# Elementary substructure and existentially closed

Definition. We say that $$\mathcal{M} \models T$$ is existentially closed if whenever $$\mathcal{N} \models T$$, $$\mathcal{M}\subseteq \mathcal{N}$$, and $$\mathcal{N}\models \exists x \phi(x, a)$$, where $$a \in M$$ and $$\phi$$ is quantifier-free,then $$\mathcal{M} \models ∃x φ(x, a)$$.

Definition. If $$\mathcal{M}$$ and $$\mathcal{N}$$ are $$\mathcal{L}$$-structures, then we say that $$\mathcal{M}$$ is an elementary substructure of $$\mathcal{N}$$ and write $$\mathcal{M}\prec \mathcal{N}$$ if

$$\mathcal{M} \models \phi(\bar{a}) \Leftrightarrow \mathcal{N} \models \phi(\bar{a})$$

for all $$\mathcal{L}$$-formulas $$\phi(\bar{x})$$ and all $$\bar{a}\in M$$.

Tarski-Vaught test. Suppose that $$\mathcal{M}$$ is a substructure of $$\mathcal{N}$$. Then, $$\mathcal{M}$$ is an elementary substructure if and only if, for any formula $$\phi(v,w)$$ and $$a \in M$$, if there is $$b \in N$$ such that $$\mathcal{N} \models \phi(b, a)$$, then there is $$c \in M$$ such that $$\mathcal{M} \models \phi(c, a)$$.

Question. Does Tarski-Vaught test imply that $$\mathcal{M}$$ is existentially closed model of $$T$$ if it is an elementary substructure of any model of $$T$$ containing $$\mathcal{M}$$? If yes, why do model theorist use different names/words?

• The definition of "existentially closed" refers only to quantifier-free formulas $\phi$, whereas the Tarski-Vaught test is about arbitrary formulas. – Andreas Blass Nov 11 '18 at 1:18