Let $X$ and $Y$ be two sets and let $F: \mathcal{P}(X) \to \mathcal{P}(Y)$ and $G: \mathcal{P}(Y) \to \mathcal{P}(X)$ be two set functions that satisfy $$F(A) \subseteq B \iff A \subseteq G(B). \tag{1}$$

That is, the pair $F$ and $G$ constitute a Galois connection. It is well known that $c = G\circ F$ is a(n abstract) closure operator and $i = F \circ G$ is an (abstract) interior operator. I.e., $c$ (resp $i$) satisfies:

  • $A \subseteq c(A)$ (resp., $i(A) \subseteq A$) for all $A \subseteq X$.
  • $c(c(A)) = c(A)$ (resp., $i(i(A)) = i(A)$), for all $A \subseteq X$.
  • $A \subseteq B$ implies $c(A) \subseteq c(B)$ (resp., $i(A) \subseteq i(B)$), for all $A, B \subseteq X$.

A topological closure operator also satisfies

  • $c(A \cup B) = c(A) \cup c(B)$ for all $A, B \subseteq X$.

A topological interior operator also satisfies

  • $i(A \cap B) = i(A) \cap i(B)$ for all $A, B \subseteq X$.

Question: are there intelligible conditions on $F$ and $G$ such that $c$ is a topological closure operator and $i$ and topological interior operator? Is there a characterization of topology via Galois connections?

A bit more motivation: if $Z$ is a topological space, and $X$ is the set of closed sets and $Y$ the set of open sets, then $F: A \mapsto int(A)$ (the interior of $A$) and $G: V \mapsto cl(B)$ (the closure), then $F$ and $G$ satisfy (1). Then $i$ sets $A$ to $int(cl(A))$, the regular interior of $A$. Now, the intersection of regular open sets is regular open, so it would seem (naively) that $F \circ G$ is a topological interior operator, despite regular opens sets not being closed under unions.

  • $\begingroup$ Do you have some examples of nontopological closure / interior operators? $\endgroup$ – Jeroen van der Meer Jan 18 at 21:49
  • $\begingroup$ I believe the map $X$ to the convex hull of $X$ is a non topological closure operator (over say, $\mathbb{R}$). $\endgroup$ – 201p Jan 19 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.