Is there any difference between the elements in $\mathbb{C}[x]$ and $C^{\omega}$? Studying Hilbert spaces, I've been told that $\{1, x, x^2,\dots, x^k,\dots\}$ form a basis (EDIT: when orthonormalised) in $\mathcal{L}^2([a,b])$ because the set of all polynomial functions is dense in $C^0([a,b])$, which is dense in $\mathcal{L}^2([a,b])$. I'm going to assume all functions here are complex functions of real variable, so $[a,b]\subseteq\mathbb{R}\to\mathbb{C}$.
That got me thinking... if $C^{\omega}([a,b])$ is the set of all analytic functions defined in $[a,b]$, which by definition of analytic can be expressed as a Taylor series (which is basically a finite or infinite polynomial), is there a natural identification between analytic functions and polynomials?, or do the elements in $\mathbb{C}[x]$ have to be polynomials of finite degree (ignoring 0)? If so, what would be the difference between a basis in $C^{\omega}([a,b])$ and a basis in $\mathbb{C}[x]$ made of all finite-degree monomials?
 A: $\mathcal{B} = \{1, x, x^2,\dots, x^k,\dots\}$ is a set of linearly independent elements of $\mathcal{L}^2([a,b])$, but it is neither a vector space basis of $\mathcal{L}^2([a,b])$ nor an orthonormal basis (= onb) of the Hilbert space $\mathcal{L}^2([a,b])$. However, the Gram-Schmidt procedure allows to transform $\mathcal{B}$ into a set of polynomials $\mathcal{P} =  \{p_n(x) \mid n = 0, 1, 2, \dots \}$ which forms an onb of $\mathcal{L}^2([a,b])$.
$\mathbb{C}[x]$ is the polynomial ring over $\mathbb{C}$ in one variable $x$. This is a purely algebraic construct characterized by a certain universal property. See for example http://math.mit.edu/~mckernan/Teaching/12-13/Spring/18.703/l_21.pdf. The monomials $x^n$, $n = 0,1,2, ...$, form a (countable) basis of $\mathbb{C}[x]$. Hence $\mathbb{C}[x]$ is vector space of countably  infinite dimension. 
Clearly $\mathbb{C}[x]$ can be regarded as a subspace $C^{\omega}([a,b])$. In fact, each $p(x) \in \mathbb{C}[x]$ can be identified with the polynomial function $t \mapsto p(t), t \in [a,b]$.
It is obvious that $\mathbb{C}[x]$ is a proper subset of $C^{\omega}([a,b])$. Even more, $C^{\omega}([a,b])$ is a vector space of uncountable dimension. However, it is impossible to explicitly describe an uncountable basis.
