What subclass of topological spaces validate the theorems of classical logic?

Question

I know that there is a topological semantics for intuitionistic logic. But that raises the question, precisely what subclass of topological spaces corresponds to classical logic? Basically, I am looking for a topological property $P$ such that the class of topological spaces with property $P$ are exactly those that validate the theorems of classical propositional logic.

Answer 1

As Qiaochu Yuan and Daniel Schepler discussed in the comments, one can prove that the only $T_0$-separable examples are the discrete spaces. In particular, a topological space validates classical logic precisely if its Kolmogorov quotient is discrete.

Consider a $T_0$ topological space $T$. For any subset $S \subseteq T$ let $\neg S$ denote $\mathrm{Int}(T \setminus S)$.

Assume that the open sets of $T$ validate the axioms of classical logic. This means that the Law of Excluded Middle holds, i.e. for all open sets $A \subseteq T$, $$A \cup \neg A= T$$ and Double Negation Elimination also holds, i.e. for all open sets $A \subseteq T$, the equality $$ \neg\neg A = A$$ obtains as well.

I claim that every singleton set is open in $T$. Take an arbitrary $x \in T$. Notice that $\neg \neg \{x\}$ is of course open, so it suffices to prove that $\neg\neg\{x\} = \{x\}$.

To do this, take any $y \in \neg\neg\{x\}$. We will show that $x$ and $y$ then belong to exactly the same open sets.

Assume that $x \in V$ for some open set $V$. Then $\{x\} \subseteq V$, so $\neg V \subseteq \neg \{x\}$ and $\neg\neg\{x\} \subseteq \neg \neg V$. Since $V$ is open, double-negation elimination gives that $\neg\neg\{x\} \subseteq V$. But $y \in \neg\neg\{x\}$, so we have $y \in V$ as desired.

For the other direction, assume that $y \in V$ for some open set $V$. By the law of excluded middle, $x \in (V \cup \neg V)$. If $x \in V$, we're done. Otherwise, $x \in \neg V$, so $\{x\} \subseteq \neg V$, so $\neg\neg V \subseteq \neg \{x\}$. By double-negation elimination, $V \subseteq \neg \{x\}$, and consequently $y \in \neg\{x\}$. But that's a contradiction, since $y \in \neg \neg \{x\}$ guarantees $y \not\in \neg \{x\}$.

This shows that $x$ and $y$ belong to exactly the same open sets. By $T_0$-separation, it follows that $x = y$. But $y$ was an arbitrary element of $\neg\neg\{x\}$, so we get that $\neg\neg\{x\} = \{x\}$. Consequently, the arbitrary singleton set $\{x\}$ is open, and $T$ is discrete as claimed.