0
$\begingroup$

In particular, if $\mathcal{N}$ is an elementary extension of $\mathcal{M}$, can we immediately conclude that if $\mathcal{M} \models T$, then $\mathcal{N} \models T$? Why or why not?

My line of reasoning

We have for all $\mathcal{L}$-formula $\phi$, $\forall \bar{a} \in \mathbb{M}$ and its interpretation in $\mathcal{N}$ $\pmb{k}(\bar{a})$: $$\mathcal{M} \models \phi(\bar{a}) \Leftrightarrow \mathcal{N} \models\phi(\pmb{k}(\bar{a})).$$

Let $T$ be comprised of such $\phi$'s, then we have $\mathcal{N} \models T$ as desired.

Update

My question is a more general version of this question, which gives more context about what $T$ is.

Suppose $T$ has $\forall\exists$-axiomatization, $(\mathbb{I},<)$ is a linear order, and $(\mathcal{M}_i : i \in \mathbb{I})$ is a chain of models of $T$. We show that $\bigcup \mathcal{M}_i$ is a model of $T$.

$\endgroup$
  • $\begingroup$ @RobArthan I reasoned this is immediate as well, but as a novice I don't trust anything I think. $\endgroup$ – chibro2 Feb 13 '15 at 16:50
  • $\begingroup$ Excuse me! The comment I just deleted wasn't helpful. What is $T$? $\endgroup$ – Rob Arthan Feb 13 '15 at 17:08
  • $\begingroup$ Yeah that's exactly my problem, I updated the question with my proof, stating what $T$ is. Please check to make sure it's sound? $\endgroup$ – chibro2 Feb 13 '15 at 17:09
  • $\begingroup$ The question doesn't seem to have changed. $\endgroup$ – Rob Arthan Feb 13 '15 at 17:10
  • $\begingroup$ @RobArthan sorry didn't know you'd respond so quickly! $\endgroup$ – chibro2 Feb 13 '15 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.