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The prescribed-pole form of Runge theorem is as follows. Let $K\subset \mathbb{C}$ be compact, let $f$ be holomorphic on a neighborhood of $K$, and let $S\subset \hat{\mathbb{C}}\setminus K$ meet every connected component of that complement. Then, for every $ϵ>0$, there is a rational function $R$, all of whose poles belong to $S$, with ${\sup }_K|f-R|<ϵ$. A pole at infinity means a polynomial part is allowed. In particular, if the complement is connected, choose $S=\{\infty \}$ and obtain the polynomial Runge theorem.

First approximate using poles anywhere off $K$. Choose a bounded polygonal neighborhood $V$ of $K$ whose closure lies in the neighborhood on which $f$ is holomorphic, and whose boundary is disjoint from $K$. Such a $V$ comes from a sufficiently fine finite grid covering a small closed neighborhood of $K$. Orient the outer boundary positively and boundaries of holes negatively. The Cauchy integral formula givesBecause $\partial V$ and $K$ have positive distance, the integrand is uniformly continuous on their product, separately along each boundary edge. Its Riemann sums converge uniformly for $z\in K$. Those sums are rational functions ${\sum }_j{c}_j/({\zeta }_j-z)$ with poles ${\zeta }_j\notin K$. Thus arbitrary exterior poles suffice.

To move them to the prescribed set, let $A$ be the uniform closure on $K$ of rational functions with poles only in $S$. It is a closed unital algebra; it contains the coordinate function when $\infty \in S$. DefineThis set is open relative to $\mathbb{C}\setminus K$. Indeed, if $a\in E$ and $|b-a|<\mathrm{dist}(a,K)$, the uniformly convergent expansionplaces the new reciprocal in $A$, using its algebra and closedness properties. It is also relatively closed: if $a_m\in E$ tends to $a\notin K$, then $(z-a_m{)}^{-1}$ tends uniformly on $K$ to $(z-a{)}^{-1}$. Each bounded component contains a finite selected pole and hence meets $E$. The unbounded component also meets $E$ if it contains a finite selected pole; if its selected pole is infinity, take $|a|>{\sup }_K|z|$ and usea uniform polynomial expansion. Being both open and closed and meeting every component, $E$ is all of $\mathbb{C}\setminus K$. Every initial Cauchy-sum reciprocal therefore belongs to $A$, and so does $f$. This proves Runge theorem, including its prescribed-pole assertion, rather than merely polynomial approximation on a special compact set.

For the requested pointwise polynomial approximation of a half-plane sign, take $n\ge 2$ andThese three compact sets are disjoint, and their union $K_n$ has connected complement: there are horizontal gaps above and below the segment, and every complementary point can be joined through these gaps or around a rectangle to the exterior of $[-n,n{]}^2$. Define $h_n$ to be $1,0,-1$ on disjoint open neighborhoods of these three sets. It is holomorphic on their union. By the polynomial Runge theorem, choose $P_n$ with ${\sup }_{K_n}|P_n-h_n|<1/n$.

For every fixed point with positive imaginary part, both coordinate bounds and $1/n\le \mathrm{Im}z$ eventually hold, so it lies in $K_n^{+}$ for all large $n$; the reflected assertion holds for negative imaginary part. Every real point eventually lies in $K_n^0$. ConsequentlyThe discontinuity of the limit does not contradict holomorphicity of the approximating functions: this convergence is pointwise, and is not locally uniform near the real axis.

Finally, identify $K^{\ast }$ as the holomorphic convex hull relative to $\Omega$. If $U$ is a component of $\hat{\mathbb{C}}\setminus K$ lying entirely in $\Omega$, it cannot contain infinity and is bounded. Its boundary lies in $K$, and its closure is compact in $\Omega$. For any holomorphic $f$ on $\Omega$, the maximum modulus principle on $U$ gives $|f(w)|\le {\sup }_{\partial U}|f|\le {\sup }_K|f|$. Thus $U\subseteq K^{\ast }$, as is $K$ itself.

Conversely let $w\in \Omega \setminus K$ lie in a complementary component $U$ that is not entirely in $\Omega$. If it contains a finite point $a\notin \Omega$, form the uniform algebra on $K$ generated by constants and $(z-a{)}^{-1}$. The same open-and-closed reciprocal argument above, now only on the connected component $U$, shows that $(z-w{)}^{-1}$ is in its uniform closure. Its approximants have poles only at $a$ and are holomorphic on $\Omega$. If the available point outside $\Omega$ is infinity, use the polynomial algebra and its large-pole expansion instead. Hence in either case choose $h\in \mathcal{O}(\Omega )$ withThe function $F(z)=1-(z-w)h(z)$ is holomorphic on $\Omega$, has $F(w)=1$, and satisfies ${\sup }_K|F|<1/2$. It separates $w$ from the hull, so $w\notin K^{\ast }$. Combining the two inclusions provesHere $\mathbb{P}$ denotes the Riemann sphere; using it ensures that the unbounded component, which contains infinity, is not accidentally filled. The empty compact set has empty hull by the usual convention; the separation argument above concerns nonempty $K$.