Closure of a set is closed proof 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
 A: 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.
A: I am new in mathematics analysis and also I was confused with the proof of this proposition, then I based on the proof of W. Rudin, Principles of mathematical analysis, 3rd ed., McGraw Hill, 1976. And I try to complement it.
My english is poor, so...
Let $X$ be a metric space, and $ E_{c} $ the closure of $ E $, and $ E_{c}^c $ the complement of $ E_{c} $.
If $p\in X$ and $p\in E_{c}^c$ then $p$ is neither a point of $E$ nor a limit point of $E$. 
Let $ E_{a} $ be the set of limit points of $ E $ such that it contains no elements of $ E $. If there exist a point $ q\in E_{c}^c $ such that it is a limit point of $ E_{c} $, then it has to be a limit point of $ E_{a} $ and not a limit point of $ E $. So there exist a neighborhood of $ q $ such that $ N_{r}(q)\subset {E_{c}^c\cup E_{a}} $ and has some elements of $ E_{a} $, but every point of $ Nr(q),  $especially the elementes of $ E_{a} $, have a neighborhood such that it is contained in $ Nr(q) $, wich implie that the elements of $ E_{a} $ inside $ Nr(q) $ aren’t limit points of $ E $. So there’s no limit points of $ E_{c} $ in $ E_{c}^c $, hence $ E_{c} $ is closed.
A: Here's an idea for proof by contradiction.  Let $p \in \overline{A}^c$.  If $p$ has no neighborhood disjoint from $\overline{A}$ that implies that $N_{\frac{1}{k}}(p) \cap A \neq \emptyset$ for all $k \in \mathbb{N}$.  This impossible though, can you think of a reason why?
A: First suppose that we have a convergent sequence $x_n\in\overline{A}$. We need to show that $x_n\rightarrow x\in\overline{A}$. 
If $x$ is in $\overline{A}^c$, then eventually all $x_n$ are in $\overline{A}\backslash A$. If there were infinitely many $x_n$ in $A$ (a convergent subsequence), then $x$ would be an accumulation point of $A$ and hence be in the closure of $A$. So what is left to be shown is that a sequence of accumulation points of $A$ always converges to an accumulation point of $A$.
Now suppose we have sequence $x_n\in\overline{A}\backslash A$ and suppose $x_n\rightarrow x\in\overline{A}^c$. For any $\epsilon>0$, $B_\epsilon(x_n)\cap A \neq \emptyset$ since each $x_n$ is an accumulation point of $A$.
Choose a sequence $\epsilon_n=1/n$ and choose $y_n\in B_{1/n}(x_n)\cap A$. Hence $y_n\rightarrow x$, but $y_n$ is a sequence in $A$, therefore $x$ is an accumulation point of $A$, and we have a contradiction. Therefore $x\in\overline{A}$ which is thus closed.
Is this argument ok? Probably not the best argument though.
