Compactness of $\ A= \{f: f$ is power series with infinite radius of convergence and is bounded by $1\}$ Consider the set $\ K=C[0,1]$, the set of continuous functions on $\ [0,1]$, with the supremum norm. Let 
$\ A= \{f: f$ is power series with infinite radius of convergence and is bounded by $1\}$.
I am asked to prove or disprove whether A is compact or not.
I tried noting a counterexample to show that the set isn't closed.
$$ \sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}=\sin(x) $$
This is a power series which converges to $\sin(x)$ which isn't in the set. Does this work?

Hint: Stone-Weierstrass theorem.
