# Statement of Bolzano-Weierstrass theorem

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?

• I'm guessing that your $\;d\;$ is the usual, Euclidean, metric...**who told you that** $\;\Bbb R\;$ is compact? It is not, – DonAntonio Oct 30 '18 at 18:39
• Please give me an example of a subsequence of $1,2,3,\ldots$ that converges to a point in $\mathbb R$. – José Carlos Santos Oct 30 '18 at 18:42
• 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 – Dair Oct 30 '18 at 18:46
• Maybe you've misinterpreted that every closed and bounded subset of $R$ is compact ? – Gabriel Romon Oct 30 '18 at 18:50
• Oops, my bad; thanks, guys. But I guess $\varnothing$ is compact since it is closed and bounded? – Siddhartha Oct 30 '18 at 19:08

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.