prove $G\cap \overline A=\emptyset $

Question

I have metric $X$ and $A\subset X$ also there is $G$ an open subset of $X$ so that $G\cap A=\emptyset $ goal:prove that $G\cap \overline A=\emptyset $

Let us suppose there is a $x\in G\cap \overline A$. So because $x\in G$ for every $ε>0$ there is$$B(x,\varepsilon)\subset G\tag1$$ for every $ x\in G $. Also because $x\in \overline A$,$$B(x,e)\cap A\neq \emptyset\tag2.$$ From $(1)$, $(2)$, $G\cap A\neq \emptyset $ which is not true meaning that $G\cap A= \emptyset $ and therefore $G\cap \overline A=\emptyset $

Is this correct?

@SolubleFish that is what i did i misstyped the question let me edit it — Edward Colens, May 18 '21 at 08:17

score 1 · Answer 1 · answered May 18 '21 at 08:25

1

Line $(1)$ is true for some $\epsilon >0$ (and therefore for all $\epsilon$ smaller than that), not for all $\epsilon >0$.

The rest is basically correct : taking one such $\epsilon >0$, you can find $y \in A\cap B(x,\epsilon) \subset G$. Then $y\in A\cap G$, which is absurd. Therefore, there can be no $x\in G\cap \bar A$.

Another way to do this : Since $G$ is open, $X-G$ is closed and $A\subset X-G$. By definition, $\bar{A}$ is the smallest closed subset of $X$ with $A \subset \bar{A}$. Therefore, $\bar{A}\subset X-G$

answered May 18 '21 at 08:25

SolubleFish

7,908

Oh ye i just noticed i had the definition of an open set wrong in my mind this whole time – Edward Colens May 18 '21 at 08:29
Nice alternative solution. – Peter Szilas May 18 '21 at 20:02

score 1 · Answer 2 · answered May 18 '21 at 08:50

Let $x \not \in A$, and $x \in G \cap \overline{A}$.

Then $x \in G$ and $x \in A'$, set of limit points of $A$.

$x \in G$, open, there is a $N_r(x) \subset G$, $r>0.$

Since $x$ is a limit point of $A$, $N_r(x) \cap A \not=\emptyset$.

A contradiction to $G \cap A=\emptyset$.

prove $G\cap \overline A=\emptyset $

2 Answers2