How is it possible and how does one prove that the least huge cardinal is less then the least compact cardinal (if both exist) but at the same time huge has higher consistency strength then the compact cardinal?

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    Here is a reason why such results are not implausible: Typically you prove consistency strength results of this kind by showing that an embedding $j\!:V\to M$ associated to a large cardinal $\kappa$ of type $A$ is such that, say, $M_{j(\kappa)}$ or some other set structure is a model of set theory plus the statement that there are large cardinals of type $B$. – Andrés E. Caicedo Sep 19 at 19:34
  • This shows that the existence of cardinals of type $A$ implies the consistency of the existence of cardinals of type $B$ (and usually much more). Note that this says nothing about there being any cardinals of type $B$ in $V$ (rather than in a set-sized structure). – Andrés E. Caicedo Sep 19 at 19:34
  • @Andres: Why are you posting an answer in the comments? :) – Asaf Karagila Sep 19 at 20:09
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    Call $\kappa$ a "silly cardinal" if it is $\aleph_\omega$, but it is regular in $L$. It is not hard to show that if a silly cardinal exists it is much below the least inaccessible cardinal if one exists at all. But the consistency strength of a silly cardinal is that of $0^\#$ which exceeds "inaccessible" by a lot. – Asaf Karagila Sep 19 at 20:12
  • @Asaf It wasn't an answer, it was truly a comment indicating what I said, that "such results are not implausible". – Andrés E. Caicedo Sep 19 at 20:14

A cardinal $\kappa$ is huge if and only if it is uncountable and there is a $\kappa$-complete normal ultrafilter $\mathcal U$ over some $\mathcal P(\lambda)$ such that $\{x\in\mathcal P(\lambda)\mid \mathrm{ot}(x\cap \lambda)=\kappa\}\in\mathcal U$. The immediate advantage of this formulation over the one in terms of elementary embeddings is that it shows that "there is a huge cardinal" is a $\Sigma_2$ statement, that is, its truth can be witnessed in some $V_\alpha$ (in fact, any $\alpha$ large enough to see the relevant sets will be correct about the fact that $\kappa$ is huge).

Being supercompact, or even strongly compact, is not a local property, it does not admit a formulation verifiable in any $V_\alpha$. It also happens that if $\kappa$ is supercompact, then $V_\kappa\prec_{\Sigma_2} V$, which gives us that, if there is a huge cardinal, then $V_\kappa$ thinks that there is one, and therefore there really is a huge cardinal below $\kappa$.

Now, if $\kappa$ is huge then there is a normal ultrafilter $U$ on $\kappa$ such that there are $U$-many $\alpha<\kappa$ such that $V_\alpha\models\mathsf{ZFC}+$''there is a supercompact cardinal". In fact, any $\rho$ such that any such $V_\alpha$ believes supercompact is actually $\mu$-supercompact for all $\mu<\alpha$, but it is not necessarily true supercompact.

In any case, the first result says that the first huge cardinal is strictly smaller than the first supercompact cardinal, should both exist. The second says that, in consistency strength, being huge is much stronger than being supercompact, because it implies the existence of many set models of $\mathsf{ZFC}+$"there is a supercompact cardinal".

All that said, when people talk of "compact cardinals" I would typically understand "strongly compact" rather than "supercompact". Any supercompact cardinal is strongly compact and "usually" both classes of cardinals essentially coincide. However, it is consistent (via an elaboration of a famous result of Magidor) that there are strongly compact cardinals and huge cardinals, and the least strongly compact is (much) smaller than the least huge cardinal, see

MR0550385 (80i:03061). Morgenstern, Carl F. On the ordering of certain large cardinals. J. Symbolic Logic 44 (1979), no. 4, 563–565.

The point is that Magidor proved that it is consistent that the least strongly compact is also the least measurable cardinal. In essence, Morgenstern proved that Magidor's argument preserves huge cardinals, if they exist, but the least huge cardinal is always larger than the least measurable.

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