# When does $V=L$ becomes inconsistent?

In a wonderful course I'm taking with Magidor we are finishing the proof of the Covering Theorem for $L$.

The theorem, in a nutshell, says that $V$ is very close to being $L$ if and only if $0^\#$ does not exist.

It is consistent that no large cardinals exist and $V\neq L$, and it is consistent that there are inaccessible, Mahlo, weakly-compact, ineffable and so on and yet $V=L$.

On the other hand, if there exists a measurable cardinal then $V\neq L$, as Scott tells us, and in fact a Ramsey cardinal is enough for that. This way we slowly fine tune the demands until we reach the definition of $0^\#$.

My question is, basically, if there is a weaker notion than $0^\#$ in terms of consistency of large cardinal axioms, which is inconsistent with $V=L$?

I am aware of the theorem that a supercompact implies $\forall A(V\neq L[A])$. Are there any other ways to measure how far are we from $L$ or $HOD$ or even $HOD[A]$ for some or all $A$?

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It's an amazing thing that you ask a question on MO which I can perfectly understand, and a few minutes later you ask a question here, in which you lose me right after the title. Usually it's the other way around... – t.b. May 18 '11 at 21:23
@Theo: :-) it is simple, this question might appear in some variation in a grad-level course about large cardinals. The one from MO will likely never to appear anywhere and would require much more experience with both researching archives of data as well non-AC intuition. – Asaf Karagila May 18 '11 at 21:30
@Theo: Also I knew there was no chance to get the other one answered here. I'm trying to somewhat enrich the set-theory specific tags here. In this case, [large-cardinals] :-) – Asaf Karagila May 18 '11 at 21:42
No worries, I didn't doubt your judgment. I just couldn't resist. – t.b. May 18 '11 at 21:49
The fact that the existence of a measurable cardinal implies $V\neq L$ is due, I believe, to Dana Scott, rather than to Kunen. – JDH May 19 '11 at 1:44

The borderline seems to be very near the $\omega_1$-Erdős cardinals.

The Wikipedia page on $0^\sharp$ explains it thus:

The existence of $\omega_1$-Erdős cardinals implies the existence of $0^\sharp$. This is close to being best possible, because the existence of $0^\sharp$ implies that in the constructible universe there is an $\alpha$-Erdős cardinal for all countable $\alpha$, so such cardinals cannot be used to prove the existence of $0^\sharp$.

In particular, an $\omega_1$-Erdős cardinal is inconsistent with $V=L$, but having $\alpha$-Erdős cardinals for countable $\alpha$ is fine with $V=L$. All the smaller large cardinals in the hierarchy (see the Wikipedia large cardinal list) also appear to relativize down to $L$.

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 Joel, many thanks. What about the additional questions about other "nicely defined" models? – Asaf Karagila May 19 '11 at 8:31

Just a supplement to Joel's answer:

The exact boundary for existence of $0^\sharp$ was pinned down by Klaus Gloede: $0^\sharp$ exists if and only if there is a cardinal $\kappa$ such that every constructible partition of $[\kappa]^{<\omega}$ has an uncountable homogeneous set.

Gloede, Klaus, Ordinals with partition properties and the constructible hierarchy. Z. Math. Logik Grundlagen Math. 18 (1972), 135–164.

As Joel points out, everything below this on the usual large cardinal hierarchy seems to be compatible with L.

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 Interesting. I will probably have to ask a few questions some folks which are more knowledgeable than me on the topic. Thanks for the reference! – Asaf Karagila Nov 17 '11 at 16:08