every normed space with Schauder basis is separable

dense in

Here we show that every normed space that has a Schauder basis is separable (http://planetmath.org/Separable). Note that we are (implicitly) assuming that the normed spaces in question are spaces over the field $K$ where $K$ is either $ℝ$ or $ℂ$. So let $(X,\parallel \cdot \parallel )$ be a normed space with Schauder basis, say $S=\{e_1,e_2,\mathrm{\dots }\}$. Notice that our notation implies that $S$ is infinite. In finite dimensional case, the same proof with a slight modification will yield the result.

Now, set $Q$ to be the set of all finite sums $q_1{e}_1+\mathrm{\cdots }+q_n{e}_n$ such that each $q_j=a_j+b_{j}i$ where $a_j,b_j\in \mathbb{Q}$. Clearly $Q$ is countable. It remains to show that $Q$ is dense (http://planetmath.org/Dense) in $X$.

Let $ϵ>0$. Let $x\in X$. By definition of Schauder basis, there is a sequence of scalars $({\alpha }_n)$ and there exists $N$ such that for all $n\ge N$ we have,

But then in particular,

Furthermore, by density of $\mathbb{Q}$ in $ℝ$, we know that there exist constants $a_1,\mathrm{\dots },a_N,b_1,\mathrm{\dots },b_N$ in $\mathbb{Q}$ such that,

By triangle inequality we obtain:

Noting that

$$\sum _{j=1}^N(a_j+b_{j}i)e_j$$

is an element of $Q$ (by construction of $Q$) and that $x$ and $ϵ$ were arbitrary, we conclude that every neighborhood of $x$ contains an element of $Q$, for all $x$ in $X$. This proves that $Q$ is dense in $X$ and completes the proof.

Title	every normed space with Schauder basis is separable
Canonical name	EveryNormedSpaceWithSchauderBasisIsSeparable
Date of creation	2013-03-22 17:36:07
Last modified on	2013-03-22 17:36:07
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	11
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 42-00
Classification	msc 15A03
Classification	msc 46B15