Question

let X b a second countable space prove that any open base for x has a countable subclass which is also an open base

Answer 1

Answer:

Another proof: let B={Bi∣i∈I} be any base for X and C={Cn∣n∈N} be a countable base for X. Consider the set of pairs

P={(n,m)∈N×N∣∃i∈I:Cn⊂Bi⊂Cm}

which is countable, as a subset of all pairs from a countable set. For each (n,m)∈P fix i=i(n,m) as the i in the definition of P. The set B′={Bi(n,m)∣(n,m)∈P} is thus a countable subfamily of B and is the required base:

Let O be any open subset of X and let x∈O. As C is a base, there exists some m∈N such that x∈Cm⊂O, and applying B is a base we find j∈I such that x∈Bj⊂Cm, and again applying that C is a base, we find Cn so that x∈Cn⊂Bj.

Now note that (n,m) is in P as witnessed by j. We don't know that j=i(n,m) but we don't care, because for i=i(n,m) we also have x∈Bi⊂O, as required. So we have found a member of B′ between every open O and its elements, showing B′ is a base.