Answer to Question #217056 in Differential Geometry | Topology for Prathibha Rose

Question

Prove that every first countable space is second countable

Answer 1

Let $(X,d)$ be a metric space and let $x\in X$. Consider the neighborhood basis

$\mathbb{B}_x=\{B_r(x) | r>0,r\in \mathbb{Q}\}$.

Clearly, this basis is countable. Hence by definition, every metric space is first countable.

Now let $(\mathbb{R},d)$ be a metric space where $d$ is the discrete metric

$d(x,y) = \begin{cases} 0 & \text{if $x=y$} \\ 1 & \text{if $x\neq y$} \end{cases}$

Clearly, a base for the topology induced by this metric must contain all singleton subsets of $\mathbb{R}$. So it is uncountable (since $\mathbb{R}$ is uncountable). Hence $(\mathbb{R},d)$ is not second countable.