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I'm working through Rudin right now, and I am stuck on one of his proofs. In theorem 2.41 he tries to prove that, if every infinite subset of a set $E$ contains a limit point in $E$, then $E$ is bounded.

To do prove this by contradiction, he attempts to construct an unbounded sequence $\{x_n\}$ such that $x_n \in E$ and $|x_n| > n$. Let $S$ denote the set of all $x_n$, which is clearly a subset of $E$.

While I understand why this set would be infinite, Rudin then asserts that this set has no limit point in $R^k$, which would be a contradiction.

I didn't really understand why this set has no limit point. Could someone help me out here?

I initially thought to prove that if there is some limit point $p \in R^k$, you would show that any neighborhood of $p$ could only contain a finite number of elements of $S$. But my proof ended up assuming that S is monotonically increasing, which is clearly not a general assumption.

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  • $\begingroup$ Did you quote Rudin correctly in the second sentence of the question? I think that, instead of "contains a limit point," it should be "has a limit point." In other words, it is not required that an infinite set $X\subseteq E$ have a limit point in $X$, but only in $E$. $\endgroup$
    – Andreas Blass
    Commented Oct 11, 2015 at 22:05

3 Answers 3

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Hint: Suppose $x_n\to x$, then $|x|\leq k$ where $k \in \mathbb{N}$. Can you solve it now?

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  • $\begingroup$ I see where that goes, but since this is before the chapter on limits of sequences, I wanted to see if I could prove thus using just basic point-set topology of limit points (which is what this chapter deals with). $\endgroup$
    – kritkally
    Commented Mar 24, 2014 at 14:42
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Suppose it has a limitpoint $x$. Then what will be the value of $|x|$? Note that |.| is continuous.

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  • $\begingroup$ Could you clarify? I don't follow this train of logic. $\endgroup$
    – kritkally
    Commented Mar 24, 2014 at 14:47
  • $\begingroup$ $x_n$ converges to limitpoint $x$. Then $n=|x_n|$ must converge to $|x|$ (continuity of |.|). $n$ however is diverging. (No value $p$ exists with $n\rightarrow p$). So the acceptance that a limitpoint exists leads to a contradiction. $\endgroup$
    – drhab
    Commented Mar 24, 2014 at 14:53
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For some $p\in \mathbf{R}^k $, and for every $n=1,2,...$; remains

$n<|x_{n}| \leq |x_{n}-p|+|p-0|=|x_{n}-p|+|p| \Longrightarrow n-|p|<|x_{n}-p|$

Then for some finite integer $m$, for instance, $m=\lceil|p|\rceil+1$ we have

$0< m-|p|\leq l-|p|<|x_{l}-p|$ for $l=m,m+1,...$.

So, $N_{m-|p|}(p)$ has just finite many points of S (or has no elements of S), then $p\in\mathbf{R}^k$ is not a limit point of S by theorem 2.20 (or by 2.18 (b)).

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  • $\begingroup$ Excuse me, what is the set $N_{m-|p|} (p)$? $\endgroup$
    – FactorY
    Commented Oct 14, 2022 at 3:29

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