Proof.

• $\Rightarrow$ Assume $M$ is ${\aleph }_0$-saturated. In particular, $M$ realises all types in $S_n(T)$ ($\varnothing$ is finite).

  Fix some finite $A\subseteq M$, and a partial elementary map $f:A\to M$. (Aim: find $d\in M$ such that $f\cup \{(c,d)\}$ is partial elementary). Given $c\in M$, define $p\in S_1^M(f(A))$ to be such that

  $$\phi (x,f(\overline{a}))⟺M\models \phi (c,\overline{a})$$

  (for all $\phi (y,\overline{x})$ $L$-formulas and $\overline{a}\in A$). Write $p=f({tp}^M(c∕\overline{a}))$.

  To show $p\in S_1^M(f(A))$, consider $\phi (x,\overline{a})\in p$. Then $M\models \phi (c,\overline{a})$, so $M\models \exists x,\phi (x,\overline{a})$, os $M\models \exists \phi (x,f(\overline{a}))$ as $f$ is partial elementary.

  So $p$ is finitely satisfiable, and completeness follows from ${tp}^M(c∕A)$ being complete. So as $M$ is $\kappa$-saturated we have $d\models p$ for some $d\in M$. Then $f\cup \{(c,d)\}$ is a partial elementary map.

• $\Leftarrow$ Fix $a_1,\dots ,a_n\in M$, $p\in S_1^M(\{a_1,\dots ,a_n\})$. We want to show $p$ is realised in $M$.

  Set

  $$q=\{\phi (y,x_1,\dots ,x_n):\phi (y,\overline{a})\in p\}\in S_{n+1}^M(T).$$

  So by assumption, we have $d,b_1,\dots ,b_n$ with $(d,\overline{b})\models q$.

  Consider $tp(\overline{a}∕\varnothing )=tp(\overline{b}∕\varnothing )$. So $f:b_i\to a_i$ is partial elementary. Let $c\in M$ such that $f\cup \{(d,c)\}$ is partial elementary. Then ${tp}^M(c,\overline{a})={tp}^M(d,\overline{b})$. So $(c,\overline{a})\models q$ and $tp(c∕\overline{a})=p$. □

Proof. First claim: For any $M\models T$, there is $N\succcurlyeq M$ with $|N|\le |M|+|L|$ and for any $\overline{a},\overline{b},c$ from $M$ such that $\overline{a}{\equiv }^M\overline{b}$ there is some $d\in N$ with $\overline{a}c{\equiv }^N\overline{b}d$.

Proof of claim: Enumerate all $(\overline{a},\overline{b},\overline{c})$ as ${({\overline{a}}_{\alpha },{\overline{b}}_{\alpha },{\overline{c}}_{\alpha })}_{\alpha \le |M|}$. Now let $M_0=M$, and use transfinite induction to form a chain ${(M_{\alpha })}_{\alpha <|M|}$.

• $\alpha$ is a limit ordinal: set $M_{\alpha }={\bigcup }_{i<\alpha }{M}_i$ (then $|M_{\alpha }|\le |\alpha |(|M|+|L|)=|M|+|L|$ as $\alpha \le |M|$).

• Given $\alpha$ (not a limit ordinal), look at $(a_{\alpha },b_{\alpha },c_{\alpha })$. Assume ${\overline{a}}_{\alpha }{\equiv }^{M_{\alpha }}{\overline{b}}_{\alpha }$ so $f_{\alpha }:{\overline{a}}_{\alpha }\to {\overline{b}}_{\alpha }$ partial elementary. Then we apply Proposition 13.4 (Note the elementary super structure construnted here is of size $\le |M|+|L|$) to find $M_{\alpha +1}\succcurlyeq M_{\alpha }$ with $|M_{\alpha +1}|\le |M_{\alpha }|+|L|\le |M|+|L|$ with a $d\in M_{\alpha +1}$ realising $f_{\alpha }(tp(c_{\alpha }∕a_{\alpha }))$. Then by construction, $({\overline{a}}_{\alpha }c){\equiv }^{M_{\alpha +1}}({\overline{b}}_{\alpha }d)$.

  Now let $N={\bigcup }_{\alpha <|M|}{M}_{\alpha }$. Note that we might have introduced new elements. Note $|N|\le |M|(|M|+|L|)=|M|+|L|$.

  We build a new chain $M=N_0\preccurlyeq N_1\preccurlyeq N_2\preccurlyeq \cdots$ of countably many steps with $|N_1|\le |M|+|L|$ and such that for any $\overline{a},\overline{b},\overline{c}\in N$, if $\overline{a}{\equiv }^M\overline{b}$ then there is $d\in N_{i+1}$ such that $\overline{a}c{\equiv }^{N_{i+1}}\overline{b}d$. We do this by iterating the claim.

  FInally let $N={\bigcup }_{i<{\aleph }_0}{N}_i$. Then:

  • $|N|\le |M|+|L|$.

  • $N$ is ${\aleph }_0$-homogeneous as any $\overline{a},\overline{b},c$ from $N$ lie in $N_i$ for some $i$.

□

Proof.

• $\Rightarrow$ $M\models T$ countable, saturated.

  • $M^n$ is countable for all $n\in \mathbb{N}$.

  • We have a map $p\to \overline{a}\models p$ ($p\in S_n(T)$, $\overline{a}$ some realisation). This is map since $M$ saturated, and injective (because complete types).

  So $S_n(T)$ is countable.

• $\Leftarrow$ Enumerate ${\bigcup }_{n\ge 1}{S}_n(T)=\{p_1,p_2,p_3,\dots \}$. Fix a countable $M_0\models T$, and build a chain $M_0\preccurlyeq M_1\preccurlyeq M_2\preccurlyeq \cdots$ such that $M_i$ realises $p_i$ and is countable, using Proposition 13.4.

  Get $N={\bigcup }_{i\in \mathbb{N}}{M}_i$. Then $N\models T$ and is countable. $N$ realises all types over $\varnothing$.

  Apply Lemma 15.6 to get $M\succcurlyeq N$ countable and ${\aleph }_0$-homogeneous structure.

  So by Proposition 15.5 is is ${\aleph }_0$-saturated. □

Proof. Exercise (use back and forth argument). □