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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.

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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 $\kappa$.

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  • $\begingroup$ I guessed that there was a provable example. I was just too lazy to think about it. :-) $\endgroup$
    – Asaf Karagila ♦
    Jul 19, 2019 at 14:42
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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.

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  • $\begingroup$ Wow. Any reference for that fact (that $L[u]\vDash``u\text{ is the unique normal measure on }\kappa"$)? $\endgroup$
    – David Fernandez-Breton
    Jul 19, 2019 at 13:34
  • $\begingroup$ I guess Jech should have a chapter about that? $\endgroup$
    – Asaf Karagila ♦
    Jul 19, 2019 at 13:34
  • $\begingroup$ Yep, found it. Thanks a lot! $\endgroup$
    – David Fernandez-Breton
    Jul 19, 2019 at 13:37
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    $\begingroup$ You should probably accept Miha's answer, it's much better than mine. $\endgroup$
    – Asaf Karagila ♦
    Jul 19, 2019 at 14:42
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    $\begingroup$ Oh, I'm not saying my answer is useless. But to quote Shelah, the best thing is to prove something; the second best thing is to disprove something; and if you can't do either, then it's good to at least prove that you cannot disprove something. I gave you the worst option of the three, Miha gave you the second best. $\endgroup$
    – Asaf Karagila ♦
    Jul 19, 2019 at 15:54

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