Muhammad Haris Rao
Fix a first-order language $\mathcal{L}$ with set of constant symbols $\mathcal{C} = \{ c_\alpha \}_{\alpha \in I_\mathcal{C}}$, set of relational symbols $\mathcal{R} = \{ R_\alpha \}_{\alpha \in I_\mathcal{R}}$ and set of function symbols $\mathcal{F} = \{ f_\alpha \}_{\alpha \in I_\mathcal{F}}$. Let $\{ \mathfrak{A}_\alpha \}_{\alpha \in I}$ be a collection of $\mathcal{L}$-structures indexed by a set $I$, and let $\mathfrak{F} \subseteq I$ be an filter. If $A_\alpha$ is the underlying universe of $\mathfrak{A}_\alpha$ for $\alpha \in I$, write \begin{align*} A^* &= \prod_{\alpha \in I} A_\alpha \end{align*} For $\alpha \in I$, define $\pi_\alpha : A^* \longrightarrow A_\alpha, x \longmapsto x_\alpha$ to be the projection maps out of the product, and define the equivalence relation \begin{align*} \sim_\mathfrak{F} &= \left\{ (x, y) \in A^* \times A^* \mid \{ \alpha \in I \mid x_\alpha = y_\alpha \} \in \mathfrak{F} \right\} \end{align*} That is, we say $x, y \in A^*$ are related by $\sim_\mathfrak{F}$ if the set of coordinates on which they agree is an element of the filter $\mathfrak{F}$. The fact that $I \in \mathfrak{F}$ corresponds to the fact that $\sim_\mathfrak{F}$ is reflexive, and the fact that $\mathfrak{F}$ is closed under intersections corresponds to $\sim_\mathfrak{F}$ being transitive. Reflexivity is obvious because the condition defining $\sim_\mathfrak{F}$ is symmetric.
Define $A = A^* / \sim_\mathfrak{F}$ to be the set of equivalence classes on $A^*$ with respect to $\sim_\mathfrak{F}$. This will be the universe of our new $\mathcal{L}$-structure. Now we move on to interpretting the language on $A$.
For a constant symbol $c \in \mathcal{C}$, define \begin{align*} c^\mathfrak{A} = \left\{ x \in A^* \mid \left\{ \alpha \in I \mid x_\alpha = c^{\mathfrak{A}_\alpha} \right\} \in \mathfrak{F} \right\} \subseteq A^* \end{align*} On first observation, the above is just a subset of $A^*$. It turns out to be an equivalence class of $\sim_\mathfrak{F}$ so that $c^\mathfrak{A} \in A$ which is what we would want.
Proposition: If $c \in \mathcal{C}$ then $c^\mathfrak{A} \in A$.
Proof. To show that $c^\mathfrak{A}$ is an equivalence class, we must show that if $x, y \in c^\mathfrak{A}$ then $x \sim_\mathfrak{F} y$. So suppose $x, y \in c^\mathfrak{A}$. Then \begin{align*} \left\{ \alpha \in I \mid x_\alpha = c^{\mathfrak{A}_\alpha} \right\} &\in \mathfrak{F} \\ \left\{ \alpha \in I \mid y_\alpha = c^{\mathfrak{A}_\alpha} \right\} &\in \mathfrak{F} \end{align*} But then it follows from the definintion of an ultrafilter that $$ \left\{ \alpha \in I \mid x_\alpha = y_\alpha \right\} \supseteq \left\{ \alpha \in I \mid x_\alpha = c^{\mathfrak{A}_\alpha} \right\} \cap \left\{ \alpha \in I \mid y_\alpha = c^{\mathfrak{A}_\alpha} \right\} \in \mathfrak{F} $$ so that $\left\{ \alpha \in I \mid x_\alpha = y_\alpha \right\} \in \mathfrak{F}$ by upwards closure. Thus, $x \sim_\mathfrak{F} y$ by definition of $\sim_\mathfrak{F}$ which is what we set out to show.
