[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. $\endgroup$ – Stefan Hamcke Sep 2 '15 at 22:01
  • 1
    $\begingroup$ 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 $A^- \subseteq B$). $\endgroup$ – darij grinberg Sep 2 '15 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? $\endgroup$ – Drew Armstrong Sep 3 '15 at 0:27

[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? $\endgroup$ – Stefan Hamcke Sep 2 '15 at 21:57
  • 1
    $\begingroup$ 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. $\endgroup$ – Martti Karvonen Sep 2 '15 at 22:14
  • $\begingroup$ True. Will fix this. $\endgroup$ – darij grinberg Sep 2 '15 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. $\endgroup$ – Drew Armstrong Sep 2 '15 at 22:24
  • $\begingroup$ What does that mean in English? $\endgroup$ – DanielWainfleet Sep 2 '15 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). $\endgroup$ – DanielWainfleet Sep 3 '15 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.)


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.