Polynomial Functions separate points

I am trying to prove that polynomial functions in space $C([0,1],\Bbb R)$ separates points. Also, whether the same holds true for polynomial functions with integral coefficients?

What I know is that a set of functions from a set $D$ is called a separating set for $D$ or said to separate the points of $D$ if for any two distinct elements $x$ and $y$ of $D$, there exists a function $f$ in so that $f(x) ≠ f(y)$ (http://en.wikipedia.org/wiki/Separating_set).

So I will need to show that polynomial functions in space $C([0,1],\Bbb R)$ admit an injective map from $[0,1]$ to $\Bbb R$.

Assume the set of polynomials $S$ of the form $a_{0}+a_{1}x+...+a_{n}x^{n} \forall x\in [0,1]$. I have a doubt here. Suppose the polynomials are of the form $x+1/4$ and $1/2$ (constant polynomial). Clearly both these belong to $S$ and at $x=1/4$, both evaluate to $1/2$. So how is an injective map possible?

Also what is the difference when polynomial coefficients are integers?

I am studying Baby Rudin. So please answer according to the material in the book in chapter sequences and series of functions. thanks

You need to show that for any two points $x, y$ there is one polynomial $f$ among the infinite possibilities such that $f(x)\ne f(y)$.
And if you read the wikipedia article carefully, then the first example contains the solution. As written in the example, already the single polynomial $f(x)=x$ separates all points. It helps of course that this and the constant functions generate the polynomial ring. In view of that, interpolation is overkill.
• So, this implies that set of polynomials defined by identity function $f_{n}(x)=x^{n}$ is the solution. But what about polynomials with integer coefficients? Feb 3, 2015 at 0:38