I got the following exercise :
Let $f_n$ be a sequence of functions defined on a countable set $A$ and such that :
$$\forall x \in A, \exists M_x > 0, \forall n, \mid f_n(x) \mid \leq M_x$$
Then prove that it’s possible to find an extraction $\phi$ such that forall $x$, $f_{\phi(n)}(x)$ converges for all $x$ in $A$. It’s not difficult to construct such an extraction.
Moreover in a recent question I’ve asked it says that Bolzano-Weiestrass is true only in finite dimensional vector spaces.
Yet, let’s say I take the space of sequence $u : \mathbb{N} \to \mathbb{R}$ then this is an infinite vector space.
But every sequence of this vector space : $a_n$ can be represented as follow : $$a_1 = (f_1(x_1), f_1(x_2),...), a_2=(f_2(x_1), ...), a_n = (f_n(x_1),....,)$$
Hence every sequence can be represented as a sequence of functions $f_n$ such that $f_i : A \to \mathbb{R}$. Hence, by the property above this infinite vector space have the Bolzano-Weirstrass property since (thank’s to the exercises Îve cited above) we can find an extraction such that the sequence $a_{\phi(n)}$ converges.
So what’s the mistake I am doing here ?
Your are working in a compact subset of an infinite-dimensional vectorspace. The (general) Bolzano-Weierstrass theorem says that every sequence in a compact metrizable space has a convergent subsequence. You work in the product $\prod_{x\in A}[-M_x,M_x]$, which is compact and metrizable.