I was trying to prove that in a topological space $X$, $A$ is closed iff $X-A$ is open using the definitions/theorems. Can I get a verification if this is right or wrong?
Theorem:A subset $U$ of $X$ is open iff for each $x \in U$, there exists an open set $U_x$, with $x \in U_x \subset U$.
Definition:A set $A$ is closed if it contains all of its limit points, that is $A= \bar A$.
Attempt:($\implies$)Assume the set $A$ is closed. Then $\bar A=A$, and $X-A=X- \bar A$. Let $x \in X-\bar A$. Then $x \in X \wedge x \notin \bar A$. Thus, there exists an open set $U$ containing $x$, such that $U \cap A=\varnothing.$ Then $U \subset X- \bar A=X-A$, hence $X-A$ is open.
( $\Longleftarrow$ )Assume the set $X-A$ is open. Then for each $x \in X-A$, there is an open set $U$ containing $x$, such that $U \subset X-A$. Let $y \in \bar A$. Assume $y$ is a limit point of $A$, but $y \notin A$.Then $y \in X-A$, and there exists an open set $V \subset X-A$ containing $y$.Then $V$ is an open set containing $y$ with $A \cap V= \varnothing$. So $y \notin \bar A$, a contradiction. Thus $\bar A=A$.
Best Answer
Your proof seems fine to me.
I would proceed as follows, using your definitions/theorems:
if $A$ is closed so $A=\overline{A}$, then for each $x \notin A$ or $x \in X - A$ we know $x \notin \overline{A}$ so there is an open set $U_x$ such that $U_x \cap A = \emptyset$. But then $x \in U_x \subseteq X-A$ and so $X-A$ is open.
If $X-A$ is open for every $x \notin A$ we have an open neighbourhood (namely $X-A$ itself!) that misses $A$, so $x \notin \overline{A}$ and so $X-A \subseteq X-\overline{A}$, hence $\overline{A} \subseteq A (\subseteq \overline{A})$ so that $A$ is closed.