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Below, given a formula $\varphi$ which $\mathsf{ZF}$ proves defines a set of reals and an inner model $W$, I'll write "$\varphi^W$" and "$L(\varphi^W)$" for "$\{x:W\models\varphi(x)\}$" and "$L(\{x: W\models\varphi(x)\})$ respectively, and I'll suppress the superscript when $W=V$.

Broadly speaking, I'm curious whether there is an inner model construction which plays an analogous role for $\mathsf{ZF}$ + "There is a Dedekind-finite infinite set of reals" that $L(\mathbb{R})$ does for $\mathsf{ZF+AD}$ - namely, that (assuming large cardinals!) it satisfies the relevant theory, is appropriately canonical and idempotent, and has a forcing-absolute theory.

Here's one way to make this precise. Say that a formula $\varphi$ is anti-Dedekindian iff the following are each provable in $\mathsf{ZFC+LC}$, where $\mathsf{LC}$ is some large cardinal hypothesis currently believed to be consistent (if one wants a specific choice here, a proper class of Woodins seems natural but I'm happy to go higher):

  1. $\varphi$ defines an infinite set of reals.

  2. $L(\varphi)\models$ "$\{x:\varphi(x)\}$ is Dedekind-finite."

  3. $\varphi^{L(\varphi)}=\varphi\cap L(\varphi)$ (so consequently $L(\varphi)^{L(\varphi)}=L(\varphi))$.

  4. For every set-generic extension $V[G]$ and every sentence $\sigma$, in the notation above we have $$L(\varphi)\models\sigma\quad\iff\quad L(\varphi)^{V[G]}\models\sigma.$$ (Of course this is really a scheme of sentences, one for each $\sigma$.)

Even without condition (4) this seems difficult, so I'm also interested in weakly anti-Dedekindian formulas (= satisfying only (1)-(3) above).

Question: is there a (weakly) anti-Dedekindian formula?

EDIT: Note that $(3)$, as far as I can tell, prevents tricks like "use $0^\sharp$ to build a canonical generic over $L$" - while this will be forcing-absolute and so satisfy $(4)$, we lose the relevant parameter (here $0^\sharp$) when we pass to the inner model and so $(3)$ breaks.

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  • $\begingroup$ Are there any results of the form “no inner model…”? $\endgroup$
    – user44143
    Commented Sep 29, 2022 at 1:01
  • $\begingroup$ @MattF. Sure! There are various inner model theoretic dichotomies (starting with Jensen's covering lemma) which can be thought of as saying that certain sorts of inner models don't exist. We also have analogues of Scott's theorem that certain large cardinals can't be witnessed in certain fine structural constructions (e.g. the eventual need for long extenders, if memory serves). $\endgroup$
    – Noah Schweber
    Commented Sep 29, 2022 at 1:09
  • $\begingroup$ @MattF. Possibly more germanely (in that it treats "uniformly definable" inner models in a sense), see the discussion starting at the bottom of page 2 of Jensen/Steel's "$K$ without the measurable" re: the optimality of their result. $\endgroup$
    – Noah Schweber
    Commented Sep 29, 2022 at 1:11
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    $\begingroup$ @MattF. “no inner model that is missing a real correctly computes 𝜔1” if AD holds. There are also various results in a similar spirit: (1) if $0^\sharp$ exists, no inner model that misses $0^\sharp$ correctly computes $\omega_2$. (2) (Assuming choice), if MM holds, no inner model that misses a real correctly computes $\omega_2$. [In fact, correctly computing cardinals, and even just correctly computing $\omega_2$, imposes very strong restrictions on inner models. On the other hand, correctly computing $\omega_1$ (under choice) does not really give us much.] $\endgroup$
    – Andrés E. Caicedo
    Commented Nov 11, 2022 at 0:08
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    $\begingroup$ @AndrésE.Caicedo: I see! There's no generic for that collapsing in $L[0^\#]$, and if there was, say that we were in $L[0^\#,r]$, where $r$ was a real collapsing $\omega_2^{L[0^\#]}$ to be countable, then we haven't computed $\omega_2^V$ correctly. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 11, 2022 at 23:16

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