Let’s say I have a vector space $V$ endowed with an order relation : $\leq$.
Then is Bolzano-Weirstrass theorem true in $V$ ?
In class we have seen Bolzano-Weirstrass theorem only in $\mathbb{R}$, that’s why I am wondering if this theorem is true in arbitrary ordered structure like for example a vector space.
In the case where it’s not true for every vector space, is it possible to classify the vector space in which Bolzano-Weirstrass theorem holds ?
Bolzano-Weirstrass holds for a normed vector space $V$ if and only if $\dim(V)<\infty$.
Here's the proof of "$\Leftarrow$":
Proof. If $V=\mathbb{R}^n$ then you take a sequence $(a_m)\subseteq V$. Write it down as $a_m=(a^1_m,\ldots, a^n_m)$ where each coordinate is a sequence of reals. Now you use induction: pick a subsequence of first coordinate that is convergent (such exists by Bolzano-Weirstrass for reals). Then you take the subsequence of the whole sequence $a_m$ indexed by the same indexes as the first coordinate subsequence we've just picked. Then you take subsequence of second coordinate in the same manner, then third, fourth, and so on. Eventually, after finitely many steps you end up with a subsequence of $(a_m)$ such that each coordinate is convergent and thus so is $(a_m)$. $\Box$
The "$\Rightarrow$" is a bit more complicated. Here you can have a look at the sketch of the proof.
For a simple counterexample consider the direct product $V=\mathbb{R}\oplus\mathbb{R}\oplus\cdots$ with Euclidean norm. Then let $e_n=(0,0,\ldots,0,1,0,\ldots)$ where $1$ is on $n$-th coordinate, i.e. $\{e_n\}$ is the standard basis. Then the sequence $(e_n)$ is bounded but it does not have a convergent subsequence because $\lVert e_n-e_m\rVert=\sqrt{2}$ for $n\neq m$.
