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 $x$.

  • $\begingroup$ P.S. I have cheated by refreshing my memory with an existing proof given in the text for the statement: A point $a$ is a limit point of the set $A$, if and only if, there exists a sequence $(a_n)$ in $A$, such that $\lim a_n = a$, with $a_n \ne a$ for all $n$. Aside: Is cheating allowed? I could not remember, how to produce a sequence $(a_n)$ in $A$. $\endgroup$ – Quasar Jan 6 at 7:28
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    $\begingroup$ The first half of the proof goes wrong at the very beginning, when you set out to construct a Cauchy sequence in $F$ with a certain property. What you have to prove is that every Cauchy sequence in $F$ converges to some point of $F$. Thus, you should either (a) start with an arbitrary Cauchy sequence in $F$ and show that it converges to a point of $F$ or (b) assume that that there is a Cauchy sequence in $F$ that does not converge to a point of $F$ and show that $F$ is not compact. $\endgroup$ – Brian M. Scott Jan 6 at 7:32
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    $\begingroup$ You’ve made a similar error in the second half. You want to show that $F$ is closed, so you should start with a limit point of $F$ and use your hypothesis to show that the point is in $F$. What you did in the first part is useful here. $\endgroup$ – Brian M. Scott Jan 6 at 7:33
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    $\begingroup$ $p\implies q$ is almost right, but you’ve overlooked a possibility: the Cauchy sequence might be eventually constant. In that case a sufficiently small $V_\epsilon(x)$ won’t necessarily contain a point of $F$ different from $x$, and $x$ need not be a limit point of $F$. (Example: $F=[0,1]\cup\{2\}$, and $x_n=2$ for all $n$.) But that can happen only when $x_n=x$ for all but finitely many $n$, in which case $x\in F$ because the $x_n$ are in $F$. \\ For the other direction you’ve done the right thing, but your explanation is just a little off. For each $n$ you’ve picked a point ... $\endgroup$ – Brian M. Scott Jan 6 at 19:21
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    $\begingroup$ ... $x_n\in V_{1/n}(x)$; that’s fine. And for any $\epsilon>0$ we have $|x_n-x|<\epsilon$ for all $n\ge\frac1{\epsilon}$. But now you can simply say that the sequence that you’ve constructed converges to $x$, and since every convergent sequence is Cauchy, we’ve shown that $x$ is the limit of a Cauchy sequence in $F$. $\endgroup$ – Brian M. Scott Jan 6 at 19:24

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 $F$.


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