I have the following question:
Prove that if a sequence of analytic polynomials converges uniformly on a region $\Omega$, then there exists a simply connected region $\Omega$ such that the sequence converges uniformly.
There is a Theorem which says that a region is simply connected iff every cycle in $\Omega$ is homologous to zero in $\Omega$ (ie. the winding number is $0$ outside of $\Omega$. I also know that if the winding number is $0$ in an unbounded component of $\mathbb{C}\backslash \gamma$. Thus, I thought that the answer might involve taking the union of $\Omega$ and the bounded components of the complement. This ensures that it is simply connected, but it may not be a region. Also, I don't even know that the sequence converges uniformly on this new set. Does this seem like the correct approach?
Any help/suggestions would be greatly appreciated.
Let $U_0=\Omega$ and $U_1$ the union of all interior domains bounded by Jordan curves that are in $U_0$ (here only the curves have to be contained in $U_0$ as that is the whole point of the construction, namely to fill the "bounded holes" of $U_0$). It follows that $U_1$ is open (by construction), connected (if $U_1=V \cup W, V \cap W$ empty, $V,W$ open, then $U_0 \subset V$ say and then $\partial W \cap U_0$ non-empty by construction as any interior domain of a Jordan curve from $U_0$ that has a point in $W$, has the full domain in $W$, hence $W \cap U_0$ non-empty which is impossible) and contains $U_0$ (for any point there take a small ball centered at it and contained in $U_0$).
By maximum modulus $\sup_{U_1}|P_n-P_m| \le \sup_{U_0}|P_n-P_m|$ since any point in $U_1$ is enclosed in a domain with boundary in $U_0$ by construction. Hence $P_n$ converges uniformly on $U_1$ too.
But now iterate and construct $U_2,...U_n..$ and take $U=\cup U_k$. $U$ is clearly open, connected and contains $\Omega$ and now if you take a Jordan curve $C$ in it, for each $z \in C$ there is $k, z \in U_k$ hence a small neighborhood is in $U_k$ and finitely many such cover $C$ so $C$ is in some $U_n$ for fixed $n$ hence it is contractible in $U_{n+1}$ so in $U$, hence $U$ simply connected.
Since $\sup_{U_k}|{P_n-P_m}| \le \sup_{\Omega}|P_n-P_m|$ it follows that $\sup_U|{P_n-P_m}| \le \sup_{\Omega}|P_n-P_m|$ so we are done!