Prove that F $ \in \mathbb{R} $ is closed if and only if every Cauchy sequences contained in F has a limit that is also an element of F. I'm a novice at proofs so I like to write out everything, so please bear with me!. I understand that this is a biconditional statement, and I will have to prove it in the forward and reverse direction. To prove that every Cauchy sequence contained in F has a limit point that is also contained in F if F is closed, I will start with the forward direction:
Assume F $ \in \mathbb{R} $. If F is closed, then the limits of the Cauchy sequences of F are contained in F. By definition, a closed set is a set that contains all of its limit points. By definition, a Cauchy sequence is a sequence ($x_n$) if for all $ \epsilon > 0 $ there exists $ N \in \mathbb{N} $ such that $ m, n \geq N $ implies that $|x_n - x_m| < \epsilon $. To prove that a Cauchy sequence's limit is contained in F, I need to find two arbitrary points, namely $ x_n $ and $ x_m $ whose distance between each other are less than $\epsilon$. I find an m and n that are greater than or equal to N. If I make N large enough that $ 1/N < \epsilon $, then $1/n$ and $1/m$ must be less than $\epsilon$ as well. The distance between $x_n$ and $x_m$ will be so arbitrarily small that it will be less than $\epsilon$.
Let $\epsilon > 0$. We have
$|x_n - x_m| < \epsilon$ in which n, m $\geq$ N and N is large enough so that $1/N < \epsilon$. 
$|x_n - x + x - x_m| < \epsilon $. Setting $|x_n - x| < \epsilon/2$ and $|x_m - x| < \epsilon/2$ and using the triangle inequality, we have
$\leq |x_n - x| + |x - x_m| < \epsilon/2 + \epsilon/2 = \epsilon$.
This shows that the sequence converges, correct? I'm a little confused at this point. I am trying to show that the Cauchy sequence converges to some limit L, and that L is contained in F, making F a closed set. Could someone help me here? I feel like I've done nothing but restate definitions...
For the reverse direction, assume F $\in \mathbb{R}$. If the limits of the Cauchy sequences of F are contained in F, then F is closed. I feel that this is very much true after having proven the forward direction, and that stating the definition of a closed set would suffice here? Please let me know.
 A: You don't really need a lot of definitions, really. 
Suppose $F$ is closed. Suppose $(x_n)$ is Cauchy, with all $x_n \in F$. Then this sequence is Cauchy in the reals (which is complete), so $x_n \rightarrow p$ for some $p \in \mathbb{R}$. Then $p$ is a limit point of $F$ (why?) and so as $F$ is closed, $p \in F$. Hence the Cauchy sequence has a limit in $F$.
On the other hand, suppose all Cauchy sequences from $F$ converge to some point from $F$. Suppose $F$ were not closed, then there would be a limit point $p$ of $F$ that is not in $F$. Now, for every $n$ pick a point $x_n \in F \cap B(p,\frac{1}{n})$ by being a limit point of $F$. Then $x_n \rightarrow p$ in the reals, so $(x_n)$ is Cauchy (all convergent sequences are Cauchy) in $F$ (as all $x_n$ are in $F$). So it has a limit $q \in F$. But limits of sequences are unique, so this would imply $p= q$, but $p \notin F, q \in F$, contradiction. So $F$ is closed.
A: Just want to add an answer that made sense for me after tackling it a bit and drawing inspiration from here. The definitions I will be going by are from "Understanding analysis" by Stephen Abbott which I believe the post author stumbled onto this.
Just for notation used a bit below
(Epsilon neighbourhood)

$V_\epsilon(a):= \{x\in\mathbb{R}: \lvert x-a\lvert <\epsilon\} = (a-\epsilon, a+\epsilon)$

(Punctured epsilon neighbourhood)

$V_\epsilon(a)\setminus\{a\}:= \{x\in\mathbb{R}: 0<\lvert x-a\lvert <\epsilon\} = (a-\epsilon, a)\cup(a, a+\epsilon)$

$(\Rightarrow)$ Suppose $F\subseteq \mathbb{R}$ is a closed set, that is it contains it's limit points. Given an arbitrary Cauchy sequence contained in $F$ by theorem 2.6.2 ( A sequence converges if and only if it is a Cauchy sequence ) this sequence, let us denote it $(a_n)_{n\in\mathbb{N}}$ is convergent and let us denote the limit by $a$.
Now either $a_n\neq a$ for all $n\in\mathbb{N}$ and then $a$ is a limit point, since given any $\epsilon>0$, we can find an index $N\in\mathbb{N}$ where for all $n\geq N$ we have that $a_n\in V_\epsilon(a)\setminus\{a\}$, thus $V_\epsilon(x)\setminus\{x\}\neq \emptyset$. By our assumption $F$ is closed, that is contains it's limit points so $a\in F$.
or alternately $a_n=a$ for some $n\in \mathbb{N}$, but then $a\in F$ since the sequence is contained in $F$.
$(\Leftarrow)$ Suppose that every Cauchy sequence contained in $F$ has a limit also contained in $F$. Given an arbitrary limit point of $F$ denoted as $x$, by theorem 3.2.5 there is a sequence $(a_n)_{n\in\mathbb{N}}$ contained in $F$ where $a_n\neq x$ for all $n\in\mathbb{N}$ and $\lim_{n\to\infty} a_n = x$. Again since the sequence is convergent by theorem 2.6.2 it is a Cauchy sequence thus the limit point $x\in F$ by assumption. $\blacksquare$
