Inaccessibility in L vs. Inaccessibility in ZFC

Question

We know that we can prove GCH using the axiom of constructibility, and so that implies the the continuum is regular and uncountable, and therefore weakly-inaccessible. However, is this "inaccessible" in the same way that, say, measurable cardinals are inaccessible in ZFC? Or is this just a definition with no actual "inaccessibility" meaning in L?

Answer 1

The continuum is not a limit cardinal if you assume that $\sf GCH$ holds. It is $\aleph_1$. An inaccessible cardinal is a limit cardinal which is also regular. As far as cardinals go, $\aleph_1$ is one of the most accessible cardinals.

Measurable cardinals are inaccessible because we can prove that if there is $\lambda<\kappa$ such that $2^\lambda\geq\kappa$, then $\kappa$ is not measurable, and similarly if $\kappa$ is singular then it cannot be measurable. These combined imply that a measurable cardinal is inaccessible, and in fact strongly inaccessible even if we do not assume $\sf GCH$.

The place where $\sf GCH$ plays a role when it comes to inaccessibility is weak and strong inaccessibility: a regular limit cardinal is weakly inaccessible, but if we assume $\sf GCH$ (e.g. in $L$), then every limit cardinal is a strong limit cardinal, so every weakly inaccessible is in fact strongly inaccessible.

Without assuming $\sf GCH$ it is consistent that the continuum is weakly inaccessible (assuming that inaccessible cardinals are at all consistent).

The key point here is that at some point in the large cardinal hierarchy, we run into properties which are strong enough to prove strong inaccessibility even when $\sf GCH$ is not assumed, like measurability, whereas some properties are just not strong enough, like just being a regular limit cardinal.