# Is the statement of theorem true: Every bounded sequence in $\Bbb R^n$ contains a convergent subsequence.

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 $\{p_n\}$.

• Your sequence $(p_n)$ should not be stationnary. the range $\{p_n\}$ must be infinite. – hamam_Abdallah Dec 3 '17 at 21:51
• @M.Winter is it true only for $\Bbb R^n$ or for all metric spaces? – Pumpkin Dec 3 '17 at 21:52
• @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. – Alex Ortiz Dec 3 '17 at 21:53
• @Pumpkin My initial claim was not true. You need an infinite range. And then it is only true in complete metric spaces. – M. Winter Dec 3 '17 at 21:55

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 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? – Pumpkin Dec 3 '17 at 21:54
• Does this answer the question? Didn't he ask, if "every bounded infinite set in $\mathbb{R}^n$ has an accumulation point? – Malte Winckler Dec 3 '17 at 21:59
• @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." – Alex Ortiz Dec 3 '17 at 22:30