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Let $P$ be a statement saying there is large cardinal of some kind. For example, $P$ can be one of

  1. There is a weakly inaccessible cardinal.
  2. There is a Mahlo cardinal.
  3. There is a weakly compact cardinal.
  4. There is a measurable cardinal.
  5. There is a strongly compact cardinal.
  6. There is a supercompact cardinal.

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Then, does $\operatorname{Con}(\mathrm{ZFC}+P)$ imply $\operatorname{Con}(\mathrm{ZFC}+\mathrm{GCH}+P)$?

Context: Traditionally, that $\mathrm{ZF}$ and $\mathrm{ZFC+GCH}$ has the same consistent strength is proved by Gödel's constructible universe $L$. However, a famous theorem from Scott says if there is a measurable cardinal then $V\ne L$, so this proof cannot work for measurable and above. The Levy–Solovay theorem implies CH is consistent with a measurable, but is there a proof for the situation where GCH holds?

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The construction of $L$ is extended to accommodate large cardinal in the study of inner model theory. The core model for large cardinal axioms provide us with a canonical inner model which captures, in a sense, "exactly the large cardinal of interest", and one of the consequences is that $\sf GCH$ holds in those models.

However, above Woodin cardinals, inner model theory starts to break down, slowly at first, but eventually more and more, to the point that we don't have good canonical inner model for very large cardinals. Woodin is working on his Ultimate-$L$ programme, which will be a canonical inner model for supercompact cardinals that actually captures all larger large cardinals as well, and will satisfy $\sf GCH$ and more. There is still much work there, though.

Still, we do have a different weapon that works just as well. Silver's theorem tells us that we can lift an elementary embedding $j\colon V\to M$ to an embedding $j\colon V[G]\to M[H]$ if and only if $j``G\subseteq H$, so we can set $j(G)=H$.

This allows us to force $\sf GCH$ while preserving large cardinal axioms such as supercompact cardinals which do not have good canonical inner models.

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  • $\begingroup$ Thank you for your answer. Can you provide a reference for how lifting elementary embeddings enables us to force GCH with large cardinal axioms? $\endgroup$
    – fantasie
    Mar 4 at 12:05
  • $\begingroup$ Also, is this an affirmative answer for all large cardinal axioms listed above, or just supercompact? $\endgroup$
    – fantasie
    Mar 4 at 12:08
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    $\begingroup$ @fantasie As Asaf said, any large cardinals that have an inner model theory will be consistent with GCH. Supercompact cardinals are well-behaved enough to allow GCH to be forced while preserving the large cardinal. However, other large cardinals are trickier. For example, I believe it is not known whether one can force GCH at a strongly compact starting from just the existence of a strongly compact. $\endgroup$ Mar 4 at 18:38

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