I'm confused about constructing a family of subsequence using a diagonalization procedure. Often I see the following argument during a proof: "using a diagonalization procedure we can assume....."
What exactly is meant by this? I just know Cantor's diagonalization method for proving some cardinality stuff. Unfortunately I don't see how this apply in this situation. I'm aware of asking a very general thing, but hopefully it's clear what I try to ask.
Just to be sure that there's no misunderstanding, here is an example:
Suppose $X$ is a separable space and $(\phi_k)_{k\in \mathbb{N}}$ is a bounded subset of the dual of $X$ (topological dual). Then there is a subsequence $\Omega \subset \mathbb{N}$ and an element $\phi \in X^* $ such that $ \phi_k $ converges weak$^*$ to $\phi$ for this subsequence.
Now in the proof, we argue: let $(x_n)$ be dense in $X$. Then looking at $ \phi_k(x_1) $. This is bounded sequence in $ \mathbb{R} $ therefore it exists a subset $\Omega_1 \subset \mathbb{N} $ such that $\phi_k(x_1)$ converges for $ k\in \Omega_1$. Repeating this for ever $n \in \mathbb{N}$ we get an decreasing family of subsets (subsequence) of $\mathbb{N}$,
$\mathbb{N}\supset \Omega_1 \supset \dots \supset \Omega_l \dots \supset \Omega$
Why can I do this? Should I choose a subsequence of $\Omega_l$ using Bolzano-Weierstrass to get $\Omega_{l+1}$? Why do I know there's a $\Omega$ ? Why do I know that $\lim_k\phi_k(x_n)$ converges for ever $n\in \mathbb{N}$ using $\Omega$? Are there any assumption to apply this diagonalization argument? How does the diagonalization argument works in this example?
I'm very thankful for your help. If you need further explanations on what I exactly want to ask, let me know.