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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.

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    @Dave: yes, I quoted correctly. Your interpretation of what he intended seems very reasonable to me. Thank you :)2011-09-29

1 Answers 1

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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.

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    @Beni This is called "pointless topology", no joke! See http://en.wikipedia.org/wiki/Pointless_topology2012-07-20