Question: We know that a bounded sequence in $\mathbb{R}$ must have a convergent sub-sequence from Bolzano-Weierstrass theorem. However, will a bounded sequence $(x_n)$ in $\mathbb{R}$ necessarily induce a compact subset of $\mathbb{R}$? By "induce", I mean the subset contains all the real numbers in $(x_n)$. Prove or provide counter-example.
My try: No, not necessary. To be compact the subset also has to be closed, which cannot be inferred from the question. Alternatively, to be compact, every sequence from the set has to have a convergent sub-sequence. This is not guaranteed by the question. For example, let $(x_n)=(1, 1.4, 1, 1.41, 1, 1.414, 1, 1.4149, 1, 1.41495, \cdots)$ without containing $\sqrt{2}$. Then if we form a sequence from this set by picking the even items, then it will not have a convergent sub-sequence. Is my answer right, please? Thank you!
The set $A=\{1,1.4,1.414,1.4149,\dots\}$ is not compact, as you say, but you need to be a little careful with your justification. If you make a sequence from every second.element of this ordered set, then it will have a convergent subsequence--in fact, every subsequence of that sequence will converge to $\sqrt 2.$ However, no subsequence converges to a point lying in $A.$ That is a way we can see that $A$ is not compact.