$\neg \text{CH}$ within an inner model

Question

I know that $\text{ZFC}$ cannot prove the consistency of $\text{ZFC} + \neg \text{CH}$ through inner models, since $\text{V} = \text{L}$ is consistent with $\text{ZFC}$ and within the constructible universe $\text{GCH}$ holds. But I was wondering whether we could prove it through inner models (of $\text{ZFC}$) by assuming the existence of some large cardinals, like for example measurable cardinals, or by simply assuming directly $\text{V} \neq \text{L}$. I know, of course, that this would be a very suboptimal relative consistency result, but I still wonder if such an inner model could exists and what should it look like.

Thanks!

Answer 1

If $0^\#$ exists, then there is some ordinal $\alpha$ such that $\alpha$ is countable in $V$, but $\alpha$ is inaccessible in $L$. In particular, $\omega_3^L$ is countable.

This means that $\operatorname{Add}(\omega,\omega_2)^L$ has only countably many dense subsets in $V$. So there is some generic filter meeting them. Therefore there is an inner model of $L[0^\#]$ in which $2^{\aleph_0}=\aleph_2$. This can be extended, wildly, as shown by Solovay. There is an inner model of $L[0^\#]$ in which $\sf GCH$ fails on a proper class.

Of course, since a lot of "very complicated" forcings in $L$ are still all countable in $V$, we can replace Cohen reals by essentially any forcing in $L$ which would force that the continuum is below $\alpha$. So in fact any "reasonable" forcing proof over $L$ becomes reality in the presence of $0^\#$ (e.g. Martin's Axiom, etc.)

The same holds if we have a measurable cardinal, etc. since it implies the existence of $0^\#$.

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Let me point out that the large cardinal axioms below $0^\#$ are in general consistent with $V=L$, so they are not useful here.