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The polynomial Runge theorem states that, for a compact set $K\subset \mathbb{C}$ with connected complement and a function $f$ holomorphic on a neighborhood of $K$, for every $\epsilon >0$ there is a polynomial $P$ such thatThe connected-complement hypothesis permits polynomial approximation rather than approximation by rational functions with poles in bounded complementary components.

Use the Runge exhaustion of a slit disk, with the principal argument $-\pi <\arg z<\pi$:These are nested compact subsets of $\Omega$. Their complements are connected: the inner disk and exterior region join through the omitted sector around the negative real radius. Every compact subset of $\Omega$ has positive distance from zero, the unit circle and the slit, so it is contained in $K_m$ for all sufficiently large $m$.

Since $f$ is holomorphic on $\Omega$, it is holomorphic on a neighborhood of each $K_m$. The polynomial Runge theorem gives $P_m$ with ${\sup }_{K_m}|P_m-f|<1/m$. Given any compact $K\subset \Omega$, eventually $K\subseteq K_m$, so the same estimate holds on $K$. ThereforeThe nested exhaustion makes this one sequence work simultaneously on all compact subsets.

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 $K$.

Choose the argument in $(0,2\pi )$ on the slit annulus and letThese compact sets increase to the slit annulus, and each has connected complement: the omitted angular corridor joins the inner disk to the outside of the outer disk. The polynomial Runge theorem says that a function holomorphic near a compact set with connected complement can be approximated uniformly there by polynomials. Choose $p_j$ with ${\sup }_{K_j}|p_j-f|<1/j$. Every compact subset $K$ of the slit annulus has positive distance from the removed positive ray, so $K\subseteq K_j$ for all sufficiently large $j$. Hence $p_j\to f$ uniformly on every such compact subset.

Let $\mathcal{A}$ be the closure of the polynomials in $C(K)$ for the uniform norm. It is a closed algebra, since products of uniformly convergent sequences converge uniformly on the compact set $K$.

Choose a point $W$ with $|W|>{\sup }_{z\in K}|z|$. The uniformly convergent geometric seriesshows that this reciprocal belongs to $\mathcal{A}$. Join $w$ to $W$ by a continuous path in $\mathbb{C}\setminus K$. Its image is compