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Yes. Let $T$ be the theory of an endless discrete order, which is a complete theory. Let $M=\mathbb{Z}\cdot\mathbb{Q}$ consist of $\mathbb{Q}$ copies of the $\mathbb{Z}$ order, which is a countable sa …

The answer is not necessarily. For a counterexample, consider the first-order structure $M$ consisting of a naked set with exactly two points $M=\{a,b\}$, in the first order language having just equ …

Regarding your edit, of course any ultrapower $\Pi\mathbb{Q}/U$ of $\mathbb{Q}$ itself is a nonstandard model of the theory of $\mathbb{Q}$ (in whatever language you choose). So this version of $\math …

Yes, but the existence of such ultrafilters is a large cardinal hypothesis; it is equivalent to the existence of a measurable cardinal. If each $A_i$ is countably infinite and $\cal U$ is countably c …

Under the global choice principle (which asserts that there is a class well-ordering of the set-theoretic universe), then every theory $T$ has rank $0$ as you define it, regardless of the number of mo …

Let me give a few examples. Example 1. Let us work in Gödel-Bernays set theory, and assume that $T\subset {}^{<\text{Ord}}2$ is a proper class tree of height Ord, but there is no cofinal branch. (Th …

Perhaps this is the kind of example for which you are searching. Let's use the logic $L_{\omega_1,\omega}$, which allows for countable conjunctions and disjunctions. Let $A\subset\mathbb{N}$ be any i …

Let me begin by mentioning that the idea of existential closure for models of set theory arises in the context of forcing axioms. Thomas Johnstone and I, for example, discuss this idea in our paper J. …

The answer to your main question is that in ZFC there is always such a proper-class elementary-extension. Theorem. In ZFC, every set-sized model in a set-sized first-order language has a proper-clas …

Your intuition is correct, for we have $L^-=L$. Inside any model $V$ of $\text{ZF}^-$, we can still build $L$. You don't need the power set axiom to construct $L$. And furthermore, the resulting inne …

A model of ZFC can indeed have two satisfaction classes. This is a result going back to Krajewski 1974, and you can find an account of it and some and some other related matters in my paper, J. D. Ham …

Yes. If they are not isomorphic, then for each bijection of $A$ with $B$, there is an atomic formula that the bijection does not respect, that is, a reason that it is not an isomorphism. Since there …

A model $M$ of a theory $T$ is existentially closed with respect to that theory, if for any quantifier-free formula $\varphi$ and any objects $\vec a$ in $M$, if there is model $N$ of the theory $T$ e …

The answer is not necessarily. Let me describe how to make a counterexample. First, notice that your requirement that $\varphi(M_i)$ is cofinal in $\varphi(M_{i+1})$ is equivalent merely to the as …

The properties are not equivalent. Let $S$ consist of (at least two) subsets of $X$, which have one point entirely in common, but which are otherwise disjoint (and which have other points than that co …