0
$\begingroup$

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 $A\subseteq X$\ $G$?

$\endgroup$
  • 1
    $\begingroup$ $G$ 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. $\endgroup$ – Ian May 17 '14 at 10:29
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.