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First thing's first, let me write the down the definition of Bolzano-Weierstrass theorem that I'm using:

Bolzano-Weierstrass theorem for sequences: Every bounded sequence in $\mathbb{R}$ has a limit point in $\mathbb{R}$

My question is— How do I visualize this intuitively? Note that I'm not asking for a proof of this theorem, I only need some intuition behind it to internalise the idea.

What I have come across so far is that "Hah, since the sequence is bounded, that is each $|x_n|<M$ for some real $M$, for all $n$ and the fact that there are infinitely many terms in a sequence, the terms must huddle around a point".

I don't find this very convincing for instance, sure there are infinitely many terms in $(x_n)$ but the terms are only countably infinite (i.e. $n \mapsto x_n$) and the cardinality of $[-M, M]$ is greater than $\aleph _0$ so I don't see why each term $x_n$ can't be assigned to some real number $[-M,M]$ without huddling around a point. I'm confused.

If you can share any other intuition that'd be even more lovely. But I'd also like to know what is wrong with my counter-reasoning while you're at it.

William
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  • Do you know the Monotone Convergence Theorem? – Mastrem Jun 11 '21 at 09:55
  • @Mastrem Every bounded monotone sequence is convergent? Yes. I think so, what about it? – William Jun 11 '21 at 10:02
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    If I'm not mistaken, the usual proof of Bolzano-Weierstrass uses the fact that every infinite sequence in $\mathbb{R}$ contains a monotone subsequence, a fact which (at least to me) is fairly intuitive. Once you have that, apply the MCT to prove that every bounded sequence contains a convergent subsequence, which is Bolzano-Weierstrass. (The limit of this subsequence is your limit point). But this is essentially Alan's answer, which you weren't satisfied with – Mastrem Jun 11 '21 at 10:55

2 Answers2

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You are guaranteed to be able to pick out at least one monotonic (Non-increasing or non-decreasing) subsequence from your sequence. All monotonic bounded sequences converge, therefore you have a subsequence converging to a point, which will make it a limit point

Alan
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  • Yes but this explanation just shifts the problem from the perspective of limits points to subsequence that converge to them which is an alternative definition of Bolzano-Weierstrass theorem. I specifically mentioned the definition because I wanted build up intuition on that definition only. – William Jun 11 '21 at 09:59
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    @William Say $(a_n){n\ge 1}$ is the bounded sequence, and $a$ is a limit point, then for all $k\ge 1$ there exists an integer $n_k$ with $|a{n_k}-a|<2^n$. Clearly $(n_k){\ge 1}$ contains a strictly increasing sequence, which gives a subsequence of $(a_n){\ge 1}$ converging to $a$. That is to say, the two definitions are obviously/intuitively equivalent. (Plot the sequence and its limit point, see how you can take points of the sequence arbitrarily close to the limit point and how these obviously make up a convergent subsequence) – Mastrem Jun 11 '21 at 11:02
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    @William: That a bounded monotone sequence converges is a straightforward consequence of the completeness property (every bounded, non-empty set of reals has a supremum and an infimum), so it seems to me Alan's answer substantially reduces the question, i.e., is not a mere shifting of perspective. :) – Andrew D. Hwang Jun 11 '21 at 13:24
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Let $(a_n)_{n\ge 1}$ be a bounded sequence in $\mathbb{R}$. Without loss of generality, we assume that the sequence takes its values in $[0,1]$.

If the sequence contains any repeated value, then per definition that repeated value is a limit point. Therefore, if the sequence has no limit point, $\{a_n:n\ge 1\}$ is infinite. It suffices to show that every infinite subset $S\subseteq [0,1]$ has a limit point.

This assertion comes down to compactness of $[0,1]$. Assume that $S$ has no limit point, then for all $a\in [0,1]$ there exists some $\epsilon(a)>0$ such that the punctured open disk $\dot{U}_{\epsilon(a)}(a)$ contains no points of $S$.

Note that $\bigcup_{a\in [0,1]}U_{\epsilon(a)}(a)=[0,1]$, so by compactness there exists a finite subset $T\subseteq[0,1]$ with $\bigcup_{a\in T}U_{\epsilon(a)}(a)=[0,1]$. Now $$S\subseteq[0,1]\setminus\bigcup_{a\in T}\dot{U}_{\epsilon(a)}(a) \subseteq T,$$ so $S$ is finite; a contradiction.

Mastrem
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