Let $F$ be a compact subset of $\mathbb{R}$. Let $p(x)=a_{n}x^{n} + ... + a_{0}$ be a polynomial. For each $a_{i}$, let $(q^{i}_{k})_{k \in \mathbb{N}}$ be a sequence of rationals such that $q^{i}_{k} \rightarrow a_{i}$. For each $k$, consider the polynomial $p_{k}= q^{n}_{k}x^{n} + ... q^{0}_{k}$. Then $p_{k} \rightarrow p$ in $C(F) = \{ f:F \rightarrow \mathbb{R} | f \text{ is continuous} \}$.
I'm trying to prove that $\forall \epsilon >0$, there is $k_{0} \in \mathbb{N}$ such that $$k > k_{0} \implies d(p_k,p) = sup\{ |p_{k}(x) - p(x)| : x\in F\} < \epsilon$$ We know that, for each $i$ and $\epsilon_{i}$, there is $k_{0}^{i}$ such that $k>k^{i}_{0} \implies |q^{i}_{k}-a{i}| < \epsilon_{i}$. then, $$ |p_{k}(x)-p(x)| = |(q^{0}_{k}-a_{0})+...+(q^{n}_{k}-a_{n})x^{n}| \leq |q^{0}_{k}-a_{0}|+...+|(q^{n}_{k}-a_{n})x^{n}| < \epsilon_{0} +...+\epsilon_{n}|x|^{n}$$
And now $$ d(p_k,p) = sup\{ |p_{k}(x) - p(x)| : x\in F\} \leq \epsilon_{0} +...+\epsilon_{n}|x|^{n} $$
Now I'm stuck trying to get rid of the powers of $|x|$. There's probably something very simple that I'm missing. Thanks in advance.
To finish this proof, I would first define $M$ to be the upper bound $\max(\{|X|:X \in F\},1)$, guaranteed to exist by the fact that F is compact. you can then select $$\epsilon_0= \epsilon_1 = ... =\epsilon_n= \frac {\epsilon}{nM^n}$$ then $\epsilon_0+\epsilon_1|X|+...\epsilon_n|X|^n \leq \epsilon_0|X|^n+\epsilon_1|X|^n+...\epsilon_n|X|^n \leq \epsilon_0nM^n=\epsilon$