Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 12F (Topics in Analysis)

The polynomial form of the Runge approximation theorem says: if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighborhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ with ${\sup }_K|f-P|<\epsilon$. Choosing errors tending to zero gives a uniformly convergent sequence of polynomials. Equivalently, on an open domain with connected complement in the Riemann sphere, every holomorphic function is the compact-uniform limit of polynomials. The connected-complement hypothesis excludes enclosed holes containing poles of the function being approximated.

A compact set $K\subset \mathbb{C}$ has exactly one unbounded complementary component, since its complement contains the outside of a sufficiently large disk. The connected, unbounded open set $\Omega$, being disjoint from $K$, lies in that component. In particular $\zeta$ belongs to the unbounded component.

Fill all bounded components of $\mathbb{C}\setminus K$ to obtain the polynomial hull $\hat{K}$. It is a compact set with connected complement, and $\zeta \notin \hat{K}$. Therefore $1/(z-\zeta )$ is holomorphic on a neighborhood of $\hat{K}$. The Runge approximation theorem gives polynomials $P_n$ converging uniformly to this function on $\hat{K}$, hence on $K$. This proves the requested uniform polynomial approximation on $K$, even when $K$ itself has holes. If $K$ is empty, the assertion is vacuous.

Take $\Omega =\{|z|<1\}$, $K=\{|z|=1\}$, and $\zeta =0$. These satisfy the stated connectedness and disjointness conditions. If polynomials $P_n$ converged uniformly to $1/z$ on $K$, convergence of the contour integrals would implya contradiction to the Cauchy integral theorem. Thus boundedness of this $\Omega$ permits the obstruction excluded in part (b).