Stone Weierstrass real applicaton? I want to prove that the span of $\{x^{2n}:n \geq0\}$ is dense in $C([0,1])$.
Furthermore, that the closure of the span of $\{x^{2n+1}:n \geq0\}$ is $\{f \in C([0,1]):f(0) = 0\}$.
My work:
I guess I can use the Stone-Weierstrass theorem. 
Define $A =$ span of $\{x^{2n}:n \geq0\}$.
I need to prove A is an algebra of real continuous functions on a compact set I = [0,1]. The set I is compact since it is continuous and bounded. 
A is an algebra since if, for example I multiply $fg = x^2x^4 = x^6 \in A$.
Next, need to show A separates points on I. So, points $x_1,x_2 \in [0,1] \Rightarrow f(x_1) \neq f(x_2), f(x_1), f(x_2) \in \{x^{2n}:n \geq0\}$.
Finally, A vanishes at no point of I, i.e. $\forall x \in [0,1]$ elements of $\{x^{2n}:n \geq0\}$ are $\neq0$.
Is this correct?
Now I do not know how to tackle the second part. Any help?
Thank you for helping the community.
 A: A polynomial with only odd coefficients is a polynomial with only even coefficients that has been multiplied by $x$. Since the span of $\{x^{2n} : n \geq 0\}$ is dense in $C([0;1])$ and $||x\cdot f- x\cdot P||_{\infty}\leq ||f-P||_\infty$ for every $f,P \in C([0;1]) $ you can deduce that the span of $\{x^{2n+1} : n \geq 0\}$ is dense in $A=\{x\cdot f : f \in C([0;1]) \}$. 
Now all you have to do is to show that $A$ is dense in $B=\{f \in C([0;1]) : f(0)=0\}$, this shouldn't be a problem for you.
If you still have trouble finishing the proof :
Take a function $f$ in $B$ and $\varepsilon>0$. For some $\eta > 0$ you have $x<\eta \Rightarrow |f(x)|<\varepsilon$ and there exists a function $g\in A$ such that $x\geq \eta \Rightarrow g(x)=f(x)$, and thus $||f-g||_\infty<\varepsilon$. 
More generally you can search for the Müntz theorem : consider a set $S\subset \mathbf{N}$ that contains $0$, the span of $\{x^n:n\in S \}$ will be dense in $C([0;1])$ iff $\displaystyle \sum_{s\in S} \frac{1}{s}=+\infty$.
