\endgroup$
3
  • 1
    $\begingroup$ No wonder you were confused about the proof. What you had written was not the construction, Jensen had in mind. I corrected it and will soon provide an answer to your question. $\endgroup$ Mar 8, 2016 at 23:33
  • 1
    $\begingroup$ @Stefan Sorry, I typed too quickly and made a few typos. Thanks for the corrections. $\endgroup$
    – Marc
    Mar 8, 2016 at 23:37
  • $\begingroup$ I will wait for Stefan's correction. There is nothing in what I see in the first sentence of the proof that prevents $\forall \alpha \in \kappa \; (S_{\alpha}=0).$ $\endgroup$ Nov 3, 2016 at 3:07

2 Answers 2

5
$\begingroup$

Regarding your first question: Say that we've constructed $((S_\alpha, C_\alpha) \mid \alpha < \beta)$ for some $\beta < \kappa$. There are two cases:

$(\dagger)$If there is some $S \subseteq \alpha$ such that $\{ \beta < \alpha \mid S_\beta \neq S \cap \beta \}$ is non-stationary, we may fix a club $C \subseteq \alpha$ such that for all $\beta \in C \colon S_\beta \neq S \cap \beta$. In this case let $C_\alpha := C$ and $S_\alpha := S$. (We need some choice to pick these $C$'s and $S$'s at every stage.)

Otherwise let $C_\alpha := S_\alpha := \alpha$.

Before we repeat the proof with more details, let us verify the following:


Lemma. Let $\kappa$ be subtle. Then for every club $C \subseteq \kappa$ and for every sequence $( (A_\alpha, B_\alpha) \mid \alpha < \kappa)$ such that $A_\alpha, B_\alpha \subseteq \alpha$ there exist $\alpha < \beta$ such that $\alpha, \beta \in C$, $A_\alpha = A_\beta \cap \alpha$ and $B_\alpha = B_\beta \cap \alpha$.

Proof. For each $\alpha < \kappa$ let $C_\alpha := \{ \langle x, y \rangle \mid x \in A_\alpha, y \in B_\alpha \} \cap \alpha$, where $\langle ., . \rangle \colon \operatorname{Ord} \times \operatorname{Ord} \to \operatorname{Ord}$ is the Gödel pairing function. Fix a club $D \subseteq \kappa$ such that for all $\alpha \in D$ we have $x,y < \alpha \to \langle x,y \rangle < \alpha$. (Since $\kappa$ is regular, such a club exists. You can use the normal function theorem to prove this.)

Now $E := D \cap C$ is a club and since $\kappa$ is subtle, there exist $\alpha < \beta$ such that $\alpha, \beta \in E$ and $C_\alpha = C_\beta \cap \alpha$. Since $\alpha$ is closed under the Gödel pairing function, we may recover all of $A_\alpha$ and all of $B_\alpha$ from $C_\alpha$. In fact, we have $$ A_\alpha = \{ x < \alpha \mid \exists y < \alpha \colon \langle x,y \rangle \in C_\alpha \} $$ and similarly for $B_\alpha, A_\beta, B_\beta$.

Hence $A_\alpha = A_\beta \cap \alpha$ and $B_\alpha \cap B_\beta \cap \alpha$. Q.E.D.


Let $((S_\alpha, C_\alpha) \mid \alpha < \kappa)$ be the sequence that we've constructed above.

Claim. $(S_\alpha \mid \alpha < \kappa)$ witnesses $\Diamond_\kappa$.

Proof. Suppose not. Then there is some $A \subseteq \kappa$ such that $\{ \alpha < \kappa \mid S_\alpha = A \cap \alpha \}$ is not stationary. Fix a club $C \subseteq \kappa$ such that for all $\alpha \in C \colon A \cap \alpha \neq S_\alpha$ and let $C^*$ be the club of all limit points of $C$. If $\alpha \in C^*$, then $C \cap \alpha$ is a club in $\alpha$ and for all $\beta \in C \cap \alpha \colon A \cap \beta \neq S_\beta$. Therefore, in our construction, we could have picked $C_\alpha := C \cap \alpha$ and $S_\alpha := A \cap \alpha$. In particular, for all $\alpha \in C^*$ we are in the "first case" $(\dagger)$ of our construction.

By the Lemma above, we may now pick $\alpha < \beta$, $ \alpha, \beta \in C^*$ such that $C_\alpha = C_\beta \cap \alpha$ and $S_\alpha = S_\beta \cap \alpha$. Since $C_\alpha$ is unbounded in $\alpha$ and $C_\beta$ is closed, we have $\alpha \in C_\beta$. But $S_\beta \cap \alpha = S_\alpha$ - contradicting the fact that we have been in the "first case" $(\dagger)$ of our construction at stage $\beta$. Q.E.D.

$\endgroup$
6
  • $\begingroup$ The Lemma above is pretty useful in these sorts of arguments. It shows that we can code $\kappa \times \kappa$ into $\kappa$ in a way such that the same coding also codes $\alpha \times \alpha$ into $\alpha$ for club many $\alpha < \kappa$. $\endgroup$ Mar 9, 2016 at 0:20
  • $\begingroup$ Why is it that "if $\alpha\in C^*$ then $C\cap\alpha$ is a club in $\alpha$"? $\endgroup$
    – Marc
    Mar 9, 2016 at 0:22
  • 1
    $\begingroup$ @Marc $C \cap \alpha$ is certainly closed and since $\alpha$ is a limit point of $C$, it is also unbounded. $\endgroup$ Mar 9, 2016 at 0:22
  • $\begingroup$ (And yes, that's exactly the argument I had in mind when I said some coding of pairs of subsets.) $\endgroup$
    – Marc
    Mar 9, 2016 at 0:22
  • $\begingroup$ That's exactly what I've been missing. Thanks! $\endgroup$
    – Marc
    Mar 9, 2016 at 0:24
-2
$\begingroup$

The proof of the lemma in the answer fails if $A_\alpha$ or $B_\alpha$ is empty; then reconstruction from $C_\alpha$ fails. A simpler argument works. Take the club consisting of $\alpha$ such that for all $\beta \leq \alpha$, $2\beta+1 < \alpha$. Then let $C_\alpha = \{2 \beta:\beta \in A_\alpha\}\cup \{ 2 \beta+1:\beta \in B_\alpha\}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .