Show that if $X$ is a bounded subset of $\textbf{R}$, then the closure $\overline{X}$ is also bounded.

Question

Show that if $X$ is a bounded subset of $\textbf{R}$, then the closure $\overline{X}$ is also bounded.

MY ATTEMPT

Since $X$ is bounded, we have that $X\subseteq[-M,M]$.

Let us consider that $x$ is an adherent point of $X$.

Then there exists a sequence $(x_{n})_{n=m}^{\infty}$ entirely contained in $X\subseteq[-M,M]$ which converges to $x$.

Since $[-M,M]$ is closed and bounded, due to the Heine-Borel theorem, the sequence $(x_{n})_{n=m}^{\infty}$ admits a subsequence which converges to some $L\in[-M,M]$.

Once a sequence converges iff each of its subsequences converges to the same value, we conclude that $L = x\in[-M,M]$.

In other words, we have just proven the $\overline{X}\subseteq[-M,M]$, which means the closure is bounded.

Could someone please verify if my proof is correct?

Answer 1

A maybe faster way:

By assumption, $X$ is bounded so it lies in some $[-M,M]$ as you said. Since $[-M,M]$ is closed, the closure $\overline{X}$ also lies in $[-M,M]$. This shows that $\overline{X}$ is bounded.

As I also said in the comment, using Heine-Borel is a bit overkill. But your solution is correct.

Answer 2

Your proof is ok, c.f.M.Wangs comment.

Another route:

$\overline {X}= X \cup X',$ where $X'$ are the limit points of $X$.

If $a \in X$ we are done (bounded), since

$X \subset [-M,+M]$, for a bound $M>0$.

Let $a \not \in X$.

Assume $X'$ is not bounded.

For $2M >0$, real, there is a $a \in X'$ s.t.

$a >2M$. Since $a$ is a limit point of $X$ there are points $x$ of $X$, in every neighbourhood of $a$. Let $x \in X$ with

$|x-a|<\epsilon$,

$a-\epsilon <x<a+\epsilon$.

For $\epsilon <M$ we have

$2M -M <a -M < x$, a contradiction.

Answer 3

If $X\subseteq [a, b] $ then you can prove that no number outside the interval $[a, b] $ can be an adherent point of $X$. Thus if $c\in\overline{X} $ then $c\in [a, b] $ so that $\overline{X} \subseteq [a, b] $.

Let's take a number $c$ outside $[a, b] $. Then either $c<a$ or $c>b$. In both cases we have a neighborhood of $c$ which lies outside the interval $[a, b] $ and therefore does not contain any point of $X$. Thus $c\notin\overline{X} $.

Note that this does not involve completeness of reals and is just a matter of using definitions.

Answer 4

Suppose $x\in X^c \land x \in \overline{X}$. Then $\forall \epsilon>0, \exists y\in X, |y-x|\le\epsilon. $

$-\epsilon\le y-x\le \epsilon$

$-\epsilon -y\le -x\le \epsilon-y$

$y-\epsilon<x<y+\epsilon$

So $\forall \epsilon >0, |x|\le M+\epsilon$.

Suppose $|x|>M$. Let $0<\epsilon < |x|-M$. Then $\exists \epsilon>0, |x|>M+\epsilon$. A contradiction.