It is well known that the Large Cardinal Axiom which asserts the existence of a measurable cardinal number is inconsistent with V=L. Do any of the well-known Large Cardinal Axioms contradict V=OD?
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asked Aug 11, 2016 at 15:03
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Almost all of the usual large cardinal axioms can be preserved by the reverse-Easton support class forcing to make every set of ordinals coded into the GCH pattern, or the $\Diamond^*$ pattern, and these axioms imply V=HOD.
For example, this is true of the supercompact cardinals by the usual Laver argument, and analogues work with many other large cardinals, including much larger cardinals. Therefore, any smaller cardinal notion that is a consequence of supercompact cardinal or these other large cardinals will also be relatively consistent with V=HOD.
Meanwhile, there are some large cardinal notions that imply V=HOD outright. For example:
Theorem. If there are a proper class of Laver-indestructible supercompact cardinals, then V=HOD.
Proof. If $\kappa$ is supercompact and Laver indestructible, then for any bounded subset $x\subset\delta<\kappa$, there is $<\kappa$-directed closed forcing $\mathbb{Q}$ that codes $x$ into the GCH pattern above $\kappa$. If $\kappa$ is still supercompact after this forcing, then it would be $\Sigma_2$-reflecting in the extension, and the existence of a coding place for $x$ would reflect below $\kappa$ into $V_\kappa$, which was not changed by the forcing. Thus, $x$ is already coded in $V$, and hence every bounded subset of $\kappa$ is coded into the GCH pattern below $\kappa$. Since we have a proper class of such $\kappa$, it follows that every set is coded, and thus V=HOD. QED
It also works with indestructible strong cardinals and many others; really all you need is that the $\Sigma_2$-correctness is indestructible.
answered Aug 11, 2016 at 16:26
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$\begingroup$ The theorem is false with only a bounded number of indestructible supercompact cardinals, since you can add a Cohen set above them all, and this preserves indestructibility while killing V=HOD. $\endgroup$ Commented Aug 11, 2016 at 17:08
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1$\begingroup$ A similar argument shows that the Shelah philosophical attitude toward the GCH --- that it holds in a "random" pattern with all possible behaviors occuring somewhere --- implies V=HOD. $\endgroup$ Commented Aug 11, 2016 at 17:22
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$\begingroup$ Many thanks for this detailed response. It was a surprise for me, since I had expected that most of the Large Cardinal axioms would contradict V=OD outright. It shows how little I know about that area of Set Theory $\endgroup$ Commented Aug 12, 2016 at 18:56
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$\begingroup$ These Large Cardinal Axioms bring many more sets into existence and I thought that quite a few of them would not be Ordinal Definable. $\endgroup$ Commented Aug 12, 2016 at 19:00