# 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?

• Hint: take first all linear combinations with rational coefficients. Jun 7 at 11:09

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 $$\Bbb C$$.

• Very clear and comprehensive. Thanks a lot :)
– L.G.
Jun 7 at 12:20

Hint: Consider the linear span of $$\{x_n\}$$ over $$\mathbb{Q}$$. This is a countable set. Try to show that it is dense.