Is there a forcing extension $M[G]$ of $M$ that adds a new $\omega$-sized subset to $\omega_2$ without adding any new subsets of $\omega$?

Solution 1:

You might want to look into Namba forcing.

This forcing changes the cofinality of $\omega_2$ to be countable, and assuming $\sf CH$ it does that without adding new reals. In particular $\omega_1$ is preserved.

Solution 2:

If you insist that both $\omega_1$ and $\omega_2$ are preserved then this is impossible. The reason is that, since $\omega_2$ is preserved and, in particular, remains regular in $M[G]$, any countable subset $X$ of it in the extension must be bounded. But given a bound $\alpha<\omega_2$, we can fix in $M$ a bijection $f\colon \omega_1\to\alpha$. Then $X\subseteq\omega_2$ is new iff $f^{-1}[X]\subseteq\omega_1$ is new, but as you state (or following a version of Noah's argument), there can be no new subsets of $\omega_1$.

If you allow $\omega_2$ to be collapsed then Asaf's Namba forcing suggestion works (at least under CH). Allowing $\omega_1$ to be collapsed doesn't really make sense, since collapsing it will just add a real.

Interesting things start to happen when you try to replace $\omega_2$ with larger cardinals $\kappa$. If you want cardinals to be preserved, the argument above tells you that $\kappa$ cannot remain regular (or even of uncountable cofinality) in the extension (in particular, this cannot work for any successor $\kappa$). Jensen covering voodoo then says that you should really have something like a measurable running around. But if we have a measurable $\kappa$ then Prikry forcing at $\kappa$ is exactly the type of thing you are looking for: it preserves cardinals, does not add bounded subsets of $\kappa$, but does add an (unbounded) countable subset to $\kappa$.