So, I've started reading about dimension theory and currently am dealing with a lemma which is used in a proof of $\dim(\mathcal{\ell}^{2}_{\mathbb{Q}})=1$. This lemma says that convergence of a sequence in a sphere in $\mathcal{\ell^2}$ to a vector in the sphere is the same as coordinatewise convergence and is stated as follows:
Let $x=(x_{i})^{\infty}_{i=1} $, $x^n=(x^{n}_{i})^{\infty}_{i=1} \in \ell^2$ for each $n=1, 2, \ldots$ such that $\left\|x^n\right\|=\left\|x\right\|$ for each $n$. Then $x^n \rightarrow x \iff x^{n}_{i} \rightarrow x_{i}$ for each $i$.
There are two dillemas I have and need help about. I hope questions are clearly stated.
I know that, in general, coordinatewise convergence does not imply convergence in infinite-dimensional spaces (or in $\ell^2$ in this case) but I can't see why would it hold in this case. I just need an explanation of idea or background.
If I want to describe an open ball in $\ell^2$ space (in topological sense) do I do that by saying that this open ball contains all square summable sequences (bounded) which limits are in closure of that ball? Problem is, I don't understand very well how open sets in topology on $\ell^2$ (which is inherited from product topology) are defined.
$\ell^2$ is Hilbert space of all square summable sequences of real numbers.
The proofs uses strongly the Hilbert space structure, and more precisely the inner product. Notice indeed that $$\lVert x^ n-x\rVert^2=2(\lVert x\rVert^2-\langle x,x^n\rangle).$$
I'm not sure which kind of description you want: the open ball of center $x$ and radius $r$ consists of all the $y$ for which $$\sum_{n=1}^{+\infty}|x_n-y_n|^2\lt r.$$