Proof about convergent sequences in a set Prove that if every convergent sequence in a set $C \subset \mathbb{R}^n$ converges to a point in $C$, then $C$ is closed.
The problem with this is that the definition my book uses is not the one with the limit points, it is this one.
A set is closed if and only if it's complement is open.
I don't know how to show it with that definition.
 A: Proof sketch:
To show that $C$ is closed, you need to show $C^c$ is open.  As usual that means you fix an $x \in C^c$ and you need to show that for some $\epsilon > 0$, the ball of radius $\epsilon$ around $x$, $B(x, \epsilon)$, is contained in $C^c$.
Suppose this were not the case.
Then $B(x, 1/n)$ would always contain a point, call it $x_n$, in $C$.
Then show $x_n$ is a sequence in $C$ but it converges to a point outside of $C$, which is a contradiction.
A: Lemma: A set contains all of its limit points if and only if the set is closed.
Proof: $\Longleftarrow$ Let $C$ be a closed set. By definition, $C^c$ is open. Let $x$ be a limit point of the set $C$. If no such point exists, $C$ contains all of its limit points and hence is closed. Assuming $x$ exists, either $x\in C$ or $x\in C^c$. If $x \in C^c$, because $C^c$ is open, then there exists a $r>0$ such that $B_r(x)\subseteq C^c$. However, this implies that $B_r(x)\cap C=\phi$. However, this violates the fact that $x$ is a limit point of $C$. Thus, $x \not\in C^c$. Thus, $x\in C$. Hence, every limit point of $C$ is contained in $C$. Hence, $C$ is closed. 
$\Longrightarrow$ Now, let $x$ be a limit point of $C$. We know that $x\in C$. Now, let $y \in C^c$. Because, $y \not \in C$, $y$ is also not a limit point of $C$, thus, there exists a $r>0$ such that $B_r(y)\cap C=\phi$ or $B_r(y)\subseteq C^c$. Thus, $C^c$ is open. Thus, $C$ is closed.

Proof (By Contraposition):
Let $C$ not be closed. Thus, there is an $x$ which is a limit point of $C$ such that $x\not \in C$. Now, because $x$ is a limit point of $C$ for every $n\in \mathbb{Z}$, $B_{1\over n}(x)\cap C$ is non-empty. Let $\beta_n$ be a an element of this intersection for a given $n$. Define a sequence with terms $\beta_n$. This sequence is convergent and converges to $x$ (By Design). However, $(x_n)$ does not converge in $C$.
$$\blacksquare$$
