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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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In set theory $0\in\omega$ :-) – Asaf Karagila Jan 11 '12 at 16:18
up vote 3 down vote accepted

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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More generally, any forcing poset that is countably closed will not "add any new reals", that is, will not add new subsets of $\omega$. – Carl Mummert Jan 11 '12 at 16:35
@Carl: Indeed. This is just a very simple example which is very understandable. – Asaf Karagila Jan 11 '12 at 16:36

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