Exercise 3.2.5 in Stephen Abbott's Understanding Analysis asks to prove the below theorem. I would like to ask, if my proof is sound.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
Theorem. A set $F \subseteq \mathbf{R}$ is closed if and only if every Cauchy sequence contained in $F$ has a limit that is also an element of $F$.
My Attempt.
($\Longrightarrow$) Assume $F \subseteq \mathbf{R}$ is closed. By definition, a set is closed, if and only if it contains all its limit points. Let $x \in F$ be an arbitrary limit point of $x$. Thus, $V_\epsilon(x)$ intersects $F$ in some point other than $x$. To produce a Cauchy sequence in $F$, we let $\epsilon = 1/n$. Then, there exists a point $x_n \in F$, where
\begin{align*}
x_n \in V_\epsilon(x) \cap F
\end{align*}
with the stipulation that $x_n \ne x$.
It is easy to see, that $(x_n) \to x$. To see this, choose $N > 1/\epsilon$. Then, for all $n \ge N$, we have,
\begin{align*}
\absval{x_n – x} < \epsilon
\end{align*}
Convergent sequences are Cauchy and Cauchy sequences are convergent. Convergent Sequence $\Longleftrightarrow$ Cauchy sequence.
Since, $F$ contains all its limit points, all Cauchy sequences in $F$ have their limiting value in $F$.
($\Longleftarrow$) Assume that every Cauchy sequence in $F$ has a limit that is also an element of $F$. Therefore, $\lim x_n = x$, $x_n \ne x$. By the definition of convergence, given any $\epsilon > 0$m there exists a term $x_N$ in the sequence satisfying $\absval{x_N – x} < \epsilon$. So, $V_\epsilon(x) \cap F$ contains elements other than $x$.
Best Answer
Your proofs in both directions are wrong. For $\implies$ assume that $F$ is closed and start with any Cauchy sequence $(x_n)$ in $F$. [This is imporatnt]. Since any Cauchy sequence of real numbers converges we see that $x =\lim x_n$ exists as a real number. Since each $x_n \in F$ and $F$ is closed it follows that $x \in F$. For the converse let $(x_n)$ be as sequence in $F$ converging to some $x$. We have to prove that $x \in F$. Now $(x_n)$ is a Cauchy sequence in $F$. By assumption the limit $x$ is in $F$.