Interior algebra

Remarks

Example: let $X$ be a topological space and let $B$ be the power set of $X$. Let $I(A)$ be the interior of $A$. The interior of $X$ is $X$, and the interior of $A$ lies in $A$. The interior of an open set is the set itself, and $I(A \cap A') = I(A) \cap I(A')$. So the four clauses hold. Non-example: on a Boolean algebra with $0 \ne 1$, the map $I(x) = 0$ for all $x$ satisfies clauses 2, 3 and 4, but fails clause 1. Clause 4 implies that $I$ is monotone: if $x \le y$, then $I(x) = I(x \wedge y) = I(x) \wedge I(y) \le I(y)$. The map $C(x) = \neg I(\neg x)$ is the dual closure operator; a Boolean algebra with such a closure operator is a closure algebra. Interior algebras are the algebraic semantics of the modal logic S4, with $I$ interpreting the necessity operator $\square$.