Why isn't every Hamel basis a Schauder basis?

Question

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?

Answer 1

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.

Answer 2

Interestingly, once once knows that not every Schauder basis is a Hamel basis, this implies almost immediately that not every Hamel basis is a Schauder basis:

A Schauder basis is clearly linearly independent, so it can be extended to a Hamel basis. If it wasn't a Hamel basis to begin with, then this new Hamel basis contains an element not contained in the original Schauder basis. But this element can be written as a countable linear combination of the Hamel basis in two different ways: Once just one times itself, and once as countable linear combination of elements from the original Schauder basis. So the new Hamel basis cannot be a Schauder basis.