We have that a set $F$ is closed if every limit point of $F$ is a point of $F$ and we say that $p$ is a limit point of a set $E$ is every neighborhood $N$ of $p$ contains a point $q \neq p$. In other words, if $N \cap E \neq \varnothing$.
${\bf TASK:}$ Prove that $U$ is open iff $U^c$ is closed.
TRY: Let $U$ be a open set. Let $x$ be a limit point of $U^c$. We need to show that $x \in U^c$. If $x \in U$, then there is a neighboorhood $N$ of $U$ such that $N \cap U = \varnothing$ so that $x$ is not limit point. Therefore, $x$ better be in $U^c$, as desired.
On the other hand, say $U^c$ is closed. Let $x \in U$. We must find a neighborhood of $x$ that is contained in $U$.
If we can show that $x$ is not limit point of $U^c$, then we will be able to have a neightborhood $N$ that is completely outside $U^c$ (that is, inside of $U$). So if $x$ is a limit point of $U^c$, then $x \in U^c$ by assumption and this contradiction proves that $x$ is not a limit point of $U^c$.
Is this a correct proof?
Thanks
Best Answer
First of all $p$ is a limit point of $E$ iff for every neighbourhood $N$ of $p$ intersects $E\setminus\{p\}$. More succinct.
If $U$ is open, let $x$ be a limit point of $U^\complement$. If $x \in U$ there is a neighbourhood $N$ of $x$ such that $N \subseteq U$ which implies that $N \cap U^\complement = \emptyset$ which a fortiori would imply that $X$ would not be a limit point of $U^\complement$, contradiction. So $x \in U^\complement$ and so $U^\complement$ is closed. ($N \cap U = \emptyset$ in your proof is a mistake!)
Suppose $U^\complement$ is closed. Let $x \in U$. Then $x$ cannot be a limit point of $U^\complement$ (or it would have been in $U^\complement$ by closedness), so there is a neighbourhood $N$ of $x$ such that $$N \cap (U^\complement\setminus\{x\}) = \emptyset = (N\setminus \{x\}) \cap U^\complement \tag{1}$$ Then $N \subseteq U$ follows. (let $y \in N$; if $y=x$ we are done right away as $x \in U$; otherwise $y \in N\setminus \{x\}$ so $y$ cannot also be in $U^\complement$, or it would contradict the empty intersection in $(1)$; in either case $y \in U$) and so all points of $x \in U$ are interior to $U$, so $U$ is open.
QED. This makes the argument stand out more, I think.