# Does an explicit formula for each member of a dense sequence of polynomials exist?

Weierstrass' Approximation Theorem implies that every continuous function on the closed interval $[0,1]$ can be uniformly approximated by polynomials. This implies the space $C[0,1]$ of continuous functions on $[0,1]$ is separable, and in particular there is a dense sequence of polynomials $\{p_n\}$ in the space. Is it possible to exhibit an explicit formula for each $p_n$?

Thank you.

• You said that "this implies the space is separable". Is you think of how it can be proved, you will know how to find the explicit formula. – user99914 Sep 23 '15 at 6:09

Every $x \in C[0,1]$ can be approximated by polynomials.
Let $\{ \phi_{m} \}$ be one of the dense countable subsets of $C[0,1]$ given by separability.
Let $p_{mn}$ be the nth polynomial approximation to $\phi_{m}$ i.e $||p_{mn} - \phi_{m} || < \frac{1}{n}$. $\{ p_{mn} \}$ is countable (countable union of countable sets) and dense in $\{ \phi_{m} \}$ and is therefore dense in $C[0,1]$.
It would take a much greater mind than I to find an explicit formula for these $\{ p_{mn} \}$, but in principal, if you have a formula for $\phi_{m}$, then you could find one...