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My textbook (Thomson, Bruckner and Bruckner) says:

Theorem 2.40 (Bolzano-Weierstrass) Every bounded sequence contains a convergent subsequence.

Since this is from the 2nd chapter, I'm assuming the textbook is referring to sequences in $\mathbb{R}$. Since $(\mathbb{R},\ d)$ is compact, every sequence in $\mathbb{R}$ has a subsequence that converges to a point in $\mathbb{R}$. Doesn't this mean the theorem stated in my textbook is unnecessarily weak?

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    $\begingroup$ I'm guessing that your $\;d\;$ is the usual, Euclidean, metric...**who told you that** $\;\Bbb R\;$ is compact? It is not, $\endgroup$ – DonAntonio Oct 30 '18 at 18:39
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    $\begingroup$ Please give me an example of a subsequence of $1,2,3,\ldots$ that converges to a point in $\mathbb R$. $\endgroup$ – José Carlos Santos Oct 30 '18 at 18:42
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    $\begingroup$ There seems to be some misunderstanding of the idea "compactness". The formal definition using open covers is a bit abstract and might be what you were introduced to first. To gather some intuition on the idea here are some notes on the subject: math.ucla.edu/%7Etao/preprints/compactness.pdf $\endgroup$ – Dair Oct 30 '18 at 18:46
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    $\begingroup$ Maybe you've misinterpreted that every closed and bounded subset of $R$ is compact ? $\endgroup$ – Gabriel Romon Oct 30 '18 at 18:50
  • $\begingroup$ Oops, my bad; thanks, guys. But I guess $\varnothing$ is compact since it is closed and bounded? $\endgroup$ – Thomas Oct 30 '18 at 19:08
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If $(u_n)$ is bounded then

$$(\forall n\in\Bbb N)\;\; u_n\in[a,b]$$

which means that $(u_n)$ is a sequence of elements in the compact $[a,b]$ and it has a convergent subsequence.

$(\Bbb R,| \; |)$ is not compact.

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