Set of polynomials with integer coefficients is dense I am trying to prove the following claim by invoking Stone Weierstrass Theorem.

Let $p$ be an integer and let $p<a<b<p+1$. Suppose that $f:[a,b]\to\mathbb{R}$ is continuous and $\varepsilon>0$ are given. The there exists a polynomial  $ P(x)=\sum_{k=0}^n a_k x^k \quad (n\geq 0,a_k\in\mathbb{Z}) $ such that $$
\mid f(x)-P(x)\mid<\varepsilon \quad \forall x\in[a,b]
$$

I try to show  closure $A^-$ of the set of all such polynomials is subalgebra of $C(X)$ that separates the points. It is easy to show if $f,g\in A$ then $f+g\in A^-$ and $fg\in A^-$.
However, I can not show $\alpha f\in A^-$ for $\alpha \in\mathbb{R}$. Also, I can not show that $A^-$ separates the points of $[a,b]$. 
Finally, does this theorem fail when $[a,b]$ contains an integer?
Any help is greatly appreciated! 
 A: Yes this result fails if $[a,b]$ contains an integer $n$ : for $\epsilon <1/2$ and $f : x \mapsto 1/2$, you can not have $\vert f-P \vert < \epsilon$ where P is a polynomial with coefficients in $\mathbb{Z}$ since $f(n)=1/2$ et $P(n) \in \mathbb{Z}$.
A: I found that what you ask is known as Chudnovsky Theorem.
One can suppose that $p=0$.
First one shows that the constant function $f=1/2$ is a uniform limit of a sequence of polynomials with coefficients in $\mathbb{Z}$ using the sequence $$ P_0=X, \: P_{n+1}=2(1-P_n)P_n. $$
Thus the closure of $\mathbb{Z} [X]$ in $\mathcal{C}([a,b],\mathbb{R})$ contains all dyadic rational and then contains $\mathbb{R}$. Also $X \in \mathbb{Z} [X]$. Then the algebra $\bar{\mathbb{Z} [X]}$ contains $\mathbb{R}[X]$ and so it is all $\mathcal{C}([a,b],\mathbb{R})$ by Weierstrass.
A: I think the following argument works. It is easy to see that $f,g\in A^-$ implies that $f+g\in A^-$ and $fg\in A^-$. 
For any $k\in\mathbb{N}$, we could write
$$ 
\frac{1}{k}=\frac{x-p}{[1-(1-k(x-p))  ]}=\sum_{j=0} (x-p) (1-k(x-p))^j,  \quad p<x<p+1\quad   (\star)
$$
Note that $\sum_{j=0}^n (x-p)(1-k(x-p))^j \in A$ for each $n$ and the convergence in $(\star)$ is uniform. So $1/k\in A^-$. In a similar way we could show that same is true for $-k$ and $1/k$. Hence $r\in A^-$ for any $r\in\mathbb{Q}$.
Take any $\alpha\in\mathbb{R}$. There exists a rational number $r$ in every neighborhood of $\alpha$, but this means that $\alpha $ is a limit point of $A^-$ so $\alpha \in A^-$. Thus $\alpha f \in A^-$ whenever $f\in A^-$.
For separation, we could take $P(x)=x$. So this proves that $A^-$ is a subalgebra in $C([a,b])$ that separates the points of $[a,b]$ and vanishes nowhere on $[a,b]$. The claim above follows from Stone-Weierstrass.
