Runge approximation with a prescribed pole set

By [L1], choose a rational function $r_0(z)={\sum }_{\nu =1}^N{c}_{\nu }\mathrm{/}(z-a_{\nu })$ whose poles $a_{\nu }$ lie in $\widehat{\mathbb{C}}\setminus K$ and such that ${\sup }_{z\in K}\mathrm{\mid }f(z)-r_0(z)\mathrm{\mid }<\epsilon \mathrm{/}2$.

If $N=0$, then $r_0$ already has no finite poles, so its poles are contained in $P$ and step 1.1 already proves the theorem. Assume from now on that $N\ge 1$.

For each pole $a_{\nu }$, let $C_{\nu }$ be the connected component of $\widehat{\mathbb{C}}\setminus K$ containing it. By [L2], choose $p_{\nu }\in P\cap C_{\nu }$. Applying [L3] to the function $c_{\nu }\mathrm{/}(z-a_{\nu })$ inside $C_{\nu }$, choose a rational function $s_{\nu }$ with poles only at $p_{\nu }$ and ${\sup }_{z\in K}\mid s_{\nu }(z)-\frac{c_{\nu }}{z-a_{\nu }}\mid <\frac{\epsilon }{2N}\mathrm{.}$

Put $r={\sum }_{\nu =1}^N{s}_{\nu }$. Then every pole of $r$ lies in $P$, and the triangle inequality together with steps 1.1 and 3.1 gives ${\sup }_{z\in K}\mathrm{\mid }f(z)-r(z)\mathrm{\mid }\le {\sup }_{z\in K}\mathrm{\mid }f(z)-r_0(z)\mathrm{\mid }+{\sum }_{\nu =1}^N{\sup }_{z\in K}\mid s_{\nu }(z)-\frac{c_{\nu }}{z-a_{\nu }}\mid <\epsilon \mathrm{.}$