[Update: I changed the question so that $-$ is only applied to closed sets and $\circ$ is only applied to open sets.]

Let $X$ be a topological space with open sets $\mathcal{O}\subseteq 2^X$ and closed sets $\mathcal{C}\subseteq 2^X$. Consider the pair of maps $-:\mathcal{O}\leftrightarrows\mathcal{C}:\circ$ where $-$ is the topological closure and $\circ$ is the topological interior. Under what conditions will it be true that for all $A\in\mathcal{O}$ and $B\in\mathcal{C}$ we have $$A\subseteq B^\circ \Longleftrightarrow A^-\subseteq B \,\,?$$

[Update: Darij showed that this condition holds for all topological spaces.] If this condition does hold then by general nonsense (the theory of abstract Galois connections), we obtain an isomorphism of lattices $$-:P\approx Q:\circ$$ where $P$ is the the lattice of sets $A\in\mathcal{O}$ such that $(A^-)^\circ=A$ and $Q$ is the lattice of sets $B\in\mathcal{C}$ such that $(B^\circ)^-=B$, where both $P$ and $Q$ are partially ordered by inclusion.

The existence of this lattice isomorphism makes me wonder: is there a nice characterization of the elements of $P$ and $Q$? Certainly not every open set is in $P$. For example, if $X=\mathbb{R}$ with the usual topology then the set $(0,1)\cup (1,2)$ is open, but $$(((0,1)\cup(1,2))^-)^\circ = ([0,2])^\circ = (0,2) \supsetneq (0,1)\cup(1,2).$$

[Update: I found the answer. See below.]

  • $\begingroup$ I think the condition is: Every open set is closed. $\endgroup$ Sep 2, 2015 at 22:01
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    $\begingroup$ Doesn't your new axiom always hold? If $B$ is closed, then $A \subseteq B^\circ$ implies $A^- \subseteq \left(B^\circ\right)^- \subseteq B^-$ (since $B^\circ \subseteq B$) and thus $A^- \subseteq B^- = B$ (since $B$ is closed). Also, if $A$ is open, then $A^- \subseteq B$ implies $A = A^\circ$ (since $A$ is open), so that $A = A^\circ \subseteq \left(A^-\right)^\circ$ (since $A \subseteq A^-$) and thus $A \subseteq \left(A^-\right)^\circ \subseteq B^\circ$ (since $A^- \subseteq B$). $\endgroup$ Sep 2, 2015 at 23:44
  • $\begingroup$ @Darij Thanks, that's good news. I guess it means I have the right setup now. Any idea which sets are in P and Q? $\endgroup$ Sep 3, 2015 at 0:27

4 Answers 4


[EDIT: This answer refers to an older version of the question, which asked for a necessary and sufficient condition on a topological space $X$ for it to satisfy the following axiom: For any two subsets $A$ and $B$ of $X$, we have the logical equivalence $A\subseteq B^{\circ}\Longleftrightarrow A^{-}\subseteq B$.]

Your $A\subseteq B^{\circ}\Longleftrightarrow A^{-}\subseteq B$ axiom holds if and only if the closed sets of $X$ are precisely the open sets of $X$.

Proof. $\Longrightarrow:$ Assume that the $A\subseteq B^{\circ }\Longleftrightarrow A^{-}\subseteq B$ axiom holds. Let $S$ be a closed set of $X$. Then, $S^{-}=S\subseteq S$. Now, applying the $A\subseteq B^{\circ }\Longleftrightarrow A^{-}\subseteq B$ axiom to $A=S$ and $B=S$ shows that $S\subseteq S^{\circ}\Longleftrightarrow S^{-}\subseteq S$. Since $S^{-}\subseteq S$ holds, we thus obtain $S\subseteq S^{\circ}$. Combined with $S^{\circ}\subseteq S$, this yields that $S=S^{\circ}$, and thus $S$ is open (since $S^{\circ}$ is open). Thus, we have shown that every closed set $S$ of $X$ is open. In other words, all closed sets of $X$ are open. Applying this to the complement set, we conclude that all open sets of $X$ are closed. Thus, the closed sets of $X$ are precisely the open sets of $X$. This proves the $\Longrightarrow$ direction of my claim.

$\Longleftarrow:$ Assume that the closed sets of $X$ are precisely the open sets of $X$. We need to show that the $A\subseteq B^{\circ}\Longleftrightarrow A^{-}\subseteq B$ axiom holds.

