Verifying proof that $F\subseteq\mathbb{R^n}$ is closed if and only if every convergent sequence $x^{(k)}\in\mathbb{R^n}$ I wish to verify my proof for the problem posed below, please. Thank you!
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Prove that $F\subseteq\R^n$ is closed if and only if every convergent sequence $\x^{(k)}\in\R^n$ such that for all $k$, $\x^{(k)}\in F$
  has its limit in $F$, that is, if $\x^{(k)}\to\x$ then $\x\in F$.

$\textbf{Solution:}$ 
$(\leftarrow)$ Assume, every sequence $\x^{(k)}$ converges in $F$. We wish to prove $F$ is closed. Let $F$ be not closed then there exists $\x$ which a limit point of $F$. Since $\x$ is a limit point of $F$, so, there exists a sequence $\x^{(n)}$ of $F$ such that $\x^{(n)} \to \x$ and $x\notin F$, which a contradiction to the supposition. Hence, $F$ is closed. 
$(\rightarrow)$ Assume that $F$ is closed. We wish to prove every sequence $\x^{(n)}$ of $F$ converges to $\x$ in $F$. By definition, $d(\x^{(n)}, \x) < \epsilon$ for all $n\ge N$; for all $n\ge N$, $\x^{(n)} \in B(\x,\epsilon)$ [*]. 
If possible, assume $\x\notin F$ then $\x \in F^c.$ Since $F$ is closed then $F^c$ is open. So $\x$ is an interior point of $F^c$. So, there exists $\epsilon > 0$ such that $B(\x, \epsilon) \subseteq F^c$ [**]. 
However, from [*] and [**], $\x^{(n)} \in B(\x, \epsilon) \subseteq F^c$ for all $n\ge N$, implying $\x^{(n)} \in F^c$ for all $n\ge N$, which is a contradiction as $\x^{(n)} \in F$ for all $n\ge N$. So our supposition is wrong, implying $\x\in F$. Thus, every convergent sequence of $F$ converges in $F$.     
 A: I suppose you are working with standard topology.
In your first implication you are assuming $F$ closed without saying it. 
The conclusion depends on your definition of closed set, which is not clear.
Another way to say this is the following: since $F$ is closed its complement is open in $\mathbb{R}^n$, so if $x \not \in F$, then $x \in \mathbb{R}^n \setminus F$, which is open. Therefore you can find a ball $B$ of radius $r>0$ small enough and centered at $x$ such that $B \subset \mathbb{R}^n \setminus F$. But this ball must then contain infinitely many elements of the sequence because of the definition of limit, which is a contradiction. Thus $x \in F$.
About the second implication I will not say much because you did not prove anything: "every point of $F$ is contained in $F$" is always true, so you cannot deduce that $F$ is closed.
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First part
Avoid to name a generic sequence with the name of a specific one. So do not say "assume every sequence $x^k$ in $F$" [...] there exists a sequence $x^n$ [...]". This is confusing. 
Usually you do not need parentheses for the indices of the elements of a sequence, so $x^k$ or $x_k$ are better than $x^{(k)}$. 
You still do not state your definition of closed set. 
That there is an element $x$ in $F$ which is the limit point of a sequence does not follow from $F$ not being closed, but from your assumption. So avoid the sentence "Let $F$ not be closed, then [...]". 
Finally, it is not clear why you get a contradiction.
Second part
Avoid to start with "if possible, assume", say instead "assume by contradiction". The rest looks like what I did above.
A: For the second part, start by saying $y$ is a limit point of $F$, but not specifying it is in $F$. Then, you are right that by definition of limit point, you can construct a sequence in $F$ that converges to $y$. By hypothesis, we conclude $y\in F$. Hence, $F$ contains all of its limit points so it's closed.
