Let $X$ be a topological space that satisfies the second axiom of countability. Show that if ...

Question

Let $X$ be a topological space that satisfies the second axiom of countability. Show that if $\cal{B}$ is a basis for $X$ then there is $\cal{B'} \subseteq \cal{B}$ countable such that $\cal{B'} $ is a basis for $X$.

A topological space $X$ is second-countable if there exists some countable collection $U = \{U_i \} _i^{\infty}$ of open subsets of $X$ such that any open subset of $X$ can be written as a union of elements of some subfamily of $U$.

If $\cal{B}$ is basis for X then there is a family $\cal{B}$ of open subsets of $X$ such that every open set of the topology is equal to the union of some subfamily of $\cal{B}$. But $X$ is also second countable, so, let $\cal{B'}$ open subset of X be the union of elements of some subfamily of the countable collection U. But $\cal{B'}$ can be written as a union of some subfamily of $\cal{B}$ so $\cal{B'} \subseteq \cal{B}$

Is it right so far? I'm stuck, how do I show that $\cal{B'}$ is base?

Answer 1

Put $\mathcal{U}=\left\{U_i\right\}_i^\infty$. For each $U_i,U_j\in \mathcal{U}$ with $U_j\subset U_i$ choose one set $B_{ji}\in \mathcal{B}$ such that $U_j \subset B_{ji} \subset U_i$, if there is at least one such $B_{ji}$.

The collection $\mathcal{B'} = \left\{ B_{ji} \right\} \subset \mathcal{B}$ is countable, being indexed by a subset of $\mathbb{N}\times\mathbb{N}$.

We have:

1. For each $x\in X$ there is a set $U_i\in \mathcal{U}$ such that $x\in U_i$. Since $\mathcal{B}$ is a base, there is a set $A\in \mathcal{B}$ such that $x\in A\subset U_i$. Since $\mathcal{U}$ is a base, there is a set $U_j\in \mathcal{U}$ such that $x\in U_j \subset A \subset U_i$. Since there is at least one element between $U_i$ and $U_j$ there is $B=B_{ji}\in\mathcal{B'}$ such that $x\in U_j \subset B \subset U_i$. Therefore, for each $x\in X$ there is a set $B\in\mathcal{B'}$ such that $x\in B$.
2. For each $x\in B_1\cap B_2$, with $B_1,B_2\in\mathcal{B'}$, there are sets $V,W\in\mathcal{U}$ such that $x\in V\subset B_1$ and $x\in W\subset B_2$. Since $\mathcal{U}$ is a base, there is a set $U_i\in\mathcal{U}$ such that $x\in U_i\subset V\cap W$. By the same reasoning of the previous point, there is a set $U_j\in \mathcal{U}$ and a set $B=B_{ji}\in\mathcal{B'}$ such that $x\in U_j \subset B\subset U_i\subset V\cap W\subset B_1\cap B_2$. Therefore, for each $B_1,B_2\in\mathcal{B'}$ and for each $x\in B_1\cap B_2$ there is a set $B\in\mathcal{B'}$ such that $x\in B \subset B_1\cap B_2$.

Then $\mathcal{B'}$ satisfies the two conditions for it to be a base.