Inaccessible limit of inaccessibles without Choice

Question

In $ZFC$, every measurable cardinal $k$ is inaccessible, with $k$-many inaccessibles below $k$. Hence a measurable cardinal is an inaccessible limit of inaccessibles.

However, in $ZF$ without Choice, it is consistent that a measurable cardinal be a successor cardinal. Does it make sense to ask:

(1) In $ZF$, is there a least large cardinal that (if it exists) is provably an inaccessible limit of inaccessibles?

(2) What's the role of Choice in proving that a measurable cardinal is inaccessible?

Also, how does one go about determining an answer to (1)?

Answer 1

Well, yes and no.

Measurable cardinals, and a lot of cardinals defined by combinatorial properties can end up being just $\omega_1$, which is not the limit of any cardinals. So in that sense the answer is negative.

However, consider the following notion due to Yair Hayut and myself in an upcoming paper:

$\kappa$ is a critical cardinal if there is are transitive sets $M,N$ such that $V_{\kappa+1}\subseteq M$ and an elementary embedding $j\colon M\to N$ whose critical point is $\kappa$.

Clearly in $\sf ZFC$ critical cardinals are exactly measurable cardinals. But in $\sf ZF$, we can prove that a critical cardinal is an inaccessible cardinal which is the limit of inaccessible cardinals, and so on. So this is much more than just being measurable.

The most famous example of a critical cardinal is a supercompact cardinal in the sense of Woodin, as discussed in his work

Woodin, W. Hugh, Suitable extender models. I, J. Math. Log. 10, No. 1-2, 101-339 (2010). ZBL1247.03110.

As for the role of choice, the issue here is that choice let us construct the Ulam matrix, which tells us that no successor cardinal is measurable; and one can work out through the embedding the measure induces that $\kappa$ is the limit of strongly inaccessible cardinals. Here choice is used again, by arguing that the measure induces an embedding (and this argument can also be used directly to prove that $\kappa$ is a strongly inaccessible cardinal as well).

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Also as a side remark, let me just point out that without choice there are several different definitions for inaccessible cardinals. And one could end up with an "inaccessible limit of inaccessible" for fairly vacuous reasons. So one should be careful when talking about inaccessible cardinals without the axiom of choice.

(In the above answer I have taken this to be the definition that $V_\kappa$ is a model of second-order $\sf ZF$.)