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.
Equivalency of forms of Bolzano-Weierstrass theorem
metric-spacesreal-analysis
Best Answer
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 $x$.