$G$ is an open set in $\mathbb{R}$ and $A\subseteq\mathbb{R}$.
1. Prove that $G\cap\overline{A}\subseteq\overline{G\cap A}$.
2. Deduce that $\overline{G\cap\overline{A}}=\overline{G\cap A}$.

My attempt:
1. $G\cap\overline{A}=(G\cap A)\cup(G\cap A')$
Part 1: $G\cap A\subseteq\overline{G\cap A}$
Part 2: Let $x\in G\cap A'\Rightarrow x\in G$ and $x\in A'$
$\because G$ is open and $x\in G\Rightarrow x\in int(G)$
$\therefore\exists\delta>0\backepsilon N(x,\delta)\subseteq G\Rightarrow N(x,\delta)\cap G\ne\emptyset$
Again, $\because x\in A'$
$\therefore\forall\epsilon>0$ $N'(x,\epsilon)\cap A\ne\emptyset\Rightarrow N'(x,\delta)\cap A\ne\emptyset\Rightarrow N(x,\delta)\cap A\ne\emptyset$

Initially, I made a few wrong assumptions and concluded that $x\in(G\cap A)'\subseteq\overline{G\cap A}$ but, I don't see how. Can I conclude that $x\in int({G\cap A})\Rightarrow x\in G\cap A$?


Your proof of 1) is not correct. Suppose $x \in G\cap \overline {A}$ and $U$ is an open set containing $x$. Then $U\cap G$ is an open set containing $x$. Since $x \in\overline {A}$ this implies that $U\cap G \cap A$ is not empty. Hence $x$ is in the closure of $G\cap A$.

For 2) it is better to use simple properties of closures. Using 1) and the fact that $\overline {G\cap A}$ is closed we get $\overline {G\cap \overline {A}} \subset \overline {G\cap A}$. The reverse inclusion follows from the fact $G\cap A \subset G\cap \overline {A}$ and $E \subset F$ implies $\overline {E} \subset \overline {F}$

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  • $\begingroup$ In 1, we used the fact that $U$ is an arbitrary open set containing $x$ and that $U\cap G\cap A\ne\emptyset$ to conclude that $x\in\overline{G\cap A}$, right? $\endgroup$ – zaira Nov 13 '19 at 9:05
  • 1
    $\begingroup$ @zaira Yes, that is right. $\endgroup$ – Kavi Rama Murthy Nov 13 '19 at 9:06

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