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Are the following propositions equivalent ?

  • $\kappa$ is a weakly inaccessible cardinal
  • $\aleph_\kappa = \kappa \land cof(\kappa) = \kappa$
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  • $\begingroup$ Can you please explain why you think this holds? In particular, why the condition $\aleph_\kappa = \kappa$? $\endgroup$ Commented Oct 29, 2019 at 11:29
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    $\begingroup$ The condition $\aleph_\kappa = \kappa$ expresses the fact that $\kappa$ is an aleph fixed point, i.e. a fixed point of the function $\alpha \mapsto \aleph_\alpha$. $cof(\kappa)=\kappa$ expresses the fact that $\kappa$ is regular. I think this could hold because in mathoverflow.net/questions/64955/… I read that the inaccessible cardinals are precisely the regular fixed points of the beth function, so I wonder if there is something equivalent for weakly inaccessible cardinals. $\endgroup$
    – JacquesB
    Commented Oct 29, 2019 at 12:12

2 Answers 2

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The answer is yes.


If $\kappa=\aleph_\kappa$, then $\kappa$ is not a successor cardinal (if $\kappa$ were the successor of $\aleph_\alpha$, then we would have $\kappa=\aleph_\alpha^+=\aleph_{\alpha+1}$ and the successor ordinal $\alpha+1$ is not a cardinal), so it is a weak limit.

When we also assume that $\mathrm{cf}(\kappa)=\kappa$, this will together with the previous imply that $\kappa$ is regular and a weak limit; that is, $\kappa$ is weakly inaccessible.


On the other hand, assume $\kappa$ is weakly inaccessible and let $\kappa=\aleph_\alpha$ for some $\alpha\leq\kappa$. We can see that $\alpha\neq\beta+1$ for any ordinal $\beta$ (otherwise $\aleph_\alpha=\aleph_{\beta+1}=\aleph_\beta^+$ means that $\aleph_\alpha$ is a successor cardinal and thus not a weak limit), thus see that $\alpha$ must be a limit ordinal. Since $\kappa$ is regular and $\alpha$ is a limit ordinal, we have $$\kappa=\aleph_\alpha=\mathrm{cf}(\aleph_\alpha)=\mathrm{cf}(\alpha)\leq\alpha\leq\kappa.$$ So $\alpha=\kappa$, and thus indeed $\kappa=\aleph_\kappa$.

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If πœ… is a regular limit cardinal, let $S$ be the set of cardinals less than πœ….
Then $\sup S= $ πœ… because πœ… is a limit cardinal.
So $S$ is cofinal in πœ…, so $|S|\ge \operatorname{cof}(πœ…)=πœ….$
So $|S|=πœ…,$ since $S\subset πœ….$
So $S$ is $\in$-order-isomorphic to $πœ….$
So $πœ…$ is the $πœ…$-th cardinal,i.e. $πœ…=\aleph_πœ….$

If $πœ…=\aleph_πœ…$ then $πœ…$ is a limit cardinal by def'n of $\aleph_πœ….$
So if also $\operatorname{cof}(πœ…)=πœ…$ then $πœ…$ is a regular limit cardinal.

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