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

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!

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 $$0^\#$$.

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