Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
@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
2  
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
1  
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
1  
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
show 6 more comments

3 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
    
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 $|\mathbf{p}| <N$. –  tetori 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. –  tetori Feb 15 '13 at 9:13
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.