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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?

AnonSubmitter85
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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 08:07
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    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. – Elchanan Solomon Feb 15 '13 at 08:09
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    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 08:09
  • 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. – Elchanan Solomon Feb 15 '13 at 08:10
  • @IsaacSolomon The variable is clearly in bold type, which means that $|\mathbf{x}_n|$ is another way to write the distance function of the metric space, so I don't see how you can read it as being a single point. – AnonSubmitter85 Feb 15 '13 at 08:16
  • @AnonSubmitter85 Sorry for deleting my comment; I didn't realize that you had already responded. Just for the record I am still very confused. What type of object is, say, $\mathbf{x}_3$, if not a point in $\mathbb{R}^k$? – Trevor Wilson Feb 15 '13 at 08:23
  • @TrevorWilson I read $|\mathbf{x}_3| > 3$ to be set of all points outside the sphere of radius of 3 centered at the origin. – AnonSubmitter85 Feb 15 '13 at 08:27
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    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 08:30
  • @TrevorWilson Thanks. If I read the proof like that, it makes sense. I will have to fault Rudin for what is at best unclear language. He must mean a sequence ${\mathbf{x}_n}$ where $|\mathbf{x}_n| > n$. – AnonSubmitter85 Feb 15 '13 at 08:33
  • Sure, I'm glad it makes sense now. It seems clear enough to me as you transcribed it, although perhaps it should be noted that the Axiom of Countable Choice is being used to get the sequence of $\mathbf{x}_n$'s. – Trevor Wilson Feb 15 '13 at 08:35
  • A couple of months late, but I asked a similar question today: http://math.stackexchange.com/questions/404818/rudin-theorem-2-41-heine-borel-theorem –  May 28 '13 at 21:52

3 Answers3

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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.

Gautam Shenoy
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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.

Hanul Jeon
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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 08:13
  • @AnonSubmitter85 For example, $N= \lfloor |\mathbf{p}| \rfloor +1$. $N$ is natural number and $|\mathbf{p}| <N$. – Hanul Jeon Feb 15 '13 at 08:17
  • 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 08:25
  • @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 09:13
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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.

Elchanan Solomon
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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 08:09