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To show that $ZFC$ is not existentially closed, we can use the following forcing argument: Let the ground model can model $V=L$ and the forcing extension model $2^{{\aleph}_{0}}=\aleph_{2}$. (Maybe there is a simpler proof of this fact, but I can't see it right now).

However suppose that I were to refine my question. Suppose that $ZFC\subseteq{T}$ is complete and consistent. Is $T$ existentially closed? If I were to guess, I would guess no. However I'm having trouble proving this.

I would guess the same for a complete theory $Th({\mathbb{N}},+, \times, <,0,1) \subseteq{T'}$, and I would like to have an argument that is applicable in both cases.

My original idea for both was to exploit the fact that for an infinite linear order where each element has a unique predecessor and a unique successor, I can show that a structure is not existentially closed by adding an element in between two elements to obtain an extension. However this idea fails for $Th({\mathbb{N}},+, \times, <,0,1)$ in the following sense:

Given $M\models{T'}$ and $a,b\in{M}$ such that $b=a+1$, I can't use compactness to add a new $c$ s.t. $a<c<b$ while also extending $T'\cup{\text{Diag}(M)}$ for the following reason; as $b=a+1$ will be in the diagram of $M$ that I'm trying to extend and $a<c<b$ would imply $b\neq{a+1}$ in the extension. So the option seems to be to construct models witnessing the failure of the criterion from scratch.

I'm not sure about $ZFC\subseteq{T}$. There will be a lot of linear orders (without any additional structure as in the above example) in a model of $T$ to play around with, but extending a model of set theory is not an easy task!

Edit: Or is my guess wrong and are models of the theories I'm interested in existentially closed (and hence model complete)?. I would be surprised by such a result as extensions of $ZFC$ tend to be badly behaved.

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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. D. Hamkins, T. Johnstone, Resurrection axioms and uplifting cardinals, Archive for Mathematical Logic, vol. 53, iss. 3-4, p. p. 463–485, 2014, which opens with the following paragraph:

Many classical forcing axioms can be viewed, at least informally, as the claim that the universe is existentially closed in its forcing extensions, for the axioms generally assert that certain kinds of filters, which could exist in a forcing extension $V[G]$, exist already in $V$. In several instances this informal perspective is realized more formally: Martin's axiom is equivalent to the assertion that $\newcommand\Hc{H_{\frak c}}\Hc$ is existentially closed in all c.c.c.~forcing extensions of the universe, meaning that $\Hc\prec_{\Sigma_1}V[G]$ for all such extensions; the bounded proper forcing axiom is equivalent to the assertion that $H_{\omega_2}$ is existentially closed in all proper forcing extensions, or $H_{\omega_2}\prec_{\Sigma_1}V[G]$; and there are other similar instances.

Meanwhile, one should take care to notice that a model of set theory is never actually existentially closed in any nontrivial forcing extension, since if $V\subseteq V[G]$ is a forcing extension adding a new set $a$ of rank less than $\alpha$, then the ground model $V_\alpha$ has all the rank $\alpha$ sets in $V$, but not in $V[G]$, violating existential closure for the statement $\varphi(\alpha,V_\alpha)=$"there is a set of rank less than $\alpha$ not in $V_\alpha$". The same argument applies to models $M\subset N$, where sets in $M$ have no new elements in $N$. This is why the existential closure assertions are made about $\Hc$ rather than $V$ itself.

From this, it follows that one can make a counterexample to your question about whether complete extensions of ZFC are existentially closed. Specifically, let $M$ be any model of set theory and consider the model $M[c]$ obtained by adding a Cohen real. Let $T$ be the theory of $M[c]$. This is a complete theory extending ZFC. It is a standard fact about forcing that adding another Cohen real gives rise to a model $M[c][d]$ with the same theory as $M[c]$. So $M[c][d]$ is a model of $T$, but has a different set of reals. This violates existential closure of $M[c]$ in $M[c][d]$.

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  • $\begingroup$ This is very close to what I want, at least for $ZFC$. But I do have one question; this does not seem to work for an arbitrary complete theory extending $ZFC$ but only certain extensions. Is there a way past this? $\endgroup$
    – user75685
    Dec 21, 2015 at 19:28
  • $\begingroup$ For example; I was thinking of an extension of that claims that it has inaccesibles and adding a Cohen real should not produce inaccessibles (unless they were in the ground model to begin with since cofinalities are preserved when adding Cohen reals). $\endgroup$
    – user75685
    Dec 21, 2015 at 19:35
  • $\begingroup$ If you start in a model M with inaccessible cardinals or any kind of large cardinal, then they will still exist in the forcing extension. $\endgroup$ Dec 21, 2015 at 19:41
  • $\begingroup$ That is true. So given an extension $T$, you should start with a model $M$ that models a sufficient part of $T$. I was wondering if you had any idea on how to make the "sufficient part of $T$" more precise? Or is it possible to add Cohen reals without breaking the complete theory of the ground model? That would be counter-intuitive (at least for me). Also, in your answer the claim: there is a set of rank less than $\alpha$ not in $V_{\alpha}$. But shouldn't the $V_{\alpha}$ be relativized? Because if I'm not mistaken, the ${V_{\alpha}}^{V[G]}$ will contain the new set that has been added. $\endgroup$
    – user75685
    Dec 21, 2015 at 19:52
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    $\begingroup$ No, because $T$ might be incompatible with being a forcing extension. For example, the assertion that the universe was not obtained by forcing to add a Cohen real is first-order expressible, and this assertion might be part of your $T$. $\endgroup$ Dec 21, 2015 at 21:21

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