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I just read a proof and, after struggling some time with a mental leap, I think that it uses tacitly the following:

Let $\kappa$ be a regular cardinal, $\theta > \kappa$ a regular cardinal too then: $ S \subset \kappa$ is stationary if and only if $\forall \mathcal{A} = (H(\theta), \in, <,..) \exists M \prec \mathcal{A}, |M| < \kappa,$ such that $sup(M \cap \kappa) \in S$.

Now my questions are:

  1. Is this statement above even true? (I think so as I have a proof, but this doesn't have to mean anything)

  2. It appears to me that the latter part of this characterization is a quite strong assumption as $\mathcal{A}$ might contain a lot of additional information, so is there a possibility to weaken it? Or could you mention any similar statements to the one above?

Thank you

EDIT: I accepted the answer of Philip, simply because he has lower points. Francois answer would have deserved it too.

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2 Answers

up vote 10 down vote accepted

(I first wanted to give an answer, but I was not quick enough. I then wanted to add a small comment and found out after 20 minutes that I had insufficient reputation.)

The comment was regarding 2) of oktan's original query: having $H(\theta)$ in the structure is overkill: it suffices to have $( \kappa, <, \in, C)$. (One does not need the structure to be able to express $C$ is closed.'')

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Thank you very much! –  Stefan Hoffelner Jun 22 '10 at 18:26
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Philip, you now have sufficient rep now to comment. But isn't "C is closed" expressible in your structure? –  Joel David Hamkins Jun 22 '10 at 18:33
    
Thanks Joel. Your are right about the expressibility. My telegraphic, and poorly put, comment was only that one does not {\em need} that "C is closed" to be internally expressible (which came up in the previous answer). Unboundedness suffices. –  Philip Welch Jun 22 '10 at 19:50
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Yes, the statement is true.

The forward direction is clear since the set $$C_{\mathcal{A}} = \{\sup(M\cap\kappa) : M \prec \mathcal{A}, |M| < \kappa\}$$ is a club. Indeed, let $\langle M_\alpha : \alpha < \kappa \rangle$ be an elementary chain of elementary submbodels of $\mathcal{A}$ with size less than $\kappa$ such that:

  • $\langle M_\alpha : \alpha < \beta \rangle \in M_{\beta+1}$ for every $\beta < \kappa$, and
  • $M_\gamma = \bigcup_{\alpha<\gamma} M_\alpha$ for every limit $\gamma < \kappa$.

Then $\langle \sup(M_\alpha \cap \kappa) : \alpha < \kappa \rangle$ enumerates a closed unbounded subset of $\kappa$ which is contained in $C_{\mathcal{A}}$.

For the converse, let $C \subseteq \kappa$ be a closed unbounded set and consider the structure $\mathcal{A} = (H(\theta),{\in},{<},C)$. If $M \prec \mathcal{A}$ then $M$ satisfies "$C$ is closed unbounded in $\kappa = \sup C$," and so $C \cap M$ is closed unbounded in $\sup(C \cap M) = \sup(\kappa \cap M)$. It follows that $C_{\mathcal{A}} \subseteq C$, where $C_{\mathcal{A}}$ is defined as above. Thus it is sufficient to consider the structures $\mathcal{A}$ as I just described.

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Thank you very much Francois! –  Stefan Hoffelner Jun 22 '10 at 17:19
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Actually, François, the set $C_{\mathcal A}$ contains a club (which is all one needs), but I do not see why it has to be itself a club. –  Andres Caicedo Jun 22 '10 at 17:30
    
Yes I noticed that too, and it does not need to be a club, but as you already said, containing a club suffices, so it doesn't really bother. –  Stefan Hoffelner Jun 22 '10 at 17:39
    
@Andres: I use club for "contains a closed unbounded set" and I spell out closed unbounded. –  François G. Dorais Jun 22 '10 at 17:43
    
@oktan: The distinction is actually important in the last paragraph. –  François G. Dorais Jun 22 '10 at 17:47
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