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This question is similar to a question I asked last year, but I'm not asking for the same thing.


Let $S$ be a set of ordinals, and consider the Levy collapse $\operatorname{Col}(\omega,S)$. Let $G$ be $\operatorname{Col}(\omega,S)$-generic. Must every unbounded subset of $\omega$ in $V[G]$ contain an unbounded further subset in the ground model $V$? I believe the answer is yes, with a proof similar to the linked question, but I am unsure (the proof cannot be brought over verbatim).

I am also interested in the more general question: Is it possible to characterise posets $\mathbb{P}$ such that for all filters $G$ which are $\mathbb{P}$-generic, every subset of $\omega$ in $V[G]$ has an unbounded further subset in $V$?

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No, this does not hold.

Given an infinite set $A$, let $\hat{A}=\{\langle n, a_1,...,a_n\rangle: n\in\mathbb{N}\}$ where $\langle\rangle$ is an appropriate "tupling" operation and $a_i$ is the $i$th smallest element of $A$. Note that any infinite subset of $\hat{A}$ lets us reconstruct all of $\hat{A}$ itself. This means in particular that if $X\in\mathcal{P}^{V[G]}(\omega)\setminus\mathcal{P}^V(\omega)$ then $\hat{X}$ is an infinite set with no infinite subset in $V$. So indeed any forcing which adds a real - including $Col(\omega,S)$ whenever $S$ is uncountable - will add infinite subsets of $\omega$ not containing any infinite subset from the ground model.

In fact, an even stronger phenomenon can occur. Identify an (infinite) set of natural numbers $A$ with its principal function $p_A$ which on input $i$ gives the $i$th smallest element of $A$. Note that if $A\subseteq B$ then $p_A$ grows at least as fast as $p_B$. Now $Col(\omega,2^{\aleph_0})$ adds a function dominating all ground model functions, which is a massive strengthening of "adding a real with no infinite ground model subset." This raises an interesting follow-up question: must $Col(\omega,\omega_1)$ add a function not escaped by any function in the ground model? This takes us into the realm of cardinal chracteristics of the continuum, and see in particular this old question of mine.

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