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By definition, a weakly Mahlo cardinal is a weakly inaccessible cardinal $\kappa$ such that $\{\alpha<\kappa : \alpha\text{ regular}\}$ is stationary in $\kappa$. The first requirement can be trivially simplified by asking that $\kappa$ be regular.

I'm interested in the case where $\kappa$ is singular of uncountable cofinality. That is,

Let $\mathrm{cf} \kappa >\omega$ and assume that $\{\alpha<\kappa : \alpha\text{ regular}\}$ is stationary in $\kappa$. Is it weakly Mahlo?

Actually, I'm mostly interested in the first such $\kappa$. I came across this while thinking of an example of a stationary $A\subseteq\kappa$ such that for all regular $\lambda<\kappa$, $A\cap \lambda$ is not stationary in $\lambda$.

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Recall that if $\kappa$ is a limit cardinal, then the cardinals below $\kappa$ form a club; since the cofinality is uncountable, then the limit cardinals in fact form a club. This means that necessarily $\kappa$ has stationarily many inaccessible cardinals below.

If $\kappa$ is singular, then we can transfer this club to a club in its cofinality, say $\mu$. But now look at a club of order type $\mu$, its limit points all have cofinality less than $\mu$. In particular, no regular cardinals can be there, since regular cardinals with cofinality $<\mu$ are all $<\mu$.

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    $\begingroup$ Funny how I fell into the trap of "$\aleph_\alpha$ is singular, then $\alpha<\aleph_\alpha$", not 12 hours ago I talked about $\aleph$-fixed points with my students. $\endgroup$
    – Asaf Karagila
    Commented Jan 9, 2017 at 5:04
  • $\begingroup$ Thank you very much! I'm still chewing the last paragraph. $\endgroup$ Commented Jan 9, 2017 at 5:15
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    $\begingroup$ The point is that if the cofinality is uncountable, then there is a cofinal club of minimal order type. Its limit points all must have even smaller cofinalities. But since the limit points of a club form a club, this means there are no regular cardinals in this club-of-limit-points. $\endgroup$
    – Asaf Karagila
    Commented Jan 9, 2017 at 8:43
  • $\begingroup$ You mean, there are no regular cardinals greater than $\mu$, isn't it? Because the observation doesn't preclude this club-of-limits to contain some weakly inaccessible less than $\mu$. $\endgroup$ Commented Jan 9, 2017 at 14:53
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    $\begingroup$ If $\mu<\kappa$, and there is a club of singular cardinals in $\kappa$, without loss of generality it starts above $\mu$. So this club is entirely of singular cardinals, therefore no inaccessible cardinals (weak or otherwise) can appear there. So the set of regular cardinals is non-stationary. $\endgroup$
    – Asaf Karagila
    Commented Jan 9, 2017 at 15:24

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