Definitions:
$A'$ is the set of all accumulation or limit points.
$\bar{A} = A \cup A'$ – this is known as the closure of $A$.
Let $A$ be a subset of $\mathbb{R}$. A point $p\in\mathbb{R}$ is an accumulation point of $A$ if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$.
Prove: $\bar{A}$ is closed
proof:
Suppose $p$ is not in $\bar{A}$. Then it has a neighborhood $N_{r}(p)$ that is included in $\bar{A}^{c}$.
This neighborhood is open, so none of its points is in $\bar{A}$. This the compliment of the closure is open, so the closure is closed
I am not sure if I am right, any suggestions would be greatly appreciated
Best Answer
Your proof is assuming that $\overline{A}^c$ is open in the assertion that $N_r(p)$ lies in it. Equivalently, you're kind of assuming that $\overline{A}$ is closed. You must do a bit more work to show that this lies in $\overline{A}^c$.
You are on the right track, though. Since $p$ is not in $\overline{A}$, it is not in $A$ nor is it a limit point of $A$. Therefore there must be some neighborhood $N$ of $p$ that does not intersect $A$ at all.
Can $N$ contain any limit points of $A$? No. If it contained one, $a$. Then by definition of limit point $N$ must contain another point of $A$. But $N$ contains no points of $A$, so this is ridiculous. Thus $N$ must be disjoint from both $A$ and its set of limit points, so $N \subseteq \overline{A}^c$, as desired.