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Let $\kappa$ be an regular uncountable cardinal carrying a $\tau$-structure for some countable language $\tau$. What can be said regarding the existence of ordinals $\alpha <\kappa$ carrying elementary substructures of $\kappa$ ?

This is an (altered) question from an exercise which I am having difficulties solving. The Löwenheim–Skolem theorem doesn't seem to provide any direct insights here.

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    Take an ordinal $\alpha_0<\kappa$, use it to generate an elementary submodel $M_0$; if its universe is an ordinal we are done. Otherwise take $\alpha_1=\sup|M_0|$; reiterate. Prove that by $\alpha=\sup\alpha_n$ is an elementary submodel. (I'm not 100% sure, and I don't have time to check this, so I'm not posting as an answer...) Note that this method [if true] guarantees a *closed and unbounded* collection of elementary structures whose domains are ordinals.2012-11-04
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    Let me know if it works. :-)2012-11-04
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    $\alpha_i \subseteq M_i\subseteq \alpha_{i+1}$, so $\alpha = \bigcup_i M_i$ is a substructure of $\kappa$. ($M_i$ can be chosen to be of card. $max(card(\alpha_i),card(\tau))<\kappa$ by Löw-Skl, so $\sup M_i<\kappa$) For $i, since both $M_i$ and $M_j$ embed elementarily in $\kappa$, $M_i$ embeds elementarily in $M_j$. By a theorem on chains, all $M_i$ emb.el. in $\alpha$. Since for each $a_1,..,a_n\in \alpha$, an index $i$ with $a_1,..,a_n\in M_i$ exists and $M_i$ emb.el. in $\kappa$, a formula $\varphi(a_1,..,a_n)$ is true in $\alpha$ iff in $\kappa$. So $\alpha$ el.emb. in $\kappa$.2012-11-04
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    Looks good. Now you should post this as an answer and accept it (note that you can only accept your own answer after two days or so, in the meantime other people slightly more fluent in model theory than me could also verify that you didn't make any mistakes there).2012-11-04
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    I'm not sure if I qualify as more fluent, but it seems okay to me. It could be perhaps made simpler if you skolemized $\kappa$ at the beginning (then you would have q.e. so all the steps (proofs of elementary inclusions, constructions of elementary extensions) would be completely automatic), but it's up to you whether or not to see it that way.2012-11-05

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Hint: Take an ordinal $\alpha_0<\kappa$, use it to generate an elementary submodel $M_0$; if its universe is an ordinal we are done. Otherwise take $\alpha_1=\sup|M_0|$; reiterate. Prove that $\alpha=\sup\alpha_n$ is an elementary submodel.