How can I determine that some particular sentence (consistent with ZFC) $\phi$ cannot be forced within a model $M$ of ZFC for no notion of forcing $P$ ? Is there a procedure that works for most of the time ? What are natural examples of such $\phi$'s ?
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2$\begingroup$ As far as I recall, $V=L$ is a natural example of a statement that can't be forced (as $L$ is absolute but forcing can only add elements to $V$; but it's been a while, and there might be some subtlety I'm missing that invalidates the argument). $\endgroup$– Daniel ScheplerCommented Dec 9, 2022 at 18:57
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$\begingroup$ @DanielSchepler So by your argument, anything that removes elements from $M$ without violating to be again a model of ZFC cannot be forced ? Is this the unique case possible ? $\endgroup$– user122424Commented Dec 9, 2022 at 19:04
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2 Answers
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You cannot force that $V=L$.
If no large cardinals exist in an inner model, then you cannot force the existence of large cardinals. Although it is possible to "resurrect" large cardinal axioms in some cases, this usually requires them to hold in an inner model (there are some caveats here, of course, we say that $\kappa$ is "Blargh" if $V_\kappa\neq L_\kappa$ and $\kappa$ is strongly inaccessible; then starting from an inaccessible cardinal we can always force the existence of a "Blargh" cardinal; but these sort of counterexamples are artificial).
Some large cardinal axioms are upwards absolute, e.g. the existence of $0^\#$, and we know that there is no (set) forcing that adds $0^\#$ to a model. So if it does not exist, then it will not exist in a generic extension. Similarly, if $0^\#$ does exist, then it will always exist further on, and so cannot be forced to not exist.
More generally, "smallness statements" about canonical inner models, like $0^\#$ or other sharps (which can be understood as "The universe is significantly larger than the canonical inner model in question), cannot be forced into inexistence once they exist, since the canonical inner models are more-or-less-by-definition are robust under generic extensions (there are some caveats, but they are not relevant to this context). At the very least this is true below Woodin cardinals.
Any statement that will ostensibly add ordinals (e.g. change $\omega$ by adding non-standard integers; add ordinals on top of the universe; add non-standard ordinals in between the existing ordinals) will also not be forceable, since forcing does not add new ordinals.
answered Dec 9, 2022 at 20:58
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Shoenfield Absoluteness Theorem says that $\Sigma^1_2$ and $\Pi^1_2$ sentences are absolute, so you cannot force them. An example is the Riemann hypotesis, which can be written as a $\Pi^1_2$ sentence, so you can't force it to be true.
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1$\begingroup$ Do you need for this that $M$ is transitive ? $\endgroup$ Commented Dec 9, 2022 at 19:06
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1$\begingroup$ You're right. But some set-theorists consider even proper class forcing. But your remark may stay true even here, if one looks at $M$ from "above" and not from "inside" ? $\endgroup$ Commented Dec 9, 2022 at 19:11
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1$\begingroup$ Perhaps also that RH doesn't need $\Pi^1_2$ sentence in the full (and hence unnecessarily strong) language of ZFC and much like P=NP is a sentence in the restricted language of Peano Arithmetic interpreted accordingly in ZFC, am I correct with this ? $\endgroup$ Commented Dec 9, 2022 at 19:47
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1$\begingroup$ @user122424: RH is equivalent to a $\Pi^0_1$ statement. $\endgroup$– Asaf Karagila ♦Commented Dec 9, 2022 at 22:41
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4$\begingroup$ Shoenfield absoluteness works between models $M\subseteq N$ of $\mathsf{ZF}$ such that $\omega_1^N\subseteq M$, and $N$ thinks $M$ is transitive. A forcing extension and its ground model always satisfy these conditions, so we cannot change $\Sigma^1_2$-statements by forcing. $\endgroup$ Commented Dec 10, 2022 at 3:49