# Is there is a real $r$ and a countable transitive model $M$ such that $r$ is not in any forcing extension of $M$?

It is a theorem that if $M[G]$ is a generic extension $M$, then for every model $N$ of ZFC with $M \subset N \subset M[G]$, $\ N$ is some generic extension of $M$ (and is, in fact, $M[G\cap D]$ for some complete subalgebra $D$ of the complete algebra $B$ over which $G$ is $M$-generic).

This made me wonder, is there a real $r$ and a countable transitive model $M$ such that $r$ is not in any forcing extension of $M$?

• You should look at a few early papers by Sy Friedman (or his book on class forcing), where he studies similar questions, for instance looking for reals that class forcing cannot add to $L$, but below $0^\sharp$ in the constructibility degrees. Commented Jan 3, 2016 at 6:04

Yes - many reals, such as $0^\#$, cannot be added by forcing to a model in which they do not already exist.

Without going that far, for any countable model $M$ the real coding $M$ is not addable by forcing. (Or, similarly, any real coding a well-ordering of length $\ge Ord(M)$.)

• Thank you. I am still only just learning forcing, and haven't encountered that yet. Commented Jan 3, 2016 at 2:42
• But since the existence of $0^\sharp$ is unprovable, I now wonder if it consistent that there is no such $r$? Commented Jan 3, 2016 at 2:47
• @vhspdfg See my edit. As long as you can prove that countable models exist in the first place, that's enough. Commented Jan 3, 2016 at 2:57
• Ah, of course, should've thought of that. Commented Jan 3, 2016 at 3:00
• @vhspdfg One way of showing $0^\sharp$ is not in $L$ or set generic extensions of $L$ is by noting that the $0^\sharp$-admissible ordinals never eventually agree with the admissible ordinals (since $0^\sharp$ admissible ordinals are $L$-cardinals). With just $\mathsf{ZFC}$, you can add a class generic real with this property. Hence a real which is not set-generic over L. See Sy Friedman's book for the details. Commented Jan 3, 2016 at 6:26

Yes, you cant force the real $0^{\#}$.

• Thank you. I am still only just learning forcing, and haven't encountered that yet. Commented Jan 3, 2016 at 2:42

While not as clever as $0^\#$ or adding ordinals, here is something a bit odd.

If $M$ is a countable transitive model of $\sf ZFC$, then it has a class forcing which adds a real $r$ such that $M[r]\models V=L[r]$. Moreover this can be minimal, so any set in $M[r]$ is either in $M$ or it codes $r$.

Such a real is not set generic over $M$. So as far as set forcing is concerned, it is not in any forcing extension of $M$.

• I think you need some assumptions on $M$ to conclude that this $r$ is not set generic. For instance, if $M\models V=L$, then we can take $r=\emptyset$. Probably a measurable is enough? Commented Jan 3, 2016 at 5:32
• Well, it's true that if $M$ satisfied $V=L$ the forcing could be trivial, but it doesn't have to be. In which case your suggestion fails. Commented Jan 3, 2016 at 5:33
• My comment was motivated by the sentence "such a real is not set generic over $M$," which I read as saying that any coding real would not be set generic over $M$. (Offhand, do you know a condition on $M$ implying that no set forcing extension of $M$ satisfies "$V=L[r]$ for some real $r$"?) Commented Jan 3, 2016 at 5:35
• "Such" as a generic for the class forcing I mentioned. As the off hand remark, if $L$ does not compute exponentiation correctly for a proper class of cardinals is more or less the only thing that comes to mind. Otherwise there is always the chance that by collapsing something large enough you get back to $L[r]$. I think. It's sufficient, that much I'm certain. Commented Jan 3, 2016 at 5:39