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In Theorem 2.41 on page 40, to show that a compact set is bounded, it is assumed that it is not. Since it is not bounded, it must contain points $\mathbf{x}_n$ with

$|\mathbf{x}_n|>n, \,\,\,n=1,2,\dots$

According to the text, the set $S$ consisting of these points $\mathbf{x}_n$ is infinite and clearly has no limit point in $\mathbb{R}^k$.

Can someone please explain to me how $S$ "clearly" has no limit point in $\mathbb{R}^k$? To give a specific example of what is confusing me, say $n=1$, then every neighborhood of $\mathbf{x}_n = (1,1,\dots,1)$ has a point $\mathbf{q}\neq \mathbf{x}_n$ such that $|\mathbf{q}|>1 \Rightarrow \mathbf{q}\in S$.So how is that $\mathbf{x}_n$, which is in $\mathbb{R}^k$, is not a limit point?

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@TrevorWilson How is it a single point? $\mathbf{x}_n > 1$ is the set of all points outside the unit ball. – AnonSubmitter85 Feb 15 '13 at 8:07
Each $x_{i}$ is a single point. $x_{n} > 1$ means that the $n$th element of the sequence is greater than $1$. It does not describe a set. – Isaac Solomon Feb 15 '13 at 8:09
I think a more direct proof is to note that $B(0,n)$ is an open cover, hence has a finite subcover. Hence the set is bounded. – copper.hat Feb 15 '13 at 8:09
Another proof is that if you know continuous functions take maximum values on compact sets, then the map $x \to |x|$ must have a maximum on a compact set, so the set must be bounded. – Isaac Solomon Feb 15 '13 at 8:10
To me, "$|\mathbf{x}_3| > 3$" is an inequality that may or may not be satisfied by the point $\mathbf{x}_3$ (and we choose $\mathbf{x}_3$ so that it is satisfied.) – Trevor Wilson Feb 15 '13 at 8:30

For a set A to be bounded, it is essential that all the points of A can be placed in a ball". But here you have a countable collection outside any open ball you can conjure. Thus it contradicts.

This is the logical flow: You want to show: Every inf subset has a limit point implies bounded (closed is afterwards).

So lets proceed via Reductio Ad absurdum. You have every inf subset has a limit point. Assume your set, say A, is unbounded. Then you can pick points $x_n \in A$ such that $|x_n| > n$ (If the set were bounded, you would stop for some n). Hence you get a countable sequence $x_n$. But this is an infinite subset of A. It must have a limit point, but it doesn't as this sequence diverges. Contradiction.

Hope that clarifies it. A key thing in the proof is that you can pick your $x_n$ distinct. If you could not, you have a finite set that is trivially bounded.

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If $\{\mathbf{x}_n\}$ has limit point, then there exists $\mathbf{p}$ such that $B(\mathbf{p} , 1)$ contains elements of $\{\mathbf{x}_n\}$ infinitely many. And there exists natural number $N$ satisfy that $|\mathbf{p}|<N$.But if $n>N+2$, then $\mathbf{x}_n $ does not cotained $B(\mathbf{p} , 1)$. So $B(\mathbf{p} , 1)$ must contain elements of $\{\mathbf{x}_n\}$ finitely many, it is contradicted that $B(\mathbf{p} , 1)$ contains elements of $\{\mathbf{x}_n\}$ infinitely many.

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I do not understand what you mean when you say, "And there exists natural number N satisfy that $|\mathbf{p}|<N$." Plus, the example I gave above has infinitely many points in its neighborhood since $\mathbf{R}^k$ is dense. – AnonSubmitter85 Feb 15 '13 at 8:13
@AnonSubmitter85 For example, $N= \lfloor |\mathbf{p}| \rfloor +1$. $N$ is natural number and $|\mathbf{p}| <N$. – Hanul Jeon Feb 15 '13 at 8:17
How are you interpreting the set $S$? Do you read its definition to mean that it is non-dense subset of $\mathbb{R}^k$? If so, how? I read $\mathbf{x}_n>1$ to be the set of all points outside the ball of radius $n$, which means that neighbourhoods in $S$ will have necessarily have infinitely many points. – AnonSubmitter85 Feb 15 '13 at 8:25
@AnonSubmitter85 correct my answer. $S$ can be a dense subset of $\mathbb{R}^k$, but $\{\mathbf{x}_n\}$ is not. – Hanul Jeon Feb 15 '13 at 9:13

If $x$ is a limit point of $S$, we can find a subsequence of $\{x_{n}\}$ converging to $x$. Then that subsequence is bounded. But no subsequence of $\{x_{n}\}$ is bounded.

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Appreciated. And I know there are other proofs that I can consult. However, I am trying to understand the specific referenced claim. – AnonSubmitter85 Feb 15 '13 at 8:09

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