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As we know the assumption $V=L$ adds a restriction on the height of the large cardinal tree. Also there is a strict border like $0^{\sharp}$ exists such that all large cardinal axioms which are equivalent or stronger than this axiom are contradictory with $V=L$ and all large cardinal assumptions below it are consistent with $V=L$.

Question: What is the situation for the weaker assumption $V=HOD$ and large cardinal tree? Precisely:

(a) Which one of the known large cardinal axioms (together with $ZFC$) does imply $V\neq HOD$?

(b) Is there any strict border in this case as same as "$0^\sharp$ exists" and "$V=L$"?

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Is the statement "all large cardinal assumptions below [$0^\sharp$] are consistent with $V=L$" intended just as a statement about the widely studied large cardinal axioms, or is it intended as a general principle (presumably based on some definition of what a "large cardinal axiom" is)? – Andreas Blass Dec 13 '13 at 21:54
See . – Andreas Blass Dec 13 '13 at 21:56
@AndreasBlass: Your question produced a question for me! Is it unknown that each (discovered and undiscovered) large cardinal axiom with strictly weaker consistency strength than $0^{\sharp}$ exists is consistent with $V=L$? – user43940 Dec 13 '13 at 22:37
(A boring tree, being linear...) – Andrés Caicedo Dec 14 '13 at 2:39
up vote 13 down vote accepted


There is no such border, because almost all the large cardinal properties, including the very strongest large cardinal axioms, are relatively consistent with $V=\HOD$. For the larger large cardinals, this is generally proved by forcing, and there are several natural ways to force $V=\HOD$.

One such way to force $V=\HOD$ is to force so as to code every set in the pattern of GCH on the cardinals. Basically, one undertakes a forcing iteration that at each cardinal stage decides generically whether to force the GCH at that cardinal or not; in other words, one uses the lottery sum of two posets, one of which forces the GCH there and the other of which forces a failure of GCH there. In the extension, a simple density argument shows that every set is coded into the GCH pattern, and so we get $V=\HOD$. (The general idea of this kind of coding is due originally to McAloon, although he used a different method. Many authors describe the coding with bookkeeping functions, but this is not actually needed, since the generic coding means that it is dense that any new set is coded.) The forcing is somewhat easier when the $\text{GCH}$ holds in the ground model, and in the general case one wants to space out the cardinals at which coding occurs. The assertion that every set is coded into the GCH pattern is named as the continuum coding axiom (CCA) in the dissertation of Jonas Reitz, who proved that this implies the ground axiom (GA). You can find further uses of the CCA in Set-theoretic geology, which also contains full details of this kind of forcing and variations.

Now, the point is that most of the usual large cardinal axioms are preserved by the $V=\HOD$ forcing, by the usual lifting arguments as for GCH. Thus, all the main large cardinals are relatively consistent with $V=\HOD$.

There are many other coding methods besides coding into the GCH pattern. Andrew Brooke-Taylor, for example, undertook coding in the $\Diamond^*_\kappa$ pattern, which allows one to force $V=\HOD+\text{GCH}$ while preserving all the usual large cardinal notions. His paper Large Cardinals and Definable Well-Orderings of the Universe exactly fits the theme of this question. See also my paper The wholeness axiom and $V=\HOD$, which is about precisely this problem in the case of the Wholeness axiom and related large cardinals.

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Let me further add that we expect the pattern to persist, to the point that the canonical inner models for large cardinals are explicitly built as subclasses of $\mathsf{HOD}$. There is also another comment that may be useful: There are "global" large cardinals (such as supercompactness) but we have local versions for all (such as: There is an inaccessible $\kappa$ such that in $V_\kappa$ there is a proper class of supercompact cardinals). For any such (local) cardinal, showing its consistency with $V=\mathsf{HOD}$ is straightforward via the GCH coding starting at a cardinal sufficiently high. – Andrés Caicedo Dec 14 '13 at 2:47
(I really like the lottery, by the way, it really simplifies many arguments.) – Andrés Caicedo Dec 14 '13 at 2:52

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