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In his talk A Condensed History of Condensation, Welch presents the following recursive sharp function, that is total when all sharps exist:

\begin{align*} \# \colon ON &\to \mathcal{P}(ON) \\ \alpha &\mapsto (\#\restriction \alpha)^\# \end{align*}

My question is, why does the co-domain need to be $\mathcal{P}(ON)$, and isn't restricted just to $\mathcal{P}(\omega)$?

I get that maybe for bigger sets $A$ the class of indiscernibles of $L[A]$ will vary, but after all its sharp will always be a set of sentences of the theory of $L[A]$. And isn't this theory always countable (and thus a member of $\mathcal{P}(\omega)$ by Gödel Numbering)? Isn't the only additional symbol of the language of that theory one predicate symbol interpreted as $A$, no matter how big the members of $A$?

I can also frame my question in a more straight forward way. Consider instead the function: \begin{align*} F \colon \mathcal{P}(ON) &\to \mathcal{P}(ON) \\ x &\mapsto x^\# \end{align*}

Assuming all sharps exist, so that $F$ is total, how com some $x^\#$ are not members of $\mathcal{P}(\omega)$?

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    $\begingroup$ I think the point is that for $L[A]$, $A$ a set of ordinals, the $#$ of $A$ is a collection of sentences in the language of set theory, with constants $c_\delta$ added, for $\delta < \sup(A)$. Hence it becomes a subset of $sup(A)$. $\endgroup$
    – Philip Welch
    Jul 24, 2022 at 10:11
  • $\begingroup$ @PhilipWelch Thank you very much for your comment, Philip! Yes, thanks to discussion with Bagaria (my supervisor) I realized exactly what you mention. In case it is of any interest to you, here's my Thesis, from which the question arised. In it (Chapter 3), I relate your results on Acceptability to some basic tools for the study of Gaps in Inner Models. $\endgroup$
    – Martín S
    Jul 28, 2022 at 15:18

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