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Say that an inner model $M$ of $V$ is generically saturated if for every forcing notion $\Bbb P\in M$, either there is an $M$-generic for $\Bbb P$ in $V$, or forcing with $\Bbb P$ over $V$ collapses cardinals.

What is the consistency strength of "$L$ is generically saturated"?

If the answer is $0^\#$ exists, is this sort of a general answer for relative constructibility (i.e. $L[A]$ is generically saturated if and only if $A^\#$ exists)? If the answer is negative, what can we conclude from this principle?

Note, Mohammad Golshani remarks (also this), that assuming $0^\#$ exists for every $\kappa$, $\operatorname{Add}(\kappa,1)^L$ collapses $\kappa$ to $\omega$ (in particular, if $\kappa$ is countable in $V$, it is just a Cohen forcing).

So in the presence of $0^\#$ at least we know that a lot of the forcings in $L$ do collapse cardinals, even if they do not admit generics in $V$ (e.g., $\operatorname{Col}(\omega,\omega_1^V)$ cannot admit a generic, although it does collapse cardinals).

(The idea here is to marry Foreman's maximality principle that states that every forcing adds a real or collapses cardinals, with inner model hypothesis-like ideas.)

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  • $\begingroup$ In light of Mohammad's answer, what about the principle, "Whatever statement can be forced is true in an inner model?" Or more specifically, if $\sigma$ is a sentence and there is $\mathbb P$ such that $\Vdash^L_{\mathbb P} \sigma$, then there is $\mathbb Q$ and an $L$-generic $G \subseteq \mathbb Q$ in $V$ such that $L[G] \models \sigma$. $\endgroup$
    – Monroe Eskew
    Commented May 20, 2019 at 8:43
  • $\begingroup$ Monroe, the first one is just Sy's IMH, if my memory serves me right. The point is that I don't want just to say that, I want to say that the universe is sort of saturated over the inner model in the sense that anything that doesn't break cardinality (and power sets, it seems) was already done. $\endgroup$
    – Asaf Karagila ♦
    Commented May 20, 2019 at 9:17
  • $\begingroup$ What about replacing powerset with “every set is countable”? $\endgroup$
    – Monroe Eskew
    Commented May 20, 2019 at 9:25
  • $\begingroup$ What ever do you mean by that, Dr. Eskew? $\endgroup$
    – Asaf Karagila ♦
    Commented May 20, 2019 at 9:25
  • $\begingroup$ “Burn them all!” $\endgroup$
    – Monroe Eskew
    Commented May 20, 2019 at 9:26

1 Answer 1

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The concept is inconsistent.

Consider $\mathbb{P}=Add(\omega, \kappa)_L=Add(\omega, \kappa),$ where $\kappa$ is $(2^{\aleph_0})^+$ of $V$. Forcing with $\mathbb{P}$ over $V$ doesn't collapse cardinals (it is ccc.c. in $V$) and there is no $M$-generic filter for $\mathbb{P}$ in $V$.

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  • $\begingroup$ Ah, of course! Do you think there is an interesting variant there? $\endgroup$
    – Asaf Karagila ♦
    Commented May 20, 2019 at 8:37
  • $\begingroup$ @Yair: Yeah, that's what I meant. $\endgroup$
    – Asaf Karagila ♦
    Commented May 20, 2019 at 9:15
  • $\begingroup$ @AsafKaragila I have no idea $\endgroup$
    – Mohammad Golshani
    Commented May 20, 2019 at 11:44
  • $\begingroup$ I imagine something like "preserving the continuum function" maybe? $\endgroup$
    – Asaf Karagila ♦
    Commented May 20, 2019 at 11:45
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    $\begingroup$ I just may say the exact version of Stanley's theorem. Suppose $0^\sharp$ exists. if $\mathbb{P} \in L$ is non-trivial and uniform and if $\kappa$ is the least cardinal such that forcing with $\mathbb{P}$ over $L$ adds a fresh subset of $\kappa$, then forcing with $\mathbb{P}$ over $V$ collapses $\kappa$ to $\omega.$ $\endgroup$
    – Mohammad Golshani
    Commented May 20, 2019 at 11:50

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