Suppose $M$ is a countable transitive model of $\mathsf{ZFC}$ (maybe more). Suppose $x, y \in {}^\omega\omega \cap M$ and $M \models y = x^\sharp$. Is it true (possible with additional assumptions), that $V \models y = x^\sharp$?

From Kanomori's book, $y = x^\sharp$ is $\Pi_2^1$. However, I can not apply Shoenfield absoluteness since $M$ is countable and does not have the countable ordinals of $V$.

An even stronger general question may be: are there conditions on $M$ or $V$ such that $\Pi_2^1$ absoluteness can hold between the universe and the countable model $M$.

Thanks for any insight on these questions.


1 Answer 1


The answer is no, not necessarily. Countable transitive models can have fake sharps.

To see this, observe simply that the assertion,

$$\hbox{There is a countable transitive model of $\text{ZFC}+x^\sharp$ exists}$$

is itself a $\Sigma^1_2(x)$ assertion, and thus it is absolute to $L[x]$, where the real $x^\sharp$ does not exist. So there will be a countable transitive model $M\in L[x]$ that thinks $x^\sharp$ exists, but it must have a fake $x^\sharp$.

In particular, if there is a transitive model of ZFC+$0^\sharp$ exists at all, then there is such a model inside $L$. And any such model inside $L$ cannot have the real $0^\sharp$.

  • 1
    $\begingroup$ Hmm, "fake sharps" should be called "dulls" or maybe "props". :-) $\endgroup$
    – Asaf Karagila
    Mar 10, 2015 at 5:27

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