# Another question on stationarity

Let $\kappa < \lambda$ be regular cardinals and let $A \subset \lambda$ be such that each $\alpha \in A$ has cofinality $\gamma < \kappa$. Then the following should hold:

$A$ is stationary iff $\lbrace$$X \in P_{\kappa} (\lambda) : sup(X) \in A \rbrace is stationary in P_{\kappa} (\lambda) The direction from the right to the left is not hard, but despite looking harmless, the \Rightarrow direction puzzles me now for a while. It should be enough to show that for any club C in P_{\kappa} (\lambda) the set \underset{\sim}{C}:= \lbrace \beta \in \lambda : \exists X \in C \quad sup(X) = \beta \rbrace has a \gamma-closed, unbounded subset of \lambda which would imply (by the assumption that each element of A has cofinality \gamma) that \underset{\sim}{C} \cap A \ne \emptyset, witnessing the stationarity of \lbrace$$X \in P_{\kappa} (\lambda) : sup(X) \in A$ $\rbrace$. However all my attempts to show this were cumbersome and dissatisfactory.

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Every club subset $C\subset P_\kappa(\lambda)$ contains the collection of size-less-than-$\kappa$ elementary substructures of some first order structure $W=\langle\lambda,\in,\ldots,\rangle$, and the collection of such elementary substructures is club. There is a Skolem function $f:\lambda^{\lt\omega}\to\lambda$ such that any set closed under $f$ is elementary in $W$. The set of ordinals $\beta\lt\lambda$ closed under $f$ is club in $\lambda$, and so there is $\beta\in A$ closed under $f$. Now pick a cofinal subset of $\beta$ and close under $f$ to find an $X$ elementary in $W$ of size less than $\kappa$ with $sup(X)=\beta\in A$. Since $X$ is elementary in $W$, it is in $C$, as desired.
One needs more than just one Skolem function when the language of $W$ is not countable, but the same idea works. Alternatively, one can use a single function $\lambda^{\lt\omega}\to P_\kappa\lambda$, and use the suitable notion of closed-under-$f$. –  Joel David Hamkins Dec 1 '10 at 19:45