Runge's theorem

Remarks

Proof idea: by the Cauchy integral formula over a cycle around $K$, $f$ is a uniform limit on $K$ of finite sums of terms $c/(z - a)$ with $a$ off $K$. Each pole $a$ can then be moved to a point of $A$ in the same component of $\C \setminus K$, by expanding in power series. For the unbounded component, the pole is moved to $\infty$ instead. A pole at $\infty$ means a polynomial.

The hypothesis on $A$ is necessary. Let $K$ be the unit circle and $f(z) = 1/z$; the bounded component of $\C \setminus K$ is the open disc. For every polynomial $p$ one has $\oint_K p(z) \, dz = 0$, while $\oint_K z^{-1} \, dz = 2\pi i$. If $\abs{p - 1/z} < \varepsilon$ on $K$, then $2\pi = \abs{\oint_K (z^{-1} - p(z)) \, dz} \le 2\pi \varepsilon$, which fails for $\varepsilon < 1$. With $A = \set{0}$ the rational function $r(z) = 1/z$ itself has its only pole in $A$ and approximates $f$ exactly.

Going deeper

For a student

Runge's theorem is the starting point of approximation theory in the complex plane. It is used to build holomorphic functions with prescribed properties, for example sequences of polynomials that converge pointwise on $\C$ to a discontinuous limit. It gives the existence of solutions of the $\bar\partial$-equation and the Mittag-Leffler theorem on general open sets. Mergelyan's theorem strengthens the polynomial case to functions that are only continuous on $K$ and holomorphic in its interior.