# Stationary sets and $\kappa$-complete normal ultrafilters

Let $$\kappa$$ be a measurable cardinal, and let $$u$$ be a normal $$\kappa$$-complete ultrafilter over $$\kappa$$. It is a standard easy fact that every closed unbounded set must belong to $$u$$ (notice that normality is key here), and therefore every element of $$u$$ is stationary. My question is about the converse:

Is it the case that, for every stationary set $$S\subseteq\kappa$$, there exists a $$\kappa$$-complete normal ultrafilter $$u$$ over $$\kappa$$ with $$S\in u$$?

Maybe there's an easy argument, but I don't see it. For example, attempting to take an existing $$\kappa$$-complete normal ultrafilter $$u$$ and considering its Rudin--Keisler image $$v=f(u)$$ under a bijection $$f:\kappa\longrightarrow X$$ does give us that $$v$$ is $$\kappa$$-complete and $$X\in v$$, but we may lose normality (e.g., if $$X$$ was not stationary). So I'd like to know if anyone is aware of a way of getting around this difficulty when dealing with a stationary set.

Asaf's answer is totally right, but let me also point out that you don't even have to go to a special model to see that your conjecture fails. The point is that sets in any normal measure on $$\kappa$$ must reflect many properties of $$\kappa$$ itself (since these sets $$X$$ are exactly the ones for which $$\kappa\in j(X)$$, where $$j$$ is the ultrapower map). For example, the set of regular (or inaccessible or Mahlo etc.) cardinals below $$\kappa$$ will be in any normal measure on $$\kappa$$. This means that the set $$S$$ of singular cardinals below $$\kappa$$ (which is stationary) is omitted from all normal measures on $$\kappa$$.
No. Work in $$L[U]$$, the canonical inner model, then $$U$$ is the unique normal measure on $$\kappa$$. Pick any $$S$$ such that $$S$$ and $$\kappa\setminus S$$ are stationary, and then only one of them can be in a normal ultrafilter.
• Wow. Any reference for that fact (that $L[u]\vDashu\text{ is the unique normal measure on }\kappa"$)? Jul 19, 2019 at 13:34