Normed Vector Space with Schauder Basis is Separable

Theorem
Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space over $\R$.

Suppose $X$ admits a Schauder basis.

Then $X$ is separable.

Proof
Let $\set {\mathbf E_i : i \in \N}$ be a Schauder basis.

Then:


 * $\ds \forall x \in X : \exists \sequence {x_i'}_{i \mathop \in \N} \in \R : x = \sum_{i \mathop = 0}^\infty x_i' \mathbf E_i$

Let $\ds \mathbf e_i = \frac {\mathbf E_i}{\norm {\mathbf E_i}}$ and $\ds x_i=x_i' \norm{\mathbf E_i}$.

Then $\ds x = \sum_{i \mathop = 0}^\infty x_i \mathbf e_i$ where $\forall i \in \N : \norm {\mathbf e_i} = 1$.

Let $\ds Y = \set {\sum_{i \mathop = 0}^n q_i \mathbf e_i : q_i \in \Q, n \in \N}$ be a countable set.

$Y$ is spanned by $\set {\mathbf e_i : i \in \N}$.

Also, Rational Numbers form Subset of Real Numbers.

Hence, $Y \subseteq X$.

We have that rationals are everywhere dense in reals.

Then for any $i \in \N$:


 * $\forall x_i \in \R: \forall \epsilon_i \in \R_{>0}: \exists q_i \in \Q: \size {x_i - q_i} < \epsilon_i$

Hence:

By the existence of Schauder basis, the limit $\ds x = \sum_{i \mathop = 0}^\infty x_i \mathbf e_i$ exists.

Thus:


 * $\ds \forall \epsilon' \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N: n > N \implies \norm {\sum_{i \mathop = 0}^\infty x_i \mathbf e_i - \sum_{i \mathop = 0}^n x_i \mathbf e_i} < \epsilon'$

Therefore:

Let $\epsilon \in \R_{> 0}$.

Let $\epsilon'$ and $\sequence {\epsilon_i}_{0 \mathop \le i \le n}$ be such that:


 * $\ds \epsilon > \epsilon' + \sum_{i \mathop = 0}^n \epsilon_i$.

Let $n > N$.

Then:


 * $\ds \norm {x - \sum_{i \mathop = 0}^n q_i \mathbf e_i} < \epsilon$

$\epsilon$ was arbitrary.

Altogether:


 * $\ds \forall x \in X : \forall \epsilon \in \R_{> 0} : \exists y_{i \mathop \in \N} \in Y : \norm {x - y_i} < \epsilon$

By definition, $X$ is separable.