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I know a cardinal is inaccessible if it is uncountable,regular,strong limit.And we cannot prove its existence in ZFC.but by axiom of choice,every infinite cardinal is an aleph.so can you write the smallest inaccessible cardinal in terms of aleph?

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    $\begingroup$ Sort of. Every strongly inaccessible cardinal is weakly inaccessible, and every weakly inaccessible cardinal is a fixed point of the aleph function. So if $\kappa$ is any inaccessible cardinal, then $\kappa$ equals the aleph number $\aleph_\kappa$. $\endgroup$ – goblin GONE Jun 7 '14 at 13:54
  • $\begingroup$ Is there a reason you keep spawning new accounts? This is you ur third question today about set theory, and it is the third different account you use. $\endgroup$ – Asaf Karagila Jun 7 '14 at 14:24
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Every inaccessible cardinal $\kappa$ is an aleph fixed point, and the limit of aleph fixed points, and the limit of limits of aleph fixed points and so on.

Not only that, because $\kappa$ is regular, it is the $\kappa$-th fixed point.

(And while we're on the subject set, the axiom of choice has nothing to do with inaccessible cardinals also being $\aleph$ numbers; but it has to do with them being strong limit cardinals. The various definitions of an inaccessible cardinal might not be equivalent with some choice, and $\aleph_1$ could have "inaccessible properties" in some failures of the axiom of choice.)

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