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Oct 15, 2013 at 19:23 comment added Noah Schweber Not at all - consider $\omega^{\mathcal{M}[G]}=\omega$ and $\omega_1^{\mathcal{M}[G]}$ as pure sets. Since $\mathcal{M}$ (and hence $\mathcal{M}[G]$) is countable, these sets are "truly" isomorphic; but of course $\mathcal{M}[G]$ does not think they are isomorphic.
Oct 15, 2013 at 12:52 comment added user38200 Thank you. If we have a ctm $\mathcal{M}[G]$ (obtained by forcing from $\mathcal{M}$ which is also a ctm) and we have two structures that are NOT isomorphic in $\mathcal{M}[G]$. Does is necessarily follow they are not isomorphic in the real world?
Oct 14, 2013 at 12:38 history answered Carlos CC BY-SA 3.0