In logic, a semantics is said to be compact iff if every finite subset of a set of sentences has a model, then so to does the entire set.

Most logic texts either don’t explain the terminology, or allude to the topological property of compactness. I see an analogy as, given a topological space X and a subset of it S, S is compact iff for every open cover of S, there is a finite subcover of S. But, it doesn’t seem strong enough to justify the terminology.

Is there more to the choice of the terminology in logic than this analogy?

**Answer**

The Compactness Theorem is equivalent to the compactness of the Stone space of the Lindenbaum–Tarski algebra of the first-order language $L$. (This is also the space of $0$-types over the empty theory.)

A point in the Stone space $S_L$ is a complete theory $T$ in the language $L$. That is, $T$ is a set of sentences of $L$ which is closed under logical deduction and contains exactly one of $\sigma$ or $\lnot\sigma$ for every sentence $\sigma$ of the language. The topology on the set of types has for basis the open sets $U(\sigma) = \{T:\sigma\in T\}$ for every sentence $\sigma$ of $L$. Note that these are all clopen sets since $U(\lnot\sigma)$ is complementary to $U(\sigma)$.

To see how the Compactness Theorem implies the compactness of $S_L$, suppose the basic open sets $U(\sigma_i)$, $i\in I$, form a cover of $S_L$. This means that every complete theory $T$ contains at least one of the sentences $\sigma_i$. I claim that this cover has a finite subcover. If not, then the set $\{\lnot\sigma_i:i\in I\}$ is finitely consistent. By the Compactness Theorem, the set consistent and hence (by Zorn’s Lemma) is contained in a maximally consistent set $T$. This theory $T$ is a point of the Stone space which is not contained in any $U(\sigma_i)$, which contradicts our hypothesis that the $U(\sigma_i)$, $i\in I$, form a cover of the space.

To see how the compactness of $S_L$ implies the Compactness Theorem, suppose that $\{\sigma_i:i\in I\}$ is an inconsistent set of sentences in $L$. Then $U(\lnot\sigma_i),i\in I$ forms a cover of $S_L$. This cover has a finite subcover, which corresponds to a finite inconsistent subset of $\{\sigma_i:i\in I\}$. Therefore, every inconsistent set has a finite inconsistent subset, which is the contrapositive of the Compactness Theorem.

**Attribution***Source : Link , Question Author : vanden , Answer Author : mrtaurho*