Runge's pole-pushing lemma

Fix one disc step of the chain, say a closed disc $\bar{D}(c,\rho )$ disjoint from $K$ and two points $a,b\in D(c,\rho )$. [given, L1] Define $S\subseteq D(c,\rho )$ to be the set of points $u$ such that $(z-a{)}^{-1}$ can be approximated uniformly on $K$ by rational functions with only pole $u$. Certainly $a\in S$.

Let $u\in S$. Choose $r>0$ so that $\bar{D}(u,r)\subseteq D(c,\rho )$ and $r<\mathrm{dist}(K,u)\mathrm{/}2$. [step 1.1, choose, algebra] If $v\in D(u,r)$ and $m\ge 1$, then for $z\in K$ one has $\mathrm{\mid }u-v\mathrm{\mid }<\mathrm{\mid }z-v\mathrm{\mid }$, so $\frac{1}{(z-u{)}^m}=\frac{1}{(z-v{)}^m}{(1-\frac{u-v}{z-v})}^{-m}={\sum }_{\nu =0}^{\mathrm{\infty }}\left(\genfrac{}{}{0px}{}{m+\nu -1}{\nu }\right)\frac{(u-v{)}^{\nu }}{(z-v{)}^{m+\nu }},$ with uniform convergence on $K$. Therefore every rational function with only pole $u$ can be approximated uniformly on $K$ by one with only pole $v$. Since $u\in S$, this shows $v\in S$, so $S$ is open. The same expansion with $u$ and $v$ exchanged shows that whenever $v\in S$ is sufficiently close to $u$, then $u\in S$ as well. Hence $S$ is also closed in $D(c,\rho )$. Because the disc is connected and $S$ is nonempty, $S=D(c,\rho )$, so in particular $b\in S$.

If $m=0$, then $a_m=a_0$, and $r(z)=(z-a_0{)}^{-1}$ proves clause 1 with zero error. Assume $m\ge 1$. Apply step 2.1 successively to the discs of the chain, choosing the $j$-th local error below $\epsilon \mathrm{/}m$. [step 2.1, choose, construct, cases, algebra] The triangle inequality then produces a rational function with only pole $a_m$ and total error below $\epsilon$ on $K$. This proves clause 1.

For clause 2, clause 1 gives a rational function $r$ with only pole $a_m$ and ${\sup }_K\mathrm{\mid }r(z)-(z-a_0{)}^{-1}\mathrm{\mid }<\epsilon \mathrm{/}2$. [step 3.1, algebra, discharge-construct] Write the principal part of $r$ at $a_m$ as ${\sum }_{\mathrm{\ell }=1}^L{d}_{\mathrm{\ell }}(z-a_m{)}^{-\mathrm{\ell }}$. Because $\mathrm{\mid }z\mathrm{\mid }<R<\mathrm{\mid }{a}_m\mathrm{\mid }$ on $K$, each factor $(z-a_m{)}^{-\mathrm{\ell }}=(-a_m{)}^{-\mathrm{\ell }}(1-z\mathrm{/}{a}_m{)}^{-\mathrm{\ell }}$ has a power series in $z\mathrm{/}{a}_m$ that converges uniformly on $K$. Truncating those finitely many series gives a polynomial $p$ with ${\sup }_K\mathrm{\mid }p-r\mathrm{\mid }<\epsilon \mathrm{/}2$. Then ${\sup }_K\mathrm{\mid }p(z)-(z-a_0{)}^{-1}\mathrm{\mid }<\epsilon$, proving the polynomial approximation.