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Suppose CH holds and $\mathbb{P}$ is a poset of size $\omega_1$, such that forcing with $\mathbb{P}$ preserves $\omega_1$. Does forcing with $\mathbb{P}$ preserve CH? If $\mathbb{P}$ is proper then the answer is yes, see this question. Is this true for improper $\omega_1$-preserving posets of size $\omega_1$?

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Meanwhile, if $\omega_1$ is not preserved, then the answer is no, not necessarily, since when you collapse $\omega_1$ to become countable, the size of $2^\omega$ becomes the ground model $2^{\omega_1}$, which can be very large. –  Joel David Hamkins Apr 25 at 19:22

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Yes. Let $X$ be a name for a subset of $\omega$. It can be describe the following way: for every $i<\omega$ $\{p^i_\alpha:\alpha<\omega_1\}$ a maximal antichain, $\{\varepsilon^i_\alpha:\alpha<\omega_1\}$ where $\varepsilon^i_\alpha\in\{0,1\}$ and $p^i_\alpha$ forces $i\in X$ iff $\varepsilon^i_\alpha=1$. In $V[G]$ the function $i\mapsto h(i)$ is an $\omega\to\omega_1$ map where $p^i_{h(i)}\in G$. As $P$ preserves $\omega_1$, the range of $h$ is bounded, i.e., some $p$ forces that $h(i)<\gamma$ for all $i$, for some $\gamma<\omega_1$. Now the structure $\langle p,\langle p^i_\alpha,\varepsilon^i_\alpha:\alpha<\gamma\rangle\rangle$ fully describes $X$.

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