Prove that $A(\Delta)$ is a separable space

Question

Let $\Delta$ be the unit closed disk in the complex plane and let $A(\Delta)$ be the family of complex functions which are continuous in $\Delta$ and analytic in the interior of $\Delta$. Now, consider in $A(\Delta)$ the following norm

\begin{equation*} \lVert f\rVert_{\infty}=\sup_{z\in\Delta} |f(z)|. \end{equation*}

I'm trying to prove that $(A(\Delta),\lVert \cdot\rVert_{\infty})$ is a separable space. I've tried to consider the countable subspace $P=\{z^k : k\geq 0\}$ and prove that it is dense but it seems to go nowhere.

Which other dense subspace or even other strategy should I follow? Thanks.

Answer 1

Fix $f(z)\in A(\Delta).$ For $0<r<1$ define the function $$f_r(z)= f(rz)$$ Then $$\lim_{r\to 1^-} f_r(z)=f(z)$$ uniformly in $\Delta.$ Every function $f_r(z)$ is analytic in the open disc $|z|<r^{-1},$ hence its MacLaurin series is uniformly convergent in $\Delta.$ In this way we have proved that the polynomials are dense in $A(\Delta).$