Is there a normed space with dense basis? Is there a normed space $(X, \lVert \cdot \rVert)$, in which one can find a (Hamel-) basis $B$ of $X$ such that $B$ is dense in $X$? And if yes: Does every normed space $X$ with $\dim(X) \geq |\mathbb{R}|$ have such a basis $B$?

In fact any infinite-dimensional separable normed space $X$ has a dense Hamel basis.

Note first we need only show there is a dense independent set, since any independent set is a subset of some basis.

Say $S$ is a countable dense set, and let $B_1,\ B_2,\dots$ be an enumeration of the balls of the form $B(x,r)$ with $x\in S$, $r\in\Bbb Q$, $r>0$. There exists an independent set $x_1,x_2,\dots$ with $x_n\in B_n$:

Let $x_1\in B_1$, $x_1\
e0$. Suppose you've chosen $x_1,\dots, x_n$. Now $$B_{n+1}\setminus span(x_1,\dots, x_n)\
e\emptyset$$ (necause _why_?), so you can choose $x_{n+1}\in B_{n+1}\setminus span(x_1,\dots x_n)$.

**Note** I believe you could remove the assumption of seqarability, constructing an appropriate $(x_\alpha)_{\alpha<\kappa}$ by transfinite induction. I leave that to you - you don't want me to have all the fun...