This is an exercise in Willard's General Topology.

A subset $B$ of a topological space is called regularly open iff $Int(Cl(B))=B$.

I need to show that if $U$ and $V$ are regularly open then $Int(Cl(U\cap V))=U\cap V$.

I've been using the facts that $Int(Y)=X\setminus Cl(X\setminus Y)$, $Cl(A\cup B)=Cl(A)\cup Cl(B)$ and $Int(A\cap B) =Int(A)\cap Int(B)$ but I always wind up getting back to where I started.

I appreciate your help.

  • 1
    $\begingroup$ (FYI comment I wrote when I assigned this in 2003, minus the references) Regularly open sets can be characterized as those open sets that arise as the interiors of closed sets. [The definition says they're the interiors of the closures of open sets. Willard's Problem 3D(3) says that we don't get anything new if "closures of open sets" is replaced with "closures of arbitrary sets". Finally, the collection of closures of arbitrary sets is equal to the collection of closed sets.] Moreover, every open set $O$ can be made regularly open by the addition (i.e. the union) of some nowhere dense set. $\endgroup$ – Dave L. Renfro Sep 28 '11 at 18:29
  • $\begingroup$ @Dave: thank you. I seize the moment and ask you another question, regarding a statement by the same author a few pages ahead: "...; but it can happen in a top. space that $y$ is in every nhood of $x$ while $x$ is in no nhood of $y$". (P. 32). Isn't that false? What would be a good example to show that the notion of closeness (specifically, its symmetry) does not carry over from metric spaces to topological spaces? $\endgroup$ – Weltschmerz Sep 29 '11 at 1:44
  • $\begingroup$ I don't have my copy of Willard with me (where I might have a note about this), but I googled "Sierpinski's space" to see if it works, and it does. This is a space that probably has no useful application outside of being a counterexample for weird things like this. (Now watch how wrong I am when several people post some applications!) Let the underlying set be $\{x,y\}$ and let the topology on this set be $\{\emptyset,\;\{y\},\;\{x,y\}\;\}.$ Wait, this doesn't work. Are you sure you quoted correctly? Isn't $x$ always going to belong to the nhood of $y$ that is the entire space? $\endgroup$ – Dave L. Renfro Sep 29 '11 at 17:14
  • $\begingroup$ @ Weltschmerz: Thinking more about this, I suspect Willard intended to say "proper nhood" in both places. Here's an example that works for proper nhoods. Let the underlying set be $\{x,y,z\}$ and let the topology on this set be $\{\emptyset,\;\{x,y,z\},\;\{y,z\}\;\}.$ $\endgroup$ – Dave L. Renfro Sep 29 '11 at 17:51
  • $\begingroup$ @ Weltschmerz: In fact, strictly speaking, I suppose the trivial topology $\{\emptyset,X\}$ on any set $X$ such that $\{x,y\} \subseteq X$ would work for proper nhoods if you can live with a vacuuous example. $\endgroup$ – Dave L. Renfro Sep 29 '11 at 17:57

One inclusion is obvious: $U\cap V \subset cl(U\cap V)$ so $U\cap V=int(U\cap V)\subset int(cl(U\cap V)$.

For the other inclusion, pick $x \in int(cl(U\cap V))$. Then there is an open set $O$ such that $x \in O \subset cl(U\cap V)$. This implies that $O\subset cl(U)$ and $O\subset cl(V)$. Then $O \subset int(cl(U))=U$ and $O\subset int(cl(V))=V$. In conclusion $O\subset U \cap V$. Because $x \in O$ it follows that $x \in U\cap V$ and we are done.

I have used that if $A \subset B$ then $cl(A) \subset cl(B)$ and $int(A) \subset int(B)$. I also used that if a set $A$ is open then $int(A)=A$.

Without mentioning points:

Denote $X=int(cl(U\cap V))$. Then $X$ is open and $X \subset cl(U),cl(V)$. Taking interiors we get $X \subset int(cl(U))=U$ and similar $X \subset V$. Therefore $X \subset U\cap V$ and we are done.

  • $\begingroup$ thanks! I think I had not been keeping in mind the fact that $U$ and $V$ are open - an obvious consequence of being regularly open. $\endgroup$ – Weltschmerz Sep 27 '11 at 22:30
  • $\begingroup$ I still wonder if there's a purely "operational" proof though, since that is Willard's approach up to that point. $\endgroup$ – Weltschmerz Sep 27 '11 at 22:54
  • $\begingroup$ @Weltschmerz: What do you mean by operational? $\endgroup$ – Beni Bogosel Sep 27 '11 at 23:06
  • $\begingroup$ I guess I mean without mentioning points, only by properties of the closure and complement (and therefore interior) operations. $\endgroup$ – Weltschmerz Sep 27 '11 at 23:10
  • $\begingroup$ @Weltschmerz: I added a proof without points for the second part. For me, reaching the result seems more important than the path, but it is interesting if it is possible to prove all topological facts using only properties of closed/open sets. $\endgroup$ – Beni Bogosel Sep 27 '11 at 23:38

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