12
$\begingroup$

What is the strongest known natural large cardinal axiom compatible with $V = L$ (strongest in the sense that it implies all known "small" large cardinal axioms, where a large cardinal axiom is said to be "small" if it doesn't imply $V \neq L$)?

One candidate might be "there is an $\alpha$-Erdos cardinal for every countable ordinal $\alpha$."

Using the ``instrumentalist dodge'' of Steel/Hamkins, it seems that we can "cheat" to obtain stronger large cardinal axioms within the confines of $\operatorname{ZF}+(V = L)$, such as "there is a transitive model of the theory ZFC+"$0^\sharp$ exists."" I will leave whether or not such axioms are "natural" for a matter of debate here.

Disclaimer: I'm an algebraist and number theorist but am currently trying to grasp the big picture in set theory, large cardinals, inner model theory, and the like.

Related MO posts:

Is there a large-cardinal completeness theorem for $L$?

If $0^{\sharp}$ exists, then every uncountable cardinal in $V$ is as large as it can be in $L$.

Erdős cardinals and $0^\sharp$

$\endgroup$
  • 3
    $\begingroup$ Your suggestion is essentially it. The modern presentation of $0^\sharp$ is as what Schimmerling calls a baby mouse, this is a structure $J_\tau$ with an external measure that is iterable (and some technical definability requirements). Asking that for each $\alpha<\omega_1$ there is such a structure, but only $\alpha$-iterable, seems as strong as possible. But this is in essence the same as considering $\alpha$-Erdős cardinals. $\endgroup$ – Andrés E. Caicedo Dec 13 '14 at 1:23
  • $\begingroup$ Another way to cheat and get slightly stronger large cardinals: it is consistent that there is an ordinal $\alpha$, $V=L$ and in some generic extension (by a set forcing notion) there is an $\alpha$-Erdos cardinal, even when $\alpha$ is not countable in $L$. $\endgroup$ – Yair Hayut Dec 13 '14 at 15:53
10
$\begingroup$

Maybe a comment:

In the paper ``A large cardinal in the constructible universe'' Silver shows that if $\kappa\to (\alpha)^{<\aleph_0}$ for all countable $\alpha,$ then the same is true for $\kappa$ in $L$.

On the other hand, by a result of Rowbottom, $\kappa \to (\omega_1)^{<\aleph_0} $ contradicts $V=L.$

Silver concludes with the following:

It does not seem extravagant, then, to assert that, for all practical purposes, $\kappa \to (\omega)^{<\aleph_0}$ is the strongest strong axiom of infinity know to be consistent with $V=L,$ and therein lies its chief interest.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.