Suppose $\kappa$ weakly inaccesible and $D$ is a normal, $\kappa$-complete and $\kappa$-saturated filter. Does this guarantees that $\{\alpha<\kappa\mid cf(\alpha)=\alpha\}\in D$?

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    $\begingroup$ @AsafKaragila I don't see how it is a club; in fact i think it ain't. Because taking a sequence $\langle \alpha_n :n<\omega\rangle$ with each term regular, obviously $\alpha=\bigcup_{n<\omega} \alpha_n$ cannot be regular. $\endgroup$ Nov 13, 2017 at 3:32
  • $\begingroup$ Oops, that was careless of me. :) I thought I read something else. $\endgroup$
    – Asaf Karagila
    Nov 13, 2017 at 6:48

1 Answer 1


I'll take a crack at it. Let $I$ be the associated normal, $\kappa$-saturated ideal on $\kappa$. Assume for the purpose of contradiction that $\{\alpha<\kappa: cf(\alpha)<\alpha\}\in I^+$ where $I^+$ is the set of $I$-positive sets. Next by normality, we must have that there exists a $\delta$ such that $\{\alpha<\kappa: cf(\alpha)=\delta\}\in I^+$.

Let's then let $X=\{\alpha<\kappa: cf(\alpha)=\delta\}$. So $X\in I^+$. For every ordinal $\alpha\in X$, let's choose by the Axiom of Choice a cofinal sequence $\langle \xi^{\alpha}_{\gamma}:\gamma<\delta\rangle$ in $\alpha$. In addition for every $\gamma<\delta$, let's define the following functions: $f_{\gamma}$ on the set $X$: $f_{\gamma}(\alpha)=\xi^{\alpha}_{\gamma}$. We then show the following claim:

For every $\gamma<\delta$, there exists a set $X_{\gamma}\subset X$ such that $X_{\gamma} \bigtriangleup X\in I$ (that is the symmetric difference is small) and there exists a closure points $\beta_{\gamma}<\kappa$ such that $f_{\gamma}"X_{\gamma}\subseteq \beta_{\gamma}$.

This is true because of the following reason: the functions $f_{\gamma}$ defined above on the set $X$, are all regressive for every $\gamma<\kappa$. Hence if we let $X_{\rho}=\{\alpha\in X: f_{\gamma}(\alpha)=\rho\}$ for $\rho<\kappa$ and for $\gamma<\kappa$ and if we let $Y=\{\rho<\kappa: X_{\rho}\in I^+\}$, then by $\kappa$-saturation we must have that $\vert Y\vert<\kappa$ and thus $sup(Y)<\kappa$. Let's then define $X_{\gamma}=\bigcup_{\rho\in Y} X_{\rho}$ using normality.

Finally using, for every $\gamma<\kappa$, the sets $X_{\gamma}\subseteq X$ such that $X_{\gamma}\bigtriangleup X\in I$ and the closure points $\beta_{\gamma}<\kappa$ such that $f_{\gamma}"X_{\gamma}\subseteq \beta_{\gamma}$, and setting $sup\{\beta_{\gamma}:\gamma<\delta\}=\beta<\kappa$ then we have $\bigcap_{\gamma<\delta}X_{\gamma}\subseteq \beta$, yet by $\kappa$-completeness we must have $\bigcap_{\gamma<\delta}X_{\gamma}\in I^+$, a contradiction.

Therefore we must have that $\{\alpha<\kappa:cf(\alpha)=\alpha\}\in D$.

In fact even more is true, every stationary set $S\subseteq \kappa$ reflects on a set in $D$ and if one assume in addition that $V=L$ then this implies that $\kappa$ must be weakly compact, this is a result of Jensen.

  • $\begingroup$ How can i see that every stationary reflects on some $X\in D$? $\endgroup$ Nov 17, 2017 at 2:05
  • $\begingroup$ I am not anymore so sure that $\bigcap_{\gamma<\delta}X_\gamma\subset\beta$, when i proved i s,howed that $\xi\in\bigcap_{\gamma<\delta}X_\gamma\to f_\gamma(\xi)<\beta$ for every $\gamma<\delta$ and i thought that was it, but clearly it is not. $\endgroup$ Feb 6, 2018 at 18:21
  • $\begingroup$ Forget it, now i remember, i missed one step. $\endgroup$ Feb 6, 2018 at 18:45

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