Prove that if $\mathbb{C}-K$ is connected, then $K$ is polynomially convex. $K $ polynomially convex means $K=\hat{K}$ where
$$\hat{K}=\{z\in\mathbb{C}:|p(z)|\le \max_{ζ\in K}|p(ζ)|\text{ for all polynomials}\; p\}$$
My Try
Let $z_0\in \mathbb{C}-K$. Then $K\cup\{z_0\}$ also has connected complement. Take a sequence $\{z_n\}$ such that $ z_n\rightarrow z_0$. Consider $f_n(z)=\frac{1}{z-z_n}$. Now $f_n(z)$ is analytic on a neighborhood of $K\cup\{z_0\}$. Fix $n$. Then by Runge's theorem there is a sequence $\{p_m\}$ of polynomials such that $p_m$ converges uniformly to $f_n$ on $K\cup\{z_0\}$. Now I wanted to pick one $p_m$ so that $|p_m(z_0)|>\max_{ζ\in K}|p(ζ)|$, but how could I do it? Can somebody please help me to complete my almost completed proof?
I think $z_n$ and $1/(z-z_n)$ do not really help here. You want a function that is greater at $z_0$ than anywhere on $K$. Among many choices, $$ f(z) = \begin{cases} 0,\quad z\in K \\ 1,\quad z=z_0\end{cases} $$ is one of simplest. Note that $f$ can be extended to a holomorphic function in a neighborhood of $K\cup \{z_0\}$. As you said, $K\cup \{z_0\}$ has connected complement, and therefore Runge's theorem provides a polynomial $p$ such that $|p-f|<1/2$ on $K\cup \{z_0\}$. This polynomial satisfies $$|p(z_0)|>\frac12>\max_K |p|$$ as required.