In sentential logic , we can define a topology over the set of models on a set of sentence symbols $\mathscr{S}$. The underlying set of the topology is $X=2^{\mathscr{S}}$, or the set of all models on $\mathscr{S}$, and the topology defined by the following basis: if $\Delta$ is a set of sentences, then the set of all models satisfying $\Delta$, denoted $[\Delta]$, is a basis element. (As a convention, if $\varphi$ is a single sentence, we denote $[{\varphi}]$ by $[\varphi]$.) To prove that this is, in fact, a basis for a topological space, we need only show that

1. For every model $A$, $A\in [\Delta]$ for some set of sentences $\Delta$.

2. For any choice of basis elements $M_1=[\Delta_1]$ and $M_2=[\Delta_2]$ and $A\in M_1\cap M_2$, there exists another basis element $M=[\Delta]$ such that $A\in [\Delta]\subset M_1\cap M_2$.

For $(1)$, we may simply notice that if $\varphi$ is some tautology, then $A\in [\varphi]$ for all $A\in 2^{\mathscr{S}}$. As for $(2)$, we may let $\Delta=\Delta_1\cup\Delta_2$, for $A$ is a model of both $\Delta_1$ and $\Delta_2$ if and only if it is a model of $\Delta_1\cup\Delta_2$.

Proposition 1.Under this topology, the closed sets are precisely the sets of the form $[\Sigma]$.

Proof.To prove this, we make use of the fact that a model $A$ satisfies a set of sentences $\Sigma$ if and only if it satisfies every set of finite sentences of $\Sigma$. If $\Sigma'\subset\Sigma$ is a set of finite sentences, then $\Sigma'$ has precisely the same models as the sentence $\sigma=\sigma_1\land\sigma_2\land...\land\sigma_n$ where $\sigma_1,...,\sigma_n$ are the sentences of $\Sigma'$. Given some set of sentences $\Sigma$, let $\mathrm{FS}(\Sigma)$ be the set of sentences in the form $\sigma=\sigma_1\land...\land\sigma_n$ where ${\sigma_1,...,\sigma_n}$ is a finite subset of $\Sigma$.First, we show that all sets of the form $[\Sigma]$ are open. Since a model $A$ satisfies $\Sigma$ if and only if it satisfies every finite subset of $\Sigma$, we have that $A$ fails to satisfy $\Sigma$ iff it fails to satisfy some finite subset of $\Sigma$, meaning that the complement of $[\Sigma]$ is given by which is a union of basis elements, making it an open set. Since $[\Sigma]^c$ is open, we have that $[\Sigma]$ is closed.

Now, we show that all closed sets take the form $[\Sigma]$, or equivalently that the complement of any open set takes the form $[\Sigma]$. The open sets consist of unions of the form which is the set of all models satisfying at least one of the sets of sentences $\Sigma_i$. The complement of this set is therefore the set of models failing to satisfy at least one set of sentences $\Sigma_i$, or failing to satisfy at least one finite subset of at least one of these sentences. It is therefore equal to the following set of models: which is, in fact, a basis element, and therefore an open set. Hence, any closed set (or the complement of any open set) takes the form $[\Sigma]$, where and the desired result is proven. $\blacksquare$

This has very significant implications on the exploration of which sets of models are sets of all models . It answers this question (in a way that may or may not be satisfactory, depending on the reader's standards) by saying that the sets of models which are sets of all models of some set of sentences are precisely the closed sets in this topology. They are also the *complements* of the sets which are arbitrary unions of sets of all models of some sets of sentences, i.e. sets of models taking the form

Theorem 1.Every open set in the model topology is the union of disjoint closed sets (assuming that the set of sentences on $\mathscr{S}$ and the set of sets of sentences on $\mathscr{S}$ can be well-ordered).

Proof.Since all closed sets take the form $[\Sigma]$ for some set of sentences $\Sigma$, all open sets take the form $[\Sigma]^c$. Additionally, since a model fails to satisfy a set of sentences if and only if it fails to satisfy some finite subset of that set, it follows that where $\mathrm{FS}(\Sigma)$ is defined as in the proof of the previous proposition. By hypothesis, the set of sentences over $\mathscr{S}$ can be well-ordered. Thus, we may also well-order $\mathrm{FS}(\Sigma)$, so that it consists of a sequence of elements $(\varphi_i)_{i\in\alpha}$ indexed by ordinals, where $\alpha$ is an ordinal.We define another ordinal-indexed sequence $(\Psi_i)_{i\in\alpha}$ as follows, where each $\Psi_i$ is a set of sentences. For each $i\in\alpha$, consider the set Notice that this set is closed, because it is the intersection of two closed sets. This means that it is equal to $[\Psi]$ for some set of sentences $\Psi$, by the previous proposition. Since the set of sets of sentences of $\mathscr{S}$ can be well-ordered by hypothesis, there exists some minimal $\Psi$ such that the above intersection is equal to $[\Psi]$, and we define this to be the value of $\Psi_i$. Then we have that $[\Sigma]^c$ is the union of the subset of the sets $[\Psi_i]$ consisting of those which are nonempty, and these sets are disjoint by definition, since each one is contained in the complement of those preceding it. Hence, we have expressed an arbitrary open set as the union of disjoint closed sets, and the theorem is proven. $\blacksquare$