# Closure of 2 subsets

Let $G$ be an open subset of $X$ and $A \subset X$.

a) Prove that $\overline{G \cap \overline{A}} = \overline{G \cap A}$.

b) Show by example that $G$ is necessarily open.

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This post is identical with artofproblemsolving.com/Forum/viewtopic.php?f=549&t=504744 – Martin Sleziak Nov 1 '12 at 17:11

1. The inclusion $\supset$ is always true, even when $G$ is not open. Let $x\in \overline{G\cap \overline A}$, and $O$ an open neighborhood of $x$. It necessarily mets a point of $G\cap \overline A$, say $y$. As $y\in\overline A$, and $G$ is a neighborhood of $y$, $G\cap A\cap O\neq \emptyset$. So $x\in \overline{G\cap A}$.

2. Consider the real line with usual topology, $A:=(0,1)$ and $G:=\{0\}$. The LHS is $\{0\}$ but the RHS empty.

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Yeah I got it. thanks a lot – John Lennon Oct 31 '12 at 0:01

$\newcommand{\cl}{\operatorname{cl}}$Clearly $G\cap A\subseteq G\cap\cl A$, and therefore $\cl(A\cap A)\subseteq\cl(G\cap\cl A)$ for any sets $A$ and $G$, so the crux of the matter must be proving that $\cl(G\cap\cl A)\subseteq\cl(G\cap A)$ when $G$ is open.

Suppose that $G$ is open, and let $x\in\cl(G\cap\cl A)$ be arbitrary; we want to show that $x\in\cl(G\cap A)$. The natural way to show this is to show that if $U$ is any open nbhd of $x$, then $U\cap(G\cap A)\ne\varnothing$, so let $U$ be an open nbhd of $x$. Since $x\in\cl(G\cap\cl A)$, we know that $U\cap(G\cap\cl A)\ne\varnothing$. Let $V=U\cap G$; then we know that $V$ is open and $V\cap\cl A\ne\varnothing$, and we want to show that $V\cap A\ne\varnothing$. Pick any point $y\in V\cap\cl A$. Then $y\in\cl A$, and $V$ is an open nbhd of $y$, so $V\cap A\ne\varnothing$. Tracing back, we see that this means that $U\cap(G\cap A)\ne\varnothing$, and since $U$ was an arbitrary open nbhd of $x$, $x\in\cl(G\cap A)$. Finally, $x$ was an arbitrary element of $\cl(G\cap\cl A)$, so we’ve shown the desired inclusion: $$\cl(G\cap\cl A)\subseteq\cl(G\cap A)$$ when $G$ is open.

We now know that if the equality is to fail for some non-open $G$, it will fail because

$$\cl(G\cap\cl A)\nsubseteq\cl(G\cap A)\;.$$

To find a counterexample when $G$ is not open, go back to the proof and see just where we used the hypothesis that $G$ is open: we needed the set $V$ to be open so that $V\cap\cl A\ne\varnothing$ would imply that $V\cap A\ne\varnothing$. Your counterexample should therefore make use of the fact that if $V$ is not open, it’s possible for $V\cap A$ to be empty even when $V\cap\cl A$ is not. In $\Bbb R$, for instance, what if $A=[0,1)$? Can you find a $V$ such that $V\cap\cl A\ne\varnothing$, but $V\cap A=\varnothing$?

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Yeah I got it. thanks a lot guys. – John Lennon Oct 31 '12 at 0:01
@vercammen: You’re welcome. – Brian M. Scott Oct 31 '12 at 0:12