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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 $0\in\omega$ :-)2012-01-11

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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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    @Carl: Indeed. This is just a very simple example which is very understandable.2012-01-11