I've been interested in sharps such as $0^\sharp$, $0^{\sharp \sharp}$ and $\mathbb{R}^\sharp$, so I wondered about iterating sharps. Let $n \in \omega$ and $x \in V$. Define $x^{\sharp n}$ like so:

  • If $n = 0$, $x^{\sharp n} = x$.
  • If $n = n' + 1$ for $n' \in \omega$, $x^{\sharp n} = (x^{\sharp n'})^\sharp$.

What is the consistency strength of $\forall n: 0^{\sharp n} \textrm{ exists}$? Obviously it should be above $0^\sharp \textrm{ exists}$, $0^{\sharp \sharp} \textrm{ exists}$, etc.

  • 2
    $\begingroup$ This statement is weaker than "for all reals $x$, $x^\sharp$ exists", which itself is weaker than existence of $\omega_1$-Erdos cardinal. $\endgroup$
    – Wojowu
    May 18, 2022 at 16:57
  • $\begingroup$ Okay, I think I understand. Could you go into any more detail? $\endgroup$
    – Binary198
    May 18, 2022 at 17:02
  • $\begingroup$ Mildly related: mathoverflow.net/q/351913/7206 $\endgroup$
    – Asaf Karagila
    May 19, 2022 at 9:34


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