Let $A$ and $B$ be two subsets of $X$. Then, the set $B^{\circ}$ is open and thus closed (since the closed sets of $X$ are precisely the open sets of $X$); hence, $\left( B^{\circ}\right) ^{-}=B^{\circ}$. Similarly, $\left( A^{-}\right) ^{\circ}=A^{-}$. Now, if $A\subseteq B^{\circ}$, then $A^{-}\subseteq\left( B^{\circ}\right) ^{-}=B^{\circ}\subseteq B$. Conversely, if $A^{-}\subseteq B$, then $A\subseteq A^{-}=\left( A^{-}\right) ^{\circ}\subseteq B^{\circ}$ (since $A^{-}\subseteq B$). Hence, $A\subseteq B^{\circ}\Longleftrightarrow A^{-}\subseteq B$. This proves the $\Longleftarrow$ direction of my claim.

Thanks to Stefan Hamcke and Martti Karvonen for correcting my proof!

  • $\begingroup$ If $B=S$, don't you get $\overline S\subseteq S\iff \overline S\subseteq\mathring S$ ? Thus showing that if $S$ is closed, then it is open? $\endgroup$ Sep 2, 2015 at 21:57
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    $\begingroup$ It seems to me that applying the axiom with $A=S^-$ and $B=S$ would only show that $S^- \subseteq S^\circ \Longleftrightarrow \left(S^-\right)^- \subseteq S$, so the rest of the argument doesn't go through. $\endgroup$ Sep 2, 2015 at 22:14
  • $\begingroup$ True. Will fix this. $\endgroup$ Sep 2, 2015 at 22:17
  • $\begingroup$ I see. OK, let me change the question slightly. Consider the maps $-:\mathcal{O}\leftrightarrows\mathcal{C}:\circ$, where $\mathcal{O}$ is the poset of open sets and $\mathcal{C}$ is the poset of closed sets. Now we are not allowed to apply both $-$ and $\circ$ to a set unless we already know that it is closed and open. $\endgroup$ Sep 2, 2015 at 22:24
  • $\begingroup$ What does that mean in English? $\endgroup$ Sep 2, 2015 at 22:39

Put $A=B=S$, where $S$ is any subset of $X$. Then $$ \overline S\subseteq S \iff S\subseteq\mathring S $$ hence $$ S \text{ is closed} \iff S \text{ is open} $$ Conversely, assume that every open set is closed (which implies that any closed set is open). If $A$ and $B$ are subsets of $X$ such that $\overline A ⊆ B$, then we also have $A ⊆ \overline A ⊆ \mathring B$, and if $A ⊆ \mathring B$, then $\overline A ⊆ \mathring B ⊆ B$. Hence such a space satisfies $$ \overline A ⊆ B \iff A \subseteq \mathring B $$


Apparently the elements of $P$ are called regular open sets and the elements of $Q$ are called regular closed sets. Interestingly, the lattices $P$ and $Q$ are Boolean with the function $A^\bot:=X-A^-$ acting as a complement on $P$ and the function $B^\top:=X-B^\circ$ acting as a complement on $Q$.


I didn't find a full characterization of these sets, but apparently every convex open set in $\mathbb{R}^n$ is regular:

Is convex open set in $\mathbb{R}^n$ is regular?

  • $\begingroup$ yes. i think that q is on this site. One interesting point is that for any A we have f(f(A)) =f(A) and g(g(A))=g(A) where f(A)= Int (Cl (A)) and g(A)= Cl(Int(A). $\endgroup$ Sep 3, 2015 at 6:03

Suppose $B$ is open and $A=B$. The condition implies that $Cl(B)=Cl(A) \subset B$ (where $Cl$ denotes closure). So the condition implies that every open set is closed. It is easily verified that if every open set is closed then the condition is satisfied for all $A,B$. So the condition is equivalent to "Every open set is closed." If the space $X$ is a $T_1$ space (which means that $\{p\}$ is closed for every $ p \in X $) then $X$ is discrete ,as $X$ \ $\{p \}$ open $\implies $ $ X$ \ $\{p \}$ closed $\implies$ $\{p\}$ open for all $p \in X$. The coarse topology also satisfies the condition, and there are many others in general. For example if $F$ is a pairwise-disjoint family of subsets of $X$ with $ \cup F =X$, let $F$ be a base for a topology.(Extra : I've just noticed that a $T_0$ space that satifies the condition is a $T_1$ space and hence discrete.)


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