Why isn't every Hamel basis a Schauder basis? I seem to have tripped on the common Hamel/Schauder confusion.
If $X$ is any vector space (not necessarily finite dimension) and $B$ is a linearly independent subset that spans $X$, then $B$ is a Hamel basis for $X$.
If there exists a sequence $(e_n)$ such that for every $x \in X$ there exists a unique sequence of scalars $(\alpha_n)$ for which $\lim_{n \to \infty} || x - (\alpha_1e_1 + \cdots + \alpha_ne_n)|| = 0$, then $(e_n)$ is a Schauder basis for $X$.
So I'm tempted to think that every Hamel basis is also a Schauder basis; just extened the finite linear combination into an infinite one by adding zero coeeficients. I know I'm wrong, but what am I missing?

The issue is ". . . unique sequence of scalars . . .". If $B$ is a Hamel basis, then there will be at least one such sequence of scalars, but there may be more. 
