Here we choose the definition of "is a cardinal" as there is no surjective map from a smaller ordinal to it.


It's easy to prove that, if $L\_{\alpha+1} \vDash\ \alpha\text{ is inaccessible}$, then $L\_\alpha \vDash ZFC$. Also it's easy to prove that if $L\_\alpha \vDash ZFC$, then α is a cardinal in $L\_{\alpha+1}$. However, I don't know how to prove confinally many cardinals won't collapse at the next step of L.


If the answer is false, then is it true at the least such α?



If $L_\alpha \vDash ZFC$, then do we have $L_{\alpha+1} \vDash \alpha\text{ is inaccessible}$?
Yes. The elements of $L\_{\alpha+1}$ are exactly those subsets of $L\_\alpha$ which are definable from parameters over $L\_\alpha$. But $L\_\alpha\models\mathrm{ZFC}$, so from here we can just use the usual proof that second order ZFC implies inaccessibility. That is:


(i) every bounded subset of $L\_\alpha$ which is in $L\_{\alpha+1}$, is already in $L\_\alpha$, and so $L\_{\alpha+1}$ agrees with $L\_\alpha$ regarding power sets and cardinalities of elements of $L\_\alpha$; that is, for each $x\in L\_\alpha$, we have $\mathcal{P}(x)^{L\_{\alpha+1}}=\mathcal{P}(x)^{L\_\alpha}$, and $\mathrm{card}(x)^{L\_{\alpha+1}}=\mathrm{card}(x)^{L\_\alpha}$; in particular, $L\_{\alpha+1}\models$"$\alpha$ is a strong limit cardinal", and


(ii) for every $\beta<\alpha$ and every $f:\beta\to\alpha$ which is in $L\_{\alpha+1}$, $f$ is bounded in $\alpha$; that is, $L\_{\alpha+1}\models$"$\alpha$ is regular".