Now to show that if $x \in c^\mathfrak{A}$ and $x \sim_\mathfrak{F} y$ for some $y \in A^*$, then also $y \in c^\mathfrak{A}$. Indeed, in this case we have \begin{align*} \left\{ \alpha \in I \mid y_\alpha = c^{\mathfrak{A}_\alpha} \right\} \supseteq \left\{ \alpha \in I \mid x_\alpha = y_\alpha \right\} \cap \left\{ \alpha \in I \mid x_\alpha = c^{\mathfrak{A}_\alpha} \right\} \in \mathfrak{F} \end{align*} So by upwards closure, we have $\left\{ \alpha \in I \mid y_\alpha = c^{\mathfrak{A}_\alpha} \right\}$ which means that $y \in c^\mathfrak{A}$ as well. This proves that $c^\mathfrak{A}$ is an equivalence class of the relation $\sim_\mathfrak{F}$. $\blacksquare$
Given a relational symbol $R \in \mathcal{R}$ with arity $n_R \in \mathbb{Z}_{> 0}$, \begin{align*} R^* &= \left\{ \left(x^1, x^2, \cdots, x^{n_R} \right) \in \left( A^* \right)^{n_R} \mid \left\{ \alpha \in I \mid \left( x_\alpha^1, x_\alpha^2, \cdots, x_\alpha^n \right) \in R^{\mathfrak{A}_\alpha} \right\} \in \mathfrak{F} \right\} \end{align*} It turns out this will give a valid relation on the quotient set $A$ as well.
Proposition: For $i \in \{ 1, 2, \cdots, n_R \}$, let $x^i, y^i \in A^*$ be such that $x^i \sim_\mathfrak{F} y^i$. Then \begin{align*} \left( x^1, x^2, \cdots, x^{n_R} \right) \in R^* \text{ if and only if } \left( y^1, y^2, \cdots, y^{n_R} \right) \in R^* \end{align*}
Proof. To say that $x^i \sim_\mathfrak{F} y^i$ means that there exists a set $X_i \in \mathfrak{F}$ such that $x_\alpha^i = y_\alpha^i$ if and only if $\alpha \in X_i$. Take $X = X_1 \cap X_2 \cap \cdots \cap X_{n_R} \in \mathfrak{F}$. So if $\alpha \in X$, then for all $i \in \{ 1, 2, \cdots, n_R \}$, $x_\alpha^i = y_\alpha^i$. Then for all $\alpha \in X$, \begin{align*} \left( x_\alpha^1, x_\alpha^2, \cdots, x_\alpha^{n_R} \right) = \left( y_\alpha^1, y_\alpha^2, \cdots, y_\alpha^{n_R} \right) \end{align*} Denote the sets \begin{align*} D_x &= \left\{ \alpha \in I \mid \left( x_\alpha^1, x_\alpha^2, \cdots, x_\alpha^{n_R} \right) \in R^{\mathfrak{A}_\alpha} \right\} \\ D_y &= \left\{ \alpha \in I \mid \left( y_\alpha^1, y_\alpha^2, \cdots, y_\alpha^{n_R} \right) \in R^{\mathfrak{A}_\alpha} \right\} \end{align*} Now suppose $\left( x^1, x^2, \cdots, x^{n_R} \right) \in R^*$. By definition, this means $D_x \in \mathfrak{F}$. Then also $D_x \cap X \in \mathfrak{F}$ by definition of a filter. It follows that if $\alpha \in D_x \cap X$ we have \begin{align*} \left( y_\alpha^1, y_\alpha^2, \cdots, y_\alpha^{n_R} \right) = \left( x_\alpha^1, x_\alpha^2, \cdots, x_\alpha^{n_R} \right) \in R^{\mathfrak{A}_\alpha} \end{align*} since $x_\alpha^i = y^i_\alpha$ whenever $\alpha \in X$, $i \in \{ 1, 2, \cdots, \in n_R \}$. This means that $\alpha \in D_y$, and so $D_x \cap X \subseteq D_y$. By upwards closure, we have $D_y \in \mathfrak{F}$ so that $\left( y^1, y^2, \cdots, y^{n_R} \right) \in R^*$.
