Applying Stone -Weierstrass theorem Prove or disprove that vector space generated by $\{1\}$ and $\{x^{an+b}: n \in \mathbb{N}\}$ where $a,b$ are fixed positive integers is dense in $C([0,1])$
 A: The beautiful result of Muntz-Szasz says that if $0<p_1<p_2 < \dots,$ then the linear span of $ \{1, t^{p_1}, t^{p_2}, \dots \}$ is dense in $C([0,1])$ iff
$$\sum_{n=1}^{\infty}\frac{1}{p_n}=\infty.$$
The answer to this specific problem is then yes. But there may be a more elementary way.
A: Here is an elementary proof of the statement. If $a\mid b$ then the element
$$x^{an+b}x^{am+b} = x^{a\left(n+m+\frac{b}{a}\right)+a}$$
is an element of the vectorspace for all $n,m\in\mathbb{N}$ making it a subalgebra of $C([0,1])$. Since the vectorspace separate points and contains a non-zero constant function the desired result follows from the Stone-Weierstrass theorem. This can in fact also be proven using the weaker Weierstrass approximation theorem (WA) and for the case $a\not\mid b$ proven below this is all we will need.
If $f(x)\in C([0,1])$ then $f(x^{1/a})\in C([0,1])$ so the problem is equivalent to proving that the vector space $$A_c=\text{span}\{1,x^{1+c},x^{2+c},x^{3+c},\ldots\}$$ where $c = \frac{b}{a}$ is dense in $C([0,1])$. We can without loss of generality assume $0<c<1$.
Fix $\epsilon>0$. By WA there exists a polynomial $P_\epsilon(x)$ that approximate $f(x^{1/a})$ to (uniform) accuracy $\frac{\epsilon}{2}$ on $[0,1]$. Next we have that the function $\frac{P_\epsilon(x)-P_\epsilon(0)}{x^c} \in C([0,1])$ since $c < 1$ so again by WA there exists a polynomial $Q_\epsilon(x)$ that approximate it to (uniform) accuracy $\frac{\epsilon}{2}$ on $[0,1]$. 
We now have that the function $$R_\epsilon(x) \equiv P_\epsilon(0) + x^c Q_\epsilon(x)$$ 
is an element of $A_c$ which satisfy
$$|f(x^{1/a})-R_\epsilon(x)| \leq |f(x^{1/a})-P_\epsilon(x)| + |P_\epsilon(x) - P_\epsilon(0) - x^cQ_\epsilon(x)|\\ \leq \frac{\epsilon}{2} + \left|x^c\right|\left|\frac{P_\epsilon(x)-P_\epsilon(0)}{x^c} - Q_\epsilon\right| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
where $|\cdot|$ is the uniform norm. Since $f$ and $\epsilon$ was abitrary it follows that $A_c$ is dense in $C([0,1])$.
