[proofplan] Index the finite pieces of $T$ by the set $I$ of finite subsets of $T$, and choose a model $M_\Delta$ for each finite piece $\Delta \in I$. For each sentence $\sigma \in T$, the finite subsets containing $\sigma$ form a set $A_\sigma \subset I$, and these sets have the finite intersection property. An ultrafilter containing all the sets $A_\sigma$ lets us form an ultraproduct of the structures $M_\Delta$. Łoś's Theorem then transfers truth of each sentence $\sigma \in T$ from $\mathcal U$-many factors to the ultraproduct, giving a model of all of $T$. [/proofplan] [step:Choose models for all finite fragments of $T$] Let $I$ denote the set of finite subsets of $T$: \begin{align*} I := \{\Delta \subset T : \Delta \text{ is finite}\}. \end{align*} Since $T$ is finitely satisfiable, for every $\Delta \in I$ there exists an $L$-structure $M_\Delta$ such that $M_\Delta \models \Delta$. Choose one such $L$-structure $M_\Delta$ for each $\Delta \in I$. If $\sigma \in T$ is an $L$-sentence, define \begin{align*} A_\sigma := \{\Delta \in I : \sigma \in \Delta\}. \end{align*} Thus $A_\sigma$ is the set of finite fragments of $T$ in which the sentence $\sigma$ appears. [/step] [step:Verify that the sets forcing individual sentences have the finite intersection property] We claim that the family \begin{align*} \mathcal A := \{A_\sigma : \sigma \in T\} \end{align*} has the finite intersection property. Let $\sigma_1,\dots,\sigma_m \in T$ be finitely many sentences, where $m \in \mathbb N$. Define \begin{align*} \Delta_0 := \{\sigma_1,\dots,\sigma_m\}. \end{align*} Then $\Delta_0$ is a finite subset of $T$, so $\Delta_0 \in I$. Also $\sigma_j \in \Delta_0$ for every $j \in \{1,\dots,m\}$, hence \begin{align*} \Delta_0 \in \bigcap_{j=1}^m A_{\sigma_j}. \end{align*} Therefore every finite intersection of members of $\mathcal A$ is nonempty. [guided] The goal is to build an ultrafilter that regards each sentence $\sigma \in T$ as present on a large set of indices. For that, we first need the family of desired large sets to be consistent in the finite sense. Let \begin{align*} \mathcal A := \{A_\sigma : \sigma \in T\}, \end{align*} where $A_\sigma = \{\Delta \in I : \sigma \in \Delta\}$. To prove that $\mathcal A$ has the finite intersection property, take finitely many members of this family. They have the form $A_{\sigma_1},\dots,A_{\sigma_m}$ for some sentences $\sigma_1,\dots,\sigma_m \in T$ and some $m \in \mathbb N$. Now define \begin{align*} \Delta_0 := \{\sigma_1,\dots,\sigma_m\}. \end{align*} This set is finite and contained in $T$, so $\Delta_0 \in I$ by the definition of $I$. Moreover, each sentence $\sigma_j$ belongs to $\Delta_0$, so $\Delta_0 \in A_{\sigma_j}$ for every $j \in \{1,\dots,m\}$. Hence \begin{align*} \Delta_0 \in \bigcap_{j=1}^m A_{\sigma_j}. \end{align*} This proves that the intersection is nonempty. Since the finitely many sentences were arbitrary, the family $\mathcal A$ has the finite intersection property. [/guided] [/step] [step:Extend the forcing sets to an ultrafilter on the finite fragments] By the [Ultrafilter Extension Lemma](/theorems/4284) applied to the set $I$ and the family $\mathcal A$ with the finite intersection property, there exists an ultrafilter $\mathcal U$ on $I$ such that \begin{align*} A_\sigma \in \mathcal U \end{align*} for every $\sigma \in T$. Here we are citing a result not yet in the wiki: Ultrafilter Extension Lemma. [guided] The finite intersection property says that the requirements $A_\sigma \in \mathcal U$ are mutually compatible. The Ultrafilter Extension Lemma states that if a family of subsets of a set has the