topology neighborhood base Let $d$ be a pseudo metric on a non-empty set $X$ and define 
$ B_n(x) := \{y \in X: d(x,y) \leq 1/n\}, x \in X, n \in\Bbb N$
Show that $B_x := \{B_n(x):n \in\Bbb N\}, x \in X$ define neighborhood bases 
inducing a first countable topology $\tau$ on $X$, which is separable if and
only if $d$ is a metric.
Thanks a lot!
 A: HINT: The first step is to see exactly what topology (if indeed it is one) is being defined. What’s intended is to let 
$$\tau=\left\{U\subseteq X:\forall x\in U\exists n\in\Bbb Z^+\big(B_n(x)\subseteq U\big)\right\}\;,$$ 
and the main assertion it that $\tau$ is a first countable topology on $X$. 
To prove that $\tau$ is a topology on $X$, you must show three things:


*

*$\varnothing,X\in\tau$.  

*If $\mathscr{U}\subseteq\tau$, then $\bigcup\mathscr{U}\in\tau$; i.e., $\tau$ is closed under arbitrary unions.  

*If $U,V\in\tau$, then $U\cap V\in\tau$; i.e., $\tau$ is closed under intersection. 


All of these follow very easily from the definition of $\tau$; just remember that $B_n(x)\subseteq B_m(x)$ whenever $n\ge m$. It’s also easy to see that this topology makes $\{B_n(x):n\in\Bbb Z^+\}$ a nbhd base at $x$ for each $x\in X$, though not necessarily of open nbhds.
The final assertion, that $\langle X,\tau\rangle$ is separable iff $d$ is a metric, is false in both directions. For one direction let $X=\Bbb Z$, and define the pseudometric $d$ by
$$d(m,n)=\begin{cases}1,&\text{if }|m|\ne|n|\\0,&\text{otherwise}\;.\end{cases}$$
Then $d(-1,1)=0$, so $d$ is not a metric, but $\Bbb Z$ is countable, so the space is necessarily separable. In this space a set $U$ is open iff it has the following property: if $n\in U$, then $-n\in U$.
For the other direction let $d$ be the discrete metric on $\Bbb R$ given by
$$d(x,y)=\begin{cases}1,&\text{if }x\ne y\\0,&\text{otherwise}\;.\end{cases}$$
Then $d$ is a metric, and every subset of $\Bbb R$ is open. Since $\Bbb R$ is uncountable, this space is not separable.
