I struggling to understand a partial step in the solution to an exercise:
Given a seq. of seq. $\{a^{(n)}\}_{n \in \mathbb N} \subseteq \ell^2$ such that $|a^{(n)}_k| \leq 1 \forall n,k \in \mathbb N$, there exists a subseq. $n_j$ such that $a^{(n_j)}_k \to a_k$ converges pointwise $\forall k > \in \mathbb N$.
The solution brushes over this by stating that for each $k \in \mathbb{N}$ the seq. $\{a^{(n)}_k\}_{n \in \mathbb{N}}$ is a bounded seq. in $\mathbb R$ and hence has a converging subseq. and hence there is a reordering of $n$ such that $a^{(n)}_k \to a_k$ converges pointwise $\forall k \in \mathbb N$. I understand that we can find a reordering of a particular $k$, but I do not understand how we can just conclude that there is one that works for all $k$.
If you take a reordering for a particular $k$, then throwing out all the other elements, you can now use this new ordering for the next $k$ (possibly $k+1$). On the other hand, there is no straight forward guarantee that after infinite steps, an infinite subsequence will remain. But if at every step you keep one element, then the algorithm still runs, while keeping infinintely elements.