Stone Spaces and the Compactness Theorem

16 May 2025

This post was machine translated and has not yet been proofread. It may contain minor errors or unnatural expressions. Proofreading will be done in the near future.

The following is a famous theorem from undergraduate mathematical logic.

Compactness Theorem. For a theory $T$ in language $\mathcal{L}$, if every finite subtheory of $T$ has a model, then $T$ has a model.

But why is this theorem called the compactness theorem? Upon closer examination, the statement of the above theorem can be written as follows:

If an arbitrary set of $\mathcal{L}$-sentences $T$ has no model, then there exists some finite subset $T’ \subset T$ such that $T’$ has no model.

This is certainly similar to the definition of compactness in topology.

If an arbitrary collection of open sets $\mathcal{C}$ covers the entire space, then there exists some finite subset $\mathcal{C}’ \subset \mathcal{C}$ such that $\mathcal{C}’$ covers the entire space.

This similarity can be extended beyond mere observation. The idea is to regard each complete $\mathcal{L}$-theory as a point in a space, and when we endow this collection with an appropriate topology, the compactness theorem becomes a necessary and sufficient condition for the corresponding space to be compact. First, we define the following:

Definition. For an $\mathcal{L}$-structure $\mathfrak{A}$, we define $\mathrm{Th}(\mathfrak{A})$ as the set of all $\mathcal{L}$-sentences that are true in $\mathfrak{A}$.

Incidentally, this set was defined in a previous post when explaining elementary equivalence.

It is clear that $\mathrm{Th}(\mathfrak{A})$ is a consistent and syntactically complete theory. Conversely, when a theory $T$ is consistent and syntactically complete, by Gödel’s completeness theorem, $T$ has a model $\mathfrak{A}$, and $T = \mathrm{Th}(\mathfrak{A})$. Therefore, $\mathrm{Th}$ generates all consistent and syntactically complete theories.

Now let us define the space that will serve as the background for our topological space. First, for a language $\mathcal{L}$, we define the space $\mathcal{S}$ as follows:

\[\mathcal{S} = \{ \mathrm{Th}(\mathfrak{A}) : \mathfrak{A}\text{ is a $\mathcal{L}$-structure } \}\]

As mentioned earlier, by Gödel’s completeness theorem, $\mathcal{S}$ is essentially the same as the collection of consistent and syntactically complete theories. However, as we shall see later, since our goal is to show that the topological compactness of $\mathcal{S}$ is equivalent to the compactness theorem of first-order logic without using the completeness theorem, it is better to think of $\mathcal{S}$ simply as a collection of $\mathrm{Th}(\mathfrak{A})$ for now.

Now let us define a topological basis for $\mathcal{S}$. For an $\mathcal{L}$-sentence $\phi$, we define:

\[U_\phi = \{ T \in \mathcal{S} : \phi \in T \}\]

Trivially, $U_\phi \cap U_\psi = U_{\phi \land \psi}$. That is, the collection $\mathcal{B}$ of $U_\phi$ is closed under finite intersections. Moreover, $\mathcal{B}$ covers $\mathcal{S}$. For $T \in \mathcal{S}$, if $\phi \in T$, then $T \in U_\phi$. Therefore, $\mathcal{B}$ satisfies the conditions for a topological basis, and we endow $\mathcal{S}$ with the topology generated by $\mathcal{B}$.

Theorem. $\mathcal{S}$ has the following properties:

1. It is Hausdorff.
2. It is a totally disconnected space. That is, the only connected subspaces in $\mathcal{S}$ are singleton sets.

Proof.

1. For $T_1, T_2 \in \mathcal{S}$, if $T_1 \neq T_2$, then there exists some sentence $\phi$ such that $\phi \in T_1$ and $\lnot\phi \in T_2$. Therefore, $U_\phi$ and $U_{\lnot \phi}$ separate $T_1$ and $T_2$ respectively.
2. A sufficient condition for a space to be totally disconnected is to have a basis where every element is clopen. Since $U_\phi^c =U_{\lnot \phi}$, $\mathcal{B}$ satisfies this condition. ■

Theorem. The compactness theorem of first-order logic is equivalent to $\mathcal{S}$ being a compact space.

Proof.

Lemma. A necessary and sufficient condition for $\mathcal{C} \subset \mathcal{B}$ to cover $\mathcal{S}$ is that $\Gamma = \lbrace \lnot \phi : U_\phi \in \mathcal{C} \rbrace $ has no model.

Proof. If $\Gamma$ has a model $\mathfrak{A}$, then $T = \mathrm{Th}(\mathfrak{A})$ is not covered by $\mathcal{C}$. On the other hand, if $\mathcal{C}$ does not cover $\mathcal{S}$, then there exists some structure $\mathfrak{A}$ such that for $\phi \in \Gamma$, $\phi \notin \mathfrak{A}$, which means that $\mathfrak{A}$ is a model of $\Gamma$. □

Now we prove the main theorem. A necessary and sufficient condition for $\mathcal{S}$ to be compact is that any cover of $\mathcal{S}$ consisting of basis elements has a finite subcover. For $\mathcal{C} \subset \mathcal{B}$, a necessary and sufficient condition for $\mathcal{C}$ to cover $\mathcal{S}$ is (by the lemma) that $\Gamma$ has no model. Therefore,

$\mathcal{S}$ is compact

$\Leftrightarrow$ If any collection of basis elements $\mathcal{C}$ is a cover, then there exists a finite subcover.

$\Leftrightarrow$ If any theory $\Gamma$ has no model, then there exists a finite subtheory that has no model.

$\Leftrightarrow$ Compactness theorem ■

Definition. A compact Hausdorff space that is totally disconnected is called a Stone space.

By the compactness theorem of first-order logic, $\mathcal{S}$ is a Stone space. This approach of analysing logical structures topologically is connected to a deep topic in modern mathematics called Stone duality.