= Solution
The set $K$ is <compact set>[compact], and its complement is connected: two disjoint closed disks do not enclose any bounded complementary component. Since $f$ is <holomorphic function>[holomorphic] on the open neighbourhood $\Omega$ of $K$, the <polynomial Runge theorem> gives, for each $n$, a polynomial $p_n$ such that
$$
\sup_{z\in K}|p_n(z)-f(z)|<\frac1n.
$$
Hence
$$
\boxed{p_n\longrightarrow f\text{ uniformly on }K.}
$$
Define a function on the disconnected open set $\Omega$ by
$$
f(z)=
\begin{cases}
0,&|z-2|<3/2,\\
1,&|z+2|<3/2.
\end{cases}
$$
It is holomorphic on $\Omega$. Applying the first part produces polynomials $P_n$ with
$$
\boxed{P_n\to0\text{ uniformly on }|z-2|\leq1,\qquad
P_n\to1\text{ uniformly on }|z+2|\leq1.}
$$
For an obstruction, take
$$
K_1=\{z:|z|=1\},
\qquad
K_2=\{0\}.
$$
These are disjoint nonempty bounded closed sets. If polynomials $Q_n$ converged uniformly to zero on $K_1$, the <maximum modulus principle on a bounded domain> would give
$$
|Q_n(0)|\leq\max_{|z|=1}|Q_n(z)|\longrightarrow0.
$$
They therefore cannot also satisfy $Q_n(0)\to1$. Thus \b[this pair $K_1,K_2$ has the requested non-approximation property].
Back to article page
