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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 $G: B \mapsto cl(B)$ (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.

201p
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  • Do you have some examples of nontopological closure / interior operators? – Jeroen van der Meer Jan 18 '21 at 21:49
  • I believe the map $X$ to the convex hull of $X$ is a non topological closure operator (over say, $\mathbb{R}$). – 201p Jan 19 '21 at 08:55
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    @JeroenvanderMeer There are lots of examples from algebra. For a group $G$, the closure operator $\mathrm{cl}(X) = \langle X\rangle$, the subgroup generated by $X$, is not a topological closure operator. Also linear span in vector spaces, relative algebraic closure in fields, etc. – Alex Kruckman Oct 03 '22 at 16:14
  • I don't understand the setup of the motivation section. $F$ and $G$ are supposed to act on all subsets of $X$ and all subsets of $Y$. But then you say "take $X$ to be the set of closed sets and $Y$ the set of open sets". Also, I think the claim that (1) is satisfied is wrong. Let $A = {0}$ (a closed set) and let $B = \varnothing$ (an open set). Then $\mathrm{int}(A) = \varnothing\subseteq B$, but $A\not\subseteq \mathrm{cl}(B) = \varnothing$. – Alex Kruckman Oct 03 '22 at 17:07
  • In the definition of Galois connection, you're also missing that $F$ and $G$ are monotone functions. – Alex Kruckman Oct 03 '22 at 17:08

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