Weierstrass Approximation Theorem: Suppose $f$ is a continuous real-valued function defined on the real interval $[a, b]$. For every ε > 0, there exists a polynomial $p$ such that for all $x \in [a, b]$, we have $|f(x)−p(x)| < \epsilon$, or equivalently, the supremum norm $\|f−p\| < \epsilon$.
So, the set of polynomials is a countable dense set in $C[a,b]$.
I have also learnt that an inner product space is separable (has a countable dense subset) iff it has a countable basis. Since $C[a,b]$ is dense in $L^2$, we can prove that the set of polynomials with rational coefficients is countable and dense in $L^2$, hence that $L^2$ should have a countable basis.
But the book ("Mathematics for Physicists - Dennery & Krzywicki) already says that, Weierstrass Theorem means $\{x^n\}$ is a basis for $L^2$. If that is the case, "separable iff has a countable basis" theorem is useless for this case (I am sure it has other uses).
I think there are two different definitions of basis here:
$E=\{e_i\}$ is a basis for $V$ if $\forall v\in V (\forall e\in E \ \langle v|e\rangle = 0$ implies $v=0$) or equivalently, $\forall v\in V\exists !\{c_i\}$ s.t. $ v=\sum_{i=1}^\infty c_i e_i$ (Schauder basis)
$\forall v\in V\exists \{c_i\} $ s.t. $v=\sum_{i=1}^\infty c_i e_i$
where infinite series above converge in $L^2$ norm. The only difference between these definitions is that in the second one, we do not require uniqueness of the coefficients. So polynomials are basis in the second sense only. Is my understanding alright? Why do we bother with Classical Orthogonal polynomials or with $\{e^{ni\pi}\}$ if we already have a countable basis, namely $\{x^n\}$?