If $V \neq L$, i.e. if nonconstructible sets exist, does it necessarily follow that $\omega=\lbrace 0,1,2,3, \ldots \rbrace$ has nonconstructible subsets?
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No. It need not be the case. Suppose we start with $V=L$ and we add a new subset of $\omega_1$ by using functions from countable subsets of $\omega_1$ into $2$. Every countable subset of $\omega_1$ (and so of $\omega$) is in the ground model, however the generic extension of the model is not $L$. |
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