According to theorem stated here Bolzano-Weierstrass theorem it says: Every bounded infinite set in $\Bbb R^n$ has an accumulation point.

Also some states it as: Every bounded sequence in $\Bbb R^n$ contains a convergent subsequence.

Is the first statement true? I doubt it because as far as I know having accumulation point is different from having a convergent sequence. E.g sequence $\{p_n\}_{n=1}^\infty$ may converge to a point $p$ ( $\lim_{n \to \infty} p_n = p$ ) but p may not be a limit (accumulation) point of range of $\{p_n\}$.

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    $\begingroup$ Your sequence $(p_n) $ should not be stationnary. the range $\{p_n\}$ must be infinite. $\endgroup$ – hamam_Abdallah Dec 3 '17 at 21:51
  • $\begingroup$ @M.Winter is it true only for $\Bbb R^n$ or for all metric spaces? $\endgroup$ – Pumpkin Dec 3 '17 at 21:52
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    $\begingroup$ @Pumpkin The claim is not true for all metric spaces. Consider $(0,1]$ with the inherited metric from $\Bbb R$ and the sequence $\{1/n\}$. The sequence is clearly bounded, but it has no convergent subsequence. $\endgroup$ – Alex Ortiz Dec 3 '17 at 21:53
  • $\begingroup$ @Pumpkin My initial claim was not true. You need an infinite range. And then it is only true in complete metric spaces. $\endgroup$ – M. Winter Dec 3 '17 at 21:55

The statements are equivalent so long as we restrict attention to sequences with infinite range.

Let $\{p_n\}$ be a bounded sequence. If the range of $\{p_n\}$ is finite, then the sequence must assume some value in its range infinitely many times, which gives a convergent subsequence. Otherwise, the range of $\{p_n\}$ is an infinite subset of $\Bbb R^n$ that is bounded, so it has an accumulation point, and hence a convergent subsequence again.

If every bounded sequence has a convergent subsequence, in particular, the limit of that convergent subsequence is an accumulation point for the sequence as long as the range of the sequence is infinite (no finite set has an accumulation point). Thus given a bounded infinite set, find a countable subset, and pick a convergent subsequence. The limit of that convergent subsequence is a limit point of the infinite set.

  • $\begingroup$ "If the range of $\{p_n\}$ is finite, then the sequence must assume some value in its range infinitely many times" is it because of pigeonhole principle? $\endgroup$ – Pumpkin Dec 3 '17 at 21:54
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    $\begingroup$ @Pumpkin Sure, something like that. If it only assumed each of its finitely many points finitely many times, then it would not be a sequence. $\endgroup$ – Alex Ortiz Dec 3 '17 at 21:54
  • $\begingroup$ Does this answer the question? Didn't he ask, if "every bounded infinite set in $\mathbb{R}^n$ has an accumulation point? $\endgroup$ – Malte Winckler Dec 3 '17 at 21:59
  • $\begingroup$ @MalteWinckler Actually, it's not clear to me now what the OP is asking. $\endgroup$ – Alex Ortiz Dec 3 '17 at 22:01
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    $\begingroup$ @Pumpkin, typically the Bolzano-Weierstrass theorem is stated as "Every bounded sequence in $\Bbb R^n$ has a convergent subsequence," not as "Every infinite subset of a bounded set in $\Bbb R^n$ has an accumulation point." $\endgroup$ – Alex Ortiz Dec 3 '17 at 22:30

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