Apparently the result "every bounded infinite set in $\mathbb{R}^n$ has a limit point" is equivalent to the Bolzano-Weierstrass theorem (every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence): see e.g. Proving every bounded infinite set has a limit point without using Bolzano-Weierstrass or https://uccs.edu/goman/sites/goman/files/inline-files/A%20short%20proof%20of%20the%20Bolzano-Weierstrass%20Theorem.pdf. I am not sure how this can be, given that the Bolzano-Weierstrass theorem gives you a convergent sequence in the bounded infinite set, but there is no guarantee that its limit would be a limit point (as here: Is the limit of a convergent sequence always a limit point of the sequence or the range of the sequence?). So can someone explain please? Thanks very much.
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The counter examples provided concern sequences that are eventually constant. But if they are eventually constant, then the range of values of that sequence is finite, not infinite. So it does not contradict the statement. – Christopher A. Wong Dec 09 '20 at 22:35
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So is there a way to show that any such counterexample would have to be eventually constant? – Prasiortle Dec 09 '20 at 22:36
1 Answers
As pointed out in the comments, the difference has to do with sequences which are eventually constant. Let $(x_n)$ be a bounded sequence. If we consider the set $A = \{x\in\mathbb{R}^d: x=x_n\text{ for some $n$}\}$, we cannot immediately apply the limit-point formulation of Bolzano-Weierstrass. Indeed, this set $A$ is bounded since $(x_n)$ is a bounded sequence, but $A$ may only have finitely many points and thus not satisfy the "infinite set" part of the hypotheses.
I claim that if $A$ is a finite set, then any convergent subsequence $(x_{n_k})$ of $(x_n)$ is eventually constant. To do this, just pick $\epsilon>0$ so small that $B(a,\epsilon)\cap B(a',\epsilon)=\emptyset$ for any $a,a'\in A$ with $a\neq a'$ (which is possible since $A$ is finite). If $x_{n_k}\to x$ as $k\to\infty$, then $x\in A$ since $A$ is closed (it is a finite union of closed singletons). With $\epsilon$ as above, there should be $K$ such that $k\geq K$ implies $x_{n_k}\in B(x,\epsilon)$. Since $x_{n_k}\in A$ though, our choice of $\epsilon$ forces $x_{n_k} = x$. Thus $x_{n_k}=x$ for $k\geq K$, i.e., the subsequence was eventually constant.
If the set $A$ happens to be infinite, then we infer there is a limit point $x$, and by definition of limit point we may construct a (sub)sequence of points from $A$ which converges to $x$.
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This shows that every bounded sequence where the set of all terms is infinite has a convergent subsequence. What about the reverse direction? If we know there is a convergent subsequence, how can we find a limit point? – Prasiortle Dec 09 '20 at 23:43
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You can't in general. If you have a convergent subsequence $x_{n_k}$ and you put $A':={x: x=x_{n_k}}$, then the cardinality of $A'$ exactly determines whether the limit is a topological limit point. If it is finite, it is not. If it is infinite, then it is. – Glare Dec 09 '20 at 23:48
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1To be clear, finite cardinality of $A'$ rules out the limit of the $x_{n_k}$ from being a topological limit point of $A'$. It could still be a limit point of the original $A$ though. Something like $(x_n) = (2,2+1/2, 2, 2+1/3, 2, 2+1/4, 2,\dotsc)$ illustrates this point. The odd subsequence has $A'={2}$, but $2$ is still a topological limit point of $A = {x:x=x_n}$. – Glare Dec 09 '20 at 23:58