Assume we showed that, in a certain transitive model of set theory, we have an isomorphism between two structures $M_1$ and $M_2$. Does the same result still holds in the real world?
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2 Answers
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The papers "Forcing isomorphism" and "Forcing isomorphism II" might be relevant.
Review for the first paper: As the authors explain in their introduction, for any theory $T$ it is easy to find two non-isomorphic models of $T$ that become isomorphic via a forcing which collapses cardinals. By contrast, if $T$ is classifiable, two non-isomorphic models of $T$ of cardinal less than lambda remain non-isomorphic after a forcing which preserves cardinals under lambda and adds no countable subset to lambda. In this paper they show that (i) the condition of preservation of cardinals is not sufficient in the result just stated, since the classifiable theory of countably many nested equivalence relations, with binary splitting, has non-isomorphic models which become isomorphic after a ccc-forcing; (ii) all non-classifiable theories have such a pair of models. The basis of the argument is the construction of a family of trees which can be made isomorphic by forcing.
Review for the second paper: This is the second paper justifying Shelah's theory of classification, that is, showing that one cannot do better. As in Part I [J. T. Baldwin, M. C. Laskowski and S. Shelah, J. Symbolic Logic 58 (1993), no. 4, 1291–1301; MR1253923 (94m:03054)], only c.c.c. forcings, which preserve cardinalities and cofinalities, are considered. This time it is shown that, given a small superstable non-$\omega$-stable theory $T$, there exists an extension by forcing of the universe with two non-isomorphic models of $T$, with the property that they can be made isomorphic by forcing; the question of whether such a pair of models must exist in the ground universe is left open. A corollary is that the invariants for the model of a non-$\omega$-stable classifiable theory cannot be substantially simplified; in particular, the models of $T$ cannot be classified by independent trees of finite subsets. In such a theory $T$ there is a type $p$ of infinite multiplicity; each of the two models constructed realizes a suitably chosen generic subset of the set of strong types extending $p$. In a second step new automorphisms of this set are added to make the two models isomorphic. For this construction, a measure on this space of strong types is defined. The last section contains examples and counterexamples, and in particular one showing the fragility of the membership relation in a pseudo-elementary class.
answered Oct 14, 2013 at 11:19
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$\begingroup$ Thanks. Does the same result holds for elementary equivalence rather than isomorphism? $\endgroup$ Commented Oct 14, 2013 at 11:31
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2$\begingroup$ Mohammad, it is even simpler than that: Consider sets as structures in an empty language. Isomorphism is just bijection. Two sets can be in bijection in an outer model without being in bijection in the ground model. Or further: If $M$ is a countable transitive model of (enough) set theory, then $\omega_1^M$ is a countable ordinal yet $M$ thinks it is uncountable. $\endgroup$ Commented Oct 14, 2013 at 15:46
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Being an isomorphism between two (first order) structures is a $\Delta_0$-property, and hence, it is absolute for transitive models.
Thus, being isomorphic is a $\Sigma_1$-property, and it is upwards absolute for transitive models. This means that the answer to the question is "yes".
The papers mentioned in the previous answer show that "being isomorphic" is not downwards absolute.
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$\begingroup$ 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? $\endgroup$ Commented Oct 15, 2013 at 12:52
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1$\begingroup$ 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. $\endgroup$ Commented Oct 15, 2013 at 19:23