Prove that if $K$ is polynomially convex, then $\mathbb{C}-K$ is connected.

Question

Prove that if $K$ is polynomially convex, then $\mathbb{C}-K$ is connected.

My Try:

Suppose $K$ is polynomially convex. Then $K=\hat{K}$. So, given $z\in \mathbb{C}-K$, there is a polynomial $p(z)$ such that $|p(z)|>max_{\zeta\in K}|p(\zeta)|$. If we assume that $\mathbb{C}-K$ is not connected, then it has at least 2 components. After that, I was stuck. Can somebody please help me to proceed?

Answer 1

If $K \subset \mathbb{C}$ is compact then it is contained in $\{ |z| \le R \} $ for some $R > 0$. The component of $\mathbb{C}-K$ containing $\{ |z| > R \}$ is the "unbounded component". All other components of $\mathbb{C}-K$ are contained in $\{ |z| \le R \} $and therefore bounded.

Assume that $\mathbb{C}-K$ has a bounded component $D$. Any polynomial $p$ is holomorphic in $D$ and continuous on $\overline D$. It follows from the maximum modulus principle that for all $z \in D$, $$ |p(z)| \le \max_{\zeta\in \partial D}|p(\zeta)| \le \max_{\zeta\in K}|p(\zeta)| $$ since $\partial D \subset K$ and therefore $z \in \hat K$. This shows:

If $K \subset \mathbb{C}$ is compact then any bounded component of $\mathbb{C}-K$ is contained in $\hat K$.

In particular, if $K$ is polynomially convex ($\hat K = K$) then $\mathbb{C}-K$ cannot have a bounded component, so that $\mathbb{C}-K$ is simply-connected.