The other direction follows by symmetry of the relation $\sim_\mathfrak{F}$. This completes the proof.$\blacksquare$
This proposition shows that the relation $R^* \subseteq \left( A^* \right)^{n_R}$ factors through the equivalence relation $\sim_\mathfrak{F}$ to given an equivalence relation \begin{align*} R^\mathfrak{A} \subseteq \left( A^* / \sim_\mathfrak{F} \right)^{n_R} = A^{n_R} \end{align*}
Now to interpret the function symbols. For this, we need the result:
Proposition: For $i \in \{ 1, 2, \cdots, n_R \}$, let $x^i, y^i \in A^*$ be such that $x^i \sim_\mathfrak{F} y^i$. Then \begin{align*} \left\{ \alpha \in I \mid f^{\mathfrak{A}_\alpha} \left( x_\alpha^1, x_\alpha^2, \cdots, x_\alpha^{n_f} \right) = f^{\mathfrak{A}_\alpha} \left( y_\alpha^1, y_\alpha^2, \cdots, y_\alpha^{n_f} \right) \right\} \in \mathfrak{F} \end{align*}
Proof. Since $x^i \sim_\mathfrak{F} y^i$, there exists $X_i \in \mathfrak{F}$ such that $\alpha \in X_i$ exactly when $x_\alpha^i = y_\alpha^i$. Set $X = X_1 \cap X_2 \cap \cdots \cap X_{n_f} \in \mathfrak{F}$ so that whenever $\alpha \in X$, then $x_\alpha^1 = y_\alpha^1, x_\alpha^2 = y_\alpha^2, \cdots, x_\alpha^{n_f} = y_\alpha^{n_f}$. Then also \begin{align*} f^{\mathfrak{A}_\alpha} \left( x_\alpha^1, x_\alpha^2, \cdots, x_\alpha^{n_f} \right) = f^{\mathfrak{A}_\alpha} \left( y_\alpha^1, y_\alpha^2, \cdots, y_\alpha^{n_f} \right) \end{align*} because the arguments are equal when $\alpha \in X$. Hence, \begin{align*} \left\{ \alpha \in I \mid f^{\mathfrak{A}_\alpha} \left( x_\alpha^1, x_\alpha^2, \cdots, x_\alpha^{n_f} \right) = f^{\mathfrak{A}_\alpha} \left( y_\alpha^1, y_\alpha^2, \cdots, y_\alpha^{n_f} \right) \right\} \supseteq X \in \mathfrak{F} \end{align*} which implies the desired result by upwards closure. $\blacksquare$
In other words, if we define $f^* : \left(A^*\right)^{n_f} \longrightarrow A^*$ using the projection maps by \begin{align*} f^* \left( x^1, x^2, \cdots, x^{n_f} \right)_\alpha &= f^{\mathfrak{A}_\alpha} \left( x_\alpha^1, x_\alpha^2, \cdots, x_\alpha^{n_f} \right) \end{align*} then $f^*$ factors through the equivalence relation $\sim_\mathfrak{F}$ to give a unquely determined map \begin{align*} f^\mathfrak{A} : \left( A^* / \sim_\mathfrak{F} \right)^{n_f} \longrightarrow A^* / \sim_\mathfrak{F} \end{align*}
Define $\mathfrak{A}$ to be the $\mathcal{L}$-structure with universe $A = A^* / \sim_\mathfrak{F}$, and the constant cymbol $c \in \mathcal{C}$ interpretted as $c^\mathfrak{A} \in A$, the relational symbol $R \in \mathcal{R}$ interpretted as $R^\mathfrak{A} \subseteq A^{n_R}$ and the functional symbol $f \in \mathfrak{F}$ interpretted as $f^\mathfrak{A} : A^{n_f} \longrightarrow A$. Then $\mathfrak{A}$ is called the ultraproduct of the family $\{ \mathfrak{A}_\alpha \}_{\alpha \in I}$ with respect to the filter $\mathfrak{F} \subseteq I$. It is denoted \begin{align*} \mathfrak{A} &= \prod_{\alpha \in I} \mathfrak{A}_\alpha / \mathfrak{F} \end{align*}
