every normed space with Schauder basis is separable

\\PMlinkescapephrase

dense in

Here we show that every normed space that has a Schauder basis isseparable (http://planetmath.org/Separable).Note that we are (implicitly) assuming that the normed spaces in questionare spaces over the field $K$ where $K$ is either $\\mathbb{R}$ or $\\mathbb{C}$ .So let $(X,\\left\\|\\cdot\\right\\|)$ be a normed space with Schauder basis,say $S=\\left\\{e_{1},e_{2},\\dots\\right\\}$ .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 $q_{1}e_{1}+\\cdots+q_{n}e_{n}$ 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 $\\epsilon>0$ . 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 $n\\geq N$ we have,

\\left\\|\\sum_{j=1}^{n}\\alpha_{j}e_{j}-x\\right\\|<\\epsilon/2

But then in particular,

\\left\\|\\sum_{j=1}^{N}\\alpha_{j}e_{j}-x\\right\\|<\\epsilon/2

Furthermore, by density of $\\mathbb{Q}$ in $\\mathbb{R}$ ,we know that there exist constants $a_{1},\\dots,a_{N},b_{1},\\dots,b_{N}$ in $\\mathbb{Q}$ such that,

\\left\\|\\sum_{j=1}^{N}(a_{j}+b_{j}i)e_{j}-\\sum_{j=1}^{N}\\alpha_{j}e_{j}\\right\\|%<\\epsilon/2

By triangle inequality we obtain:

\\left\\|\\sum_{j=1}^{N}(a_{j}+b_{j}i)e_{j}-x\\right\\|\\leq\\left\\|\\sum_{j=1}^{N}(a_%{j}+b_{j}i)e_{j}-\\sum_{j=1}^{N}\\alpha_{j}e_{j}\\right\\|+\\left\\|\\sum_{j=1}^{N}%\\alpha_{j}e_{j}-x\\right\\|<\\epsilon

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 $\\epsilon$ 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.