# Prove that a set $F \subseteq \mathbf{R}$ is closed $\Longleftrightarrow$ every Cauchy sequence in $F$ has a limit in $F$

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. $$\newcommand{\absval}{\left\lvert #1 \right\rvert}$$

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 $$x$$.

• 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 $A$. – Quasar Jan 6 at 7:28
• 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. – Brian M. Scott Jan 6 at 7:32
• 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. – Brian M. Scott Jan 6 at 7:33
• $p\implies q$ 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 ... – Brian M. Scott Jan 6 at 19:21
• ... $x_n\in V_{1/n}(x)$; 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 $F$. – 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 $$F$$.