This is a repost of the [same question on math.SE](https://math.stackexchange.com/questions/4227470/unbounded-set-in-vg-has-an-unbounded-subset-in-v), which received several comments but no answers/comments on the first question.




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Suppose $\kappa$ is a cardinal preserved in the generic extension $V[G]$. Let $Y \subseteq \kappa$ be an unbounded set in $V[G]$. Does there always exist an $X \in V$ such that $X \subseteq Y$ and $X$ is unbounded?


This question comes from Section 18 of James Cummings' [*Singular Cardinal Arithmetic*](https://projecteuclid.org/journalArticle/Download?urlId=10.1305%2Fndjfl%2F1125409326). In this context, he forced with the Levy collapses $\operatorname{Col}(\omega,\delta^{+\omega}) \times \operatorname{Col}(\delta^{+\omega+2},<\kappa)$, and fixed "$X \subseteq \gamma$ unbounded of order type $\delta\_V^{+\omega+1}$...". In the next line, he then says that "However, it is easy to see that there is $Y \in V$ with $Y \subseteq X$ unbounded".


This observation is certainly not easy for me, so my questions are:


1. Why do such a $Y$ exist?
2. Is the existence of such a $Y$ limited to forcing with Levy collapse?


Any help is appreciated.


**Note:** Andreas Blass provided a counterexample to the second question, which is a Mathias real.



Unbounded set in $V[G]$ has an unbounded subset in $V$?
Let $\mathbb{P} = \textsf{Col}(\delta^{+ \omega + 2}, < \kappa)$ and $\mathbb{Q} = \textsf{Col}(\omega, \delta^{+ \omega})$. Then $\mathbb{P}$ is $< \delta^{+ \omega + 2}$-closed so it doesn't add any new set of ordinals of order type $\leq \delta^{+ \omega + 1}$. Put $V\_1 = V^{\mathbb{P}}$ and note that all cardinals $\leq \delta^{+ \omega + 2}$ are preserved in $V\_1$. Suppose $V\_1^{\mathbb{Q}} \models \mathring{Y}$ has order type $\delta^{+ \omega + 1}$. Since $|\mathbb{Q}| < \delta^{+ \omega + 1}$, we can find a condition $p \in \mathbb{Q}$ such that $p$ forces $ \delta^{+ \omega + 1}$ many ordinals into $\mathring{Y}$. Put $X = \{\alpha: p \Vdash\_{\mathbb{Q}} \alpha \in \mathring{Y}\}$. Then $X \in V\_1$ and therefore by the previous observation, $X \in V$.


