Proving Urysohn's Metrization Theorem on X

[math8]

the question is to prove Urysohn's metrization theorem. But there some steps I need to show first.
Assuming X is normal, second countable. We show there is a homeomorphism of X onto a subspace of [0,1]^w (the Hilbert cube which is metrizable), so X is metrizable.

We first show we can assume that X is infinite.
Suppose X is finite. Since X is normal, it is T1, so every singleton is closed. Hence we can see that every subset of X is closed so every subset is open. So X has the discrete topology. So from here, how can I show that X is metrizable (what would be a homeomorphism between X and [0,1]^w ?).

Now, I can show that X has a countably infinite basis {B1,B2,...} where each Bn is neither X nor the empty set. How do I show that the set of all ordered pairs (Bi,Bk) such that closure(Bi) C Bk is countably infinite?
I can show it's countable because that set is a subset of {B1,B2,...} X {B1,B2,...} which is countable. But how do I show it is infinite?

 

[lippyka]

To show that the set of all ordered pairs (Bi,Bk) such that closure(Bi) C Bk is countably infinite, you will need to use the fact that X is normal and second countable. Since X is normal, we know that for each Bn there is a open set Un containing its closure. Since X is second countable, we can assume that there is a countable basis {Un1,Un2,...} of open sets containing the closures of Bn. So, given any Bn, its closure is contained in some Unm and so (Bn,Unm) is an element of the set of all ordered pairs (Bi,Bk) such that closure(Bi) C Bk. Thus, this set contains at least one element for each Bn and so it is countably infinite.