How close is a countable generator related to separability? Let $(\Omega, \tau)$ be a topological space. If the space is separable then exists countable $S\subseteq \Omega$ so that for any $V \in \tau$ and $v \in V$ there exists a $W \in \tau$ and $s \in S$ and $s \in W$ so that $v \in W \subseteq V$
hence I could generate $V$ via countably many open balls around elements from $S$ so $S$ has to be a countable generator 
And if there is a countable generator (say $\mathcal{S})$ then for any $v \in \Omega$ and any neighborhood $V\in \tau$ of $v$ there must exist $S \in \mathcal{S}$ so that $v \in S \subseteq V$ and hence separability?
So the $2$ statements are equivalent? 
 A: A space is called 2nd-countable iff it has a countable base (basis). A separable metrizable space is 2nd-countable but a non-metrizable separable space need not be 2nd-countable.
(1). Suppose the topology $T$ on the set $X$ is generated by the metric $d.$ Let $D$ be a dense subset of $X.$ Then $B=\{B_d(y,q):y\in D\land q\in \Bbb Q^+\}$ is a base for $T.$
Proof: For $x\in U\in T$ take $r\in \Bbb R^+$ such that $B_d(x,r)\subset U.$ Now (because $D$ is dense) take $q\in \Bbb Q\cap (0,r/2)$ and $y\in D \cap B_d(x,q).$  The Triangle Inequality shows that $B_d(y,q)\subset B_d(x,r).$ So $x\in B_d(y,q)\in B$ and $B_d(y,q)\subset U.$
Now if $D$ is countable then $B$ is countable.
(2). An example of a separable non-2nd-countable space is the Sorgenfrey Line $S$: Let $C=\{[x,y):x,y\in \Bbb R\land x<y\}$ be a base for a topology $T_S$ on the set $\Bbb R. $ Clearly $\Bbb Q$ has non-empty intersection with every $c\in C$ so $\Bbb Q$ is dense with respect to $T_S.$
Now let  $B$ be any base for $T_S$. For each $x\in \Bbb R$ take $U(x)\in B$ such that $x\in U(x)\subset [x,x+1).$ If $x\ne x'$ then $\min U(x)=x\ne x'=\min U(x')$ so $U(x)\ne U(x').$ Hence $U:\Bbb R \to B$ is injective, so $B$ cannot be countable.
