# Is a linearly independent set whose span is dense a Schauder basis?

If $$X$$ is a Banach space, then a Schauder basis of $$X$$ is a subset $$B$$ of $$X$$ such that every element of $$X$$ can be written uniquely as an infinite linear combination of elements of $$B$$. My question is, if $$A$$ is a linearly independent subset of $$X$$ such that the closure of the span of $$A$$ equals $$X$$, then is $$A$$ necessarily a Schauder basis of $$X$$?

If not, does anyone know of any counterexamples?

No, certainly not. The linearly independent set $$\{1, x, x^2, x^3, \dots\}$$ has span dense in $$C[0,1]$$ by the Weierstrass approximation theorem. But it is not a Schauder basis of that space, since not every continuous function is given by a power series.