Non-complete and non-reflexive normed space I am searching for an examples of a normed space that is non-complete and non-reflexive, but separable. My idea was to somehow build an infinite dimensional space out of the rationals, such that we get separability and non-completeness from the rationals with norm the absolute value. But I do not really know how to do that. Can anyone provide such an example?
 A: You can show that $\Bbb{R}^* = \bigcup_{i=1}^{\infty} \Bbb{R}^i$ viewed as an inner product space with the induced norm is not complete as follows.
View $\Bbb{R}^*$ as the set of all sequences $\mathbf{v} = (v_1, v_2, \ldots)$ where $v_i\in \Bbb{R}$ for every $i$ and where $v_i = 0$ for all but a finite number of $i$.
$\Bbb{R}^*$ has an orthormal basis comprising the vectors $\mathbf{e}_1 = (1, 0, \ldots), \mathbf{e}_2 = (0, 1, 0, \ldots), \ldots$. Consider the sequence of vectors $(\mathbf{v}_n)_{n=1}^{\infty}$ defined by:
$$\mathbf{v}_n = \sum_{i=1}^{n} {\mathbf{e}_i \over i!}
$$
Then for $n \ge m$ we have:
$$
\|\mathbf{v}_n - \mathbf{v}_m\|^2 = \sum_{i = m + 1}^n{1 \over (i!)^2}
$$
Since the sum $\sum_{i=1}^{\infty} {1 \over (i!)^2}$ converges (by comparison with the standard series representation for $e = \exp(1)$) this shows that the sequence  $(\mathbf{v}_n)_{n=1}^{\infty}$ is a Cauchy sequence. But it can have no limit in $\Bbb{R}^*$: if $\mathbf{v} = (v_1, v_2, \ldots) \in \Bbb{R}^*$, there is an $n$ such that $v_i = 0$ for all $i \ge n$, and then, with $\epsilon = {1 \over 2n!}$, $\|\sum_{i=1}^m\mathbf{v}_i - \mathbf{v}\| > \epsilon$ for every $m \ge n$, so $\mathbf{v}$ is not the limit of the sequence $(\mathbf{v}_n)_{n=1}^{\infty}$.
As pointed out in the comments, as $\Bbb{R}^*$ is not complete it cannot be reflexive. See Show reflexive normed vector space is a Banach space for a proof. 
