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

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?

• Maybe an easier way: $X \subset [-M,M]$, which is closed. So the closure $\overline{X} \subset [-M,M]$, which implies that $\overline{X}$ is also bounded. About your proof, I don't think Heine-Borel is necessary. You have a sequence contained in $[-M,M]$, which converges to $x$. Since $[-M,M]$ is closed, you know that $x$ lies in $[-M,M]$. Your proof is correct though. – M. Wang May 16 '20 at 19:27
• Thanks for the contribution! I really liked the proposed approach. Would you mind to write it as a full answer so I can upvote it? – BrickByBrick May 16 '20 at 19:30

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.

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 .

For $$\epsilon we have

$$2M -M , a contradiction.

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 $$c or $$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.