Question #217056

Prove that every first countable space is second countable


1
Expert's answer
2021-07-18T18:19:43-0400

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

Bx={Br(x)r>0,rQ}\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 (R,d)(\mathbb{R},d) be a metric space where dd is the discrete metric

d(x,y)={0if x=y1if xyd(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 R\mathbb{R}. So it is uncountable (since R\mathbb{R} is uncountable). Hence (R,d)(\mathbb{R},d) is not second countable.


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