# Can we always add sets without collapsing cardinals or adding [very] bounded sets?

Given a model of $\sf ZFC$, and an infinite ordinal $\alpha$. Can we prove that there is always a cardinal $\kappa$, and a forcing $\Bbb P$, such that:

1. $\Bbb P$ does not add sets of rank $\leq\alpha$.
2. $\Bbb P$ adds sets to $\kappa$.
3. $\Bbb P$ does not collapse cardinals.

Any two of the three can be easily done. 1+2 can be done with $\operatorname{Add}(\kappa,1)$ which may collapse $\kappa^+$; 1+3 is satisfied by the trivial forcing; and 2+3 is satisfied by adding a single Cohen real.

Satisfying all three would violate Foreman's Maximality Principle (any nontrivial forcing adds a real or collapses cardinals). But the consistency of the principle is an open problem.

Assume that $V$ and $L$ are "close enough" (e.g. $0^\#$ does not exist, or $V$ is set generic over $L$, or something of this sort). Can we prove that such $\kappa$ and $\Bbb P$ exist?

• I think it is still open whether "Foreman's maximality principle" is consistent, which states that every nontrivial forcing either adds reals or collapses cardinals. – Monroe Eskew Jul 29 '15 at 21:02
• I think you're right. Which is why I added the fallback question. Since Foreman's MP implies large cardinals, I think. – Asaf Karagila Jul 29 '15 at 21:04
• If $V$ is a set forcing extension of $L$ then GCH holds eventually in $V$. Wouldn't that allow you to take $\mathbb{P}=\mathrm{Add}(\kappa,1)$ for some $\kappa$ large enough? – Miha Habič Jul 29 '15 at 22:40
• Also, I'm not sure if this is true, but I think that $0^\sharp$ not existing implies SCH. If so, if $\lambda$ is the first $\beth$-fixed point above $\alpha$ then $\mathbb{P}=\mathrm{Add}(\lambda^+,1)$ works again, I think. – Miha Habič Jul 29 '15 at 22:48
• @Miha: That works. And indeed $0^\#$ not existing implies that successor of singulars are the same as in $L$ (because they are still singular in $L$), and therefore SCH indeed holds. – Asaf Karagila Jul 29 '15 at 22:52

As was confirmed in the comments, if $0^\sharp$ does not exist then SCH must hold. If we now let $\lambda$ be the least $\beth$-fixed point above $\alpha$, we can consider the forcing $\mathbb{P}=\mathrm{Add}(\lambda^+,1)$. This has $\lambda^{++}$-cc precisely because SCH holds, ensuring $2^\lambda=\lambda^+$. It follows that $\mathbb{P}$ does not collapse cardinals and it clearly does not add sets at rank $\alpha$.
The situation is a bit weird, since all we really needed for the argument is that GCH (or SCH, I guess) holds unboundedly often. So what you are asking for will be true in models which are close to canonical inner models (i.e. things like $0^\sharp$, $0^\dagger$ etc. do not exist) but also in models with many large cardinals (at least a strongly compact).