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Let inaccessible cardinal mean uncountable regular strong limit cardinal. Consider $\mathsf{ZFC}$ with an additional axiom: For every set $x$ there is an inaccessible cardinal $\kappa$ such that $\kappa\notin x$ (in other words, inaccessible cardinals form a proper class). Let $\lambda_\alpha$ be the $\alpha^{\text{th}}$ inaccessible cardinal. Note that $\alpha\mapsto\lambda_\alpha$ is not a normal function, because $\lambda_\omega\ne\bigcup\limits_{\alpha<\omega}\lambda_\alpha$ (the rhs is singular). Is it possible to prove the function $\alpha\mapsto\lambda_\alpha$ has a fixed point? a proper class of fixed points?

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    $\begingroup$ Not in ZFC + there is a proper class of inaccessibles. Any fixed point will model ZFC + there is a proper class of inaccessibles. It does follow from ZFC + the ordinals are Mahlo, however. $\endgroup$ – GME Feb 23 '16 at 18:09
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No, of course not. If $\lambda$ is the least fixed point of such function, then $V_\lambda$ must satisfy that there is a proper class of inaccessible cardinals without a fixed point of the enumeration.

To see this, simply note that if $\lambda$ is a fixed point it has to have $\lambda$ inaccessible cardinals below it. On the other hand, if $\lambda$ is an inaccessible which is a limit of inaccessible cardinals, then it has to be a fixed point. So this gives you that fixed points are $1$-inaccessible cardinals (or $2$-, depending whether or not $0$- means an inaccessible or not).

If there is a proper class of fixed points, it means that in some sense the ordinals are $2$-inaccessible (or $3$-, depending who taught you how to count). And so on. So from the assumption mentioned by GME that "$\rm Ord$ is Mahlo" you get that there are proper classes of fixed points of every possible order.

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