finite intersection property, then there is an ultrafilter on that set containing every member of the family. We apply this lemma to the underlying set $I$ and to the family \begin{align*} \mathcal A = \{A_\sigma : \sigma \in T\}. \end{align*} The hypothesis of the lemma is satisfied by the previous step. Therefore there exists an ultrafilter $\mathcal U$ on $I$ such that \begin{align*} A_\sigma \in \mathcal U \end{align*} for every $\sigma \in T$. Here we are citing a result not yet in the wiki: Ultrafilter Extension Lemma. [/guided] [/step] [step:Form the ultraproduct of the finite-fragment models] Let \begin{align*} M := \prod_{\Delta \in I} M_\Delta / \mathcal U \end{align*} be the ultraproduct of the family of $L$-structures $(M_\Delta)_{\Delta \in I}$ with respect to the ultrafilter $\mathcal U$. This is an $L$-structure by the standard ultraproduct construction. [/step] [step:Transfer truth of each sentence in $T$ to the ultraproduct] Fix a sentence $\sigma \in T$. Define the truth set of $\sigma$ in the family $(M_\Delta)_{\Delta \in I}$ by \begin{align*} B_\sigma := \{\Delta \in I : M_\Delta \models \sigma\}. \end{align*} If $\Delta \in A_\sigma$, then $\sigma \in \Delta$. Since $M_\Delta \models \Delta$, it follows that $M_\Delta \models \sigma$. Hence \begin{align*} A_\sigma \subset B_\sigma. \end{align*} Because $A_\sigma \in \mathcal U$ and $\mathcal U$ is upward closed, we have $B_\sigma \in \mathcal U$. By Łoś's Theorem applied to the $L$-sentence $\sigma$, the family $(M_\Delta)_{\Delta \in I}$, and the ultrafilter $\mathcal U$, we obtain \begin{align*} M \models \sigma \iff \{\Delta \in I : M_\Delta \models \sigma\} \in \mathcal U. \end{align*} The set on the right-hand side is $B_\sigma$, and $B_\sigma \in \mathcal U$, so $M \models \sigma$. Here we are citing a result not yet in the wiki: Łoś's Theorem. [guided] We now show that an arbitrary sentence $\sigma \in T$ holds in the ultraproduct $M$. Define \begin{align*} B_\sigma := \{\Delta \in I : M_\Delta \models \sigma\}. \end{align*} This is the set of indices at which the chosen finite-fragment model satisfies $\sigma$. The reason for introducing $A_\sigma$ earlier is that membership in $A_\sigma$ forces satisfaction of $\sigma$. Indeed, if $\Delta \in A_\sigma$, then by definition $\sigma \in \Delta$. Since $M_\Delta$ was chosen so that $M_\Delta \models \Delta$, every sentence in $\Delta$ is true in $M_\Delta$, and therefore $M_\Delta \models \sigma$. Thus \begin{align*} A_\sigma \subset B_\sigma. \end{align*} We already know $A_\sigma \in \mathcal U$. Since $\mathcal U$ is a filter, it is upward closed: whenever $C \in \mathcal U$ and $C \subset D \subset I$, then $D \in \mathcal U$. Applying this with $C = A_\sigma$ and $D = B_\sigma$ gives \begin{align*} B_\sigma \in \mathcal U. \end{align*} Now apply Łoś's Theorem. Its relevant sentence-level form says that for an $L$-sentence $\sigma$, \begin{align*} \prod_{\Delta \in I} M_\Delta / \mathcal U \models \sigma \iff \{\Delta \in I : M_\Delta \models \sigma\} \in \mathcal U. \end{align*} The structure on the left is exactly $M$, and the set on the right is exactly $B_\sigma$. Since $B_\sigma \in \mathcal U$, the equivalence yields \begin{align*} M \models \sigma. \end{align*} Here we are citing a result not yet in the wiki: Łoś's Theorem. [/guided] [/step] [step:Conclude that the ultraproduct satisfies all of $T$] The sentence $\sigma \in T$ was arbitrary, and the previous step proves $M \models \sigma$ for every $\sigma \in T$. Therefore \begin{align*} M \models T. \end{align*} Thus there exists an $L$-structure satisfying $T$, as required. [/step]