Proof (sketch). Induction on

• Complexity of terms

• Formulas

(essentially Proposition 5.8). □

Proof.

• $\Rightarrow$ Suppose $M\models \sigma$. Then $\sigma (M)$ is a non-empty subset of $M^0$, i.e. its the empty tuple (). So $\{[{(\sigma (M_j))}_{j\in J}]\}=\{()\}$. So

  $$\{j\in J:()\in \sigma (M_j)\}=\{j\in J:M_j\models \sigma \}.$$

  Since the LHS is in $U$, we get that the RHS is too.

• $\Leftarrow$ Similar □

Proof.

• $\Rightarrow$ Clear.
• $\Leftarrow$ Assume every finite subset of $\mathrm{\Sigma }$ is consistent. Let $J$ be the set of all finite subsets of $\mathrm{\Sigma }$. For each $j\in J$, let

  $$ĵ=\{k\in J:j\subseteq k\}.$$

  Let $B=\{ĵ:j\in J\}$ and let

  $$F=\{A\subseteq J:\exists B\in B,B\subseteq A\}.$$

  Exercise: $F$ is a filter.

  Let $U$ be an ultrafilter extending $F$. For each $j\in J$, let $M_j\models j$. Let $M=\left({\prod }_{j\in J}{M}_j∕U\right)$.

  Claim: $M\models \mathrm{\Sigma }$.

  Let $\sigma \in \mathrm{\Sigma }$. Then $\{\sigma \}\in J$ and $\widehat{\{\sigma \}}\subseteq \{j\in J:M_j\models \sigma \}$.

  So $\{j\in J:M_j\models \sigma \}\in U$, so $M\models \sigma$.

  So: $\forall \sigma \in \mathrm{\Sigma },M\models \sigma$. □