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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?

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  • $\begingroup$ 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. $\endgroup$ – M. Wang May 16 '20 at 19:27
  • $\begingroup$ 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? $\endgroup$ – BrickByBrick May 16 '20 at 19:30
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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.

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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.

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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.

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