The central model theoretic property of the ultraproduct construction is:
Theorem (Łoś): Let $\mathcal{L}$ be a first-order language and $\{ \mathfrak{A}_\alpha \}_{\alpha \in I}$ a family of $\mathcal{L}$-structures indexed by $I$. Let $\mathfrak{F} \subseteq \mathcal{P}(I)$ be an ultrafilter, and let \begin{align*} \mathfrak{A} &= \prod_{\alpha \in I} \mathfrak{A}_\alpha / \mathfrak{F} \end{align*} Then for any $\mathcal{L}$-sentence $\varphi \in \texttt{Sent}(\mathcal{L})$, $$ \mathfrak{A} \models \varphi \text{ if and only if } \left\{ \alpha \in I \mid \mathfrak{A}_\alpha \models \varphi \right\} \in \mathfrak{F} $$
This gives a quick proof of the compactness theorem from model theory:
Theorem (Compactness): Let $\mathcal{L}$ be a first-order language and $\Sigma \subseteq \texttt{Sent}\left( \mathcal{L} \right)$. Then $\Sigma$ is satisfiable if and only if every finite subset of $\Sigma$ is satisfiable.
Proof. The fact that $\Sigma$ being satisfiably implies satisfiability for every finite subset of $\Sigma$ is trivial, so we prove the other direction. So let $I$ be the collection of finite subsets of $\Sigma$ and for each $\Gamma \in I$, let $\mathfrak{A}_\Gamma$ be an $\mathcal{L}$-structure $\mathfrak{A}_\Gamma$ such that $\mathfrak{A}_\Gamma \models \Gamma$. For $\phi \in \Sigma$, define $\Gamma_\phi = \left\{ \Gamma \in I \mid \phi \in \Gamma \right\}$ to be the set of finite subsets of $\Sigma$ which contain $\phi$. It is clear that $\mathfrak{G} = \left\{ \Gamma_\phi \mid \phi \in \Sigma \right\}$ possess the finite intersction property because for any $\phi_1, \phi_2, \cdots, \phi_n \in \Sigma$ there exists a fintie subset of $\Sigma$ containing all of them, namely $\{ \phi_1, \phi_2, \cdots, \phi_n \}$. Hence, $\mathfrak{G}$ extends to an ultrafilter $\mathfrak{F} \supseteq \mathfrak{G}$ on $I$. Now set \begin{align*} \mathfrak{A} &= \prod_{\Gamma \in I} \mathfrak{A}_\Gamma / \mathfrak{F} \end{align*} If $\phi \in \Sigma$ then we have by Łoś theorem that $\mathfrak{A} \models \phi$ if and only if $\left\{ \Gamma \in I \mid \mathfrak{A}_\Gamma \models \phi \right\} \in \mathfrak{F}$. But see that \begin{align*} \left\{ \Gamma \in I \mid \mathfrak{A}_\Gamma \models \phi \right\} \supseteq \left\{ \Gamma \in I \mid \phi \in \Gamma \right\} = \Gamma_\phi \in \mathfrak{F} \end{align*} so $\left\{ \Gamma \in I \mid \mathfrak{A}_\Gamma \models \phi \right\} \in \mathfrak{F}$ by upwards closure. Thus, we have from Łoś theorem that $\mathfrak{A} \models \phi$ for any $\phi \in \Sigma$. Hence, $\mathfrak{A} \models \Sigma$ which completes the proof.$\blacksquare$