On the Definition of a Separable Space The only type of separability definition I know that a separable topological space is one that has a countable dense subset. In particular a metric space is separable if it has a countable dense set.
But in the proof of Theorem 12.39 in Bruckner's Real Analysis book, it is assumed that if ${\{x_n}\} \subset X$ is countable and the linear space spanned by the set ${\{x_n}\}$ be dense in $X$ then $X$ is separable.
My question is : ${\{x_n}\}$ is countable but $\operatorname{Span}({\{x_n}\})$ is uncountable, because $\mathbb{R}$ and $\mathbb{C}$ is uncountable makes a linear combination of uncountable and countable, uncountable. So how $\operatorname{Span}({\{x_n}\})$ is countable such that its closure being $X$ makes $X$ separable?
 A: Hint: Consider the linear span of $\{x_n\}$ over $\mathbb{Q}$. This is a countable set. Try to show that it is dense.
A: Let us treat first case where the field is $\Bbb R$.
If ${\{x_n}\} \subset X$ is countable and the linear space spanned by the set ${\{x_n}\}$ be dense in $X$, then consider
$$ C = \left \{ \sum_{i=1}^k q_i x_i :k \in \Bbb N, k>0 \text{ and } \forall i (x_i \in \{x_n\} \text{ and } q_i \in \Bbb Q) \right \} $$
It is immediate that $C$ is countable.  Now, given any $y \in X$ and any $\varepsilon >0$, since $\operatorname{Span}({\{x_n}\})$ is dense in $X$, there is $x \in \operatorname{Span}({\{x_n}\})$ such that $\|y-x\| < \frac{\varepsilon}{2}$.
Note that $x= \sum_{i=1}^k r_i x_i $, where $k \in \Bbb N, k>0$ and $\forall i \in \{ 1,\cdots, k\}$,  $x_i \in \{x_n\}$ and $ r_i \in \Bbb R $. Since $\Bbb Q$ is dense in $\Bbb R$, it is follows that there are, for each $i \in \{ 1,\cdots, k\}$, $q_i \in \Bbb Q$, such that
$$ \left \| \sum_{i=1}^k r_i x_i  - \sum_{i=1}^k q_i x_i \right \| < \frac{\varepsilon}{2}$$
that is
$$ \left \| x  - \sum_{i=1}^k q_i x_i \right \| < \frac{\varepsilon}{2}$$
Clealy, $\sum_{i=1}^k q_i x_i  \in C$ and
$$\left \| y  - \sum_{i=1}^k q_i x_i \right \|\leq \|y-x\|+ \left \| x  - \sum_{i=1}^k q_i x_i \right \| < \varepsilon $$
So $C$ is dense in $X$. So $X$ is separable.
The case where the field is $\Bbb C$ is completely similar, using the fact that $\Bbb Q + \Bbb Q i$ is dense in $\Bbb C$.
