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Let $\sf T$ be a theory that has as axioms every axiom of $\sf ZFC$, and every theorem of $\sf ZFC + [V=L]$ that is neither provable nor disprovable by $\sf ZFC$, whose addition or addition of its negation doesn't increase consistency strength over $\sf ZFC$, and that is consistent with $\sf ZFC + [V \neq L]$.

Are there models of $\sf T$ that satisfy existence of a measurable cardinal?

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    $\begingroup$ You ask two questions, but a negative answer to the first amounts to a positive answer to the second, which makes yes/no confusing. $\endgroup$
    – Joel David Hamkins
    Commented Sep 25 at 16:08
  • $\begingroup$ @JoelDavidHamkins, I'll delete the second. Thanks! $\endgroup$
    – Zuhair Al-Johar
    Commented Sep 26 at 6:00

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The answer is no, you cannot have measurable cardinals consistently with your theory.

Your theory includes the axiom "V=L or V=L[c] for an $L$-generic Cohen real $c$". This statement is provable from V=L, but it is neither provable nor refutable in ZFC. Neither the axiom nor its negation has increased consistency strength (since we could be in some other kind of forcing extension), and it is consistent with V≠L.

But there can be no measurable cardinals in L nor in L[c] for a Cohen-real forcing extension of L.

Upate. In fact, your theory is simply equivalent to ZFC + V=L. The reason is that your theory also includes the statement "V=L or V=L[r] for an $L$-generic random real", for similar reasons as above.

And since no Cohen extension is also a random extension, it follows that these two statements together imply V=L. So your theory is the same as ZFC+V=L.

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  • $\begingroup$ I think the poster only wants to add consequences of $V=L$ that are also consistent with $V\neq L$. $\endgroup$
    – Jason Starr
    Commented Sep 25 at 21:06
  • $\begingroup$ My two axioms have that feature. And yet jointly they imply V=L. $\endgroup$
    – Joel David Hamkins
    Commented Sep 25 at 21:55
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    $\begingroup$ I get it now. Results that are separately consistent with $V\neq L$ need not be jointly consistent with $V\neq L$. $\endgroup$
    – Jason Starr
    Commented Sep 25 at 23:04
  • $\begingroup$ @JasonStarr, yes, to have my original intention I must forbid this joint action of any finite number of axioms to go against $V \neq L$, like adding that any finite conjunction of axioms of $\sf T$ is an axiom of $\sf T$ also. But that would be messy. $\endgroup$
    – Zuhair Al-Johar
    Commented Sep 26 at 5:10

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