Prove that $L$ is closed if and only if for any converging sequence $(x_n)$ with $x_n\in L$, the limit $x=\lim x_n$ is also an element of $L$ Could somebody please point me in the right direction to prove $\Leftarrow$?
Here is my proof for $\Rightarrow$:
Let $L\subseteq R$ be closed and $(x_n)$ be a converging sequence with $x_n\in L$.
Show that (1) $L$ is closed gives (2) $\lim x_n=x\in L$: suppose by contradiction that (2) does not hold given (1), i.e. $\lim x_n=x\notin L$; for any arbitrary $\epsilon>0$, there exists $N^\prime\in N$ such that for $n\geq N^\prime,|x_n-x|<\epsilon\Leftrightarrow -\epsilon<x_n-x<\epsilon\Leftrightarrow x-\epsilon<x_n<x+\epsilon$; $x_n\in L$ but $(x-\epsilon,x+\epsilon)\subseteq L^c\Rightarrow$ contradiction; thus, (1) gives (2).
 A: Proposition
Let $(X,d)$ be a metric space, let $E$ be a subset of $X$, and let $x_{0}$ be a point in $X$. Then the following statements are logically equivalent:
(a) $x_{0}$ is an adherent point of $E$.
(b) There exists a sequence $(x_{n})_{n=1}^{\infty}$ in $E$ which converges to $x_{0}$ with respect to $d$.
Proof
Let us prove the implication $(\Rightarrow)$ first.
If $x_{0}$ is an adherent point of $E$, then for every $\varepsilon > 0$ there corresponds $x_{\varepsilon}\in E$ such that $d(x_{\varepsilon},x_{0}) \leq \varepsilon$.
In particular, if we choose $\varepsilon = 1/n$, there corresponds a $x_{n}\in E$ such that $d(x_{n},x_{0})\leq 1/n$.
Taking the limit from both sides we conclude that $x_{n}\to x_{0}$, and we are done.
Let us prove now the implication $(\Leftarrow)$.
Suppose that $x_{n}\to x_{0}\in E$.
Then for every $\varepsilon > 0$, there corresponds a $n_{\varepsilon}\in\mathbb{N}$ such that $d(x_{n},x_{0})\leq \varepsilon$ whenever $n\geq n_{\varepsilon}$.
In other words, for every $\varepsilon > 0$, there corresponds a $x_{n_{\varepsilon}}\in B(x_{0},\varepsilon)\cap E$.
Consequently, $x_{0}$ is an adherent point of $E$.
Solution
At your case, $L$ is considered to be closed. This means that $L$ contains all its adherent points.
Based on the previous result, this means that every $x\in L$ is the limit of a sequence $x_{n}\in L$.
Conversely, if every convergent sequence $x_{n}\in L$ converges within $L$,we conclude that $L\supseteq\overline{L}$.
Since $L\subseteq\overline{L}$, it results that $L$ is closed.
Hopefully this helps!
A: The previous answer is perfect but i will try to give an explanation with concepts.
Cosider those two things:

*

*L is closed if and only if every point that is arbitrarily near to L is in L.

*The limit point of a sequence is arbitrarerly near to the set of points on the sequence.

Your theorem states that every point that is arbitrarely near to L it is also arbitrarely near to a sequence in L.
So the key is that the definition of limit point links 1) with 2). In fact if L were the set of points on a convergence sequence you get the equivalence. So your question states that in some sense if L is closed it is the union of all his convergent sequences and its limits points.
I hope this helps you
