# Given a topological space with $A\subseteq X$, then $x\in \overline A$ iff for each neibourhood of $x$, $U\bigcap A\neq \emptyset$.

Theorem: Given a topological space with $$A\subseteq X$$, then $$x\in \overline A$$ iff for each neibourhood of $$x$$, $$U\bigcap A\neq \emptyset$$.

Proof:
$$\rightarrow$$ : Let $$x\in \overline A$$ and let $$U$$ be a neibourhood of $$x$$; then there exists open $$G$$ with $$x\in G \subseteq U$$. If $$U\bigcap A=\emptyset$$, then $$G\bigcap A=\emptyset$$, and so $$A\subset X$$ \ $$G$$ $$\rightarrow$$ $$\overline A\subseteq X$$ \ $$G$$ whence $$x\in X$$ \ $$G$$, thereby contradicting the assumption that $$U\bigcap A=\emptyset$$.
$$\leftarrow :$$ If $$x\in X$$\ $$\overline A$$ is an open nhd of $$x$$ so that, by hypothesis, $$(X$$\ $$\overline A)\bigcap A\neq \emptyset$$, which is a contradiction.

I lost the understanding at $$\overline A\subseteq X$$\ $$G$$. Why does it hold when $$A\subseteq X$$\ $$G$$?

• $G$ is an open set so its complement $X\setminus G$ is closed. So on the one hand, $A\subset X\setminus G$ implies $\overline A\subseteq \overline{X\setminus G}$, and on the other hand $\overline{X\setminus G}=X\setminus G$ since $X\setminus G$ is closed.
– Ian
May 17, 2014 at 10:29

There are two (equivalent) definitions of $\overline A$:
• It is the smallest closed set containing $A$.
• It is the set of all adherence points of $A$, i.e. of points $x$ so that $U\cap A\ne\emptyset$ for all neighborhoods $U$ of $x$.
Since you going to prove that the closure equals the set of adherence points, the definition you must be using is that as the smallest closed set containing $A$. But then since $X\setminus G$ is closed, $G$ being open, and $A\subseteq X\setminus G$, the closure of $A$ must be in that set.