Does every saturated infinite model $A$ have a proper elementary extension B, such that $\lvert A\rvert =\lvert B\rvert$

Question

The Upward Löwenheim-Skolem theorem can give us an elementary extension of the same cardinality, but I am not sure how to prove that this is a proper elementary extension.

Answer 1

This has nothing to do with saturation.

Given an infinite structure $A$ in a language of cardinality $\le\vert A\vert$, let $\hat{A}$ be any proper elementary extension of $A$ (which exists by compactness). Now pick $a\in \hat{A}\setminus A$, and apply downward Lowenheim-Skolem to $\hat{A}$ to get an elementary substructure $A\cup\{a\}\subseteq B\preccurlyeq \hat{A}$ with $\vert B\vert=\vert A\cup\{a\}\vert$ (note that $\vert A\cup\{a\}\vert=\vert A\vert$).

Since $a\in B$ we have $A\subsetneq B$, and by coherence we have $A\preccurlyeq B$. So we're done.