# Prove $G\cap\overline{A}\subseteq\overline{G\cap A}$ where $G$ is open

$$G$$ 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 $$\overline {E} \subset \overline {F}$$
• 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? – zaira Nov 13 '19 at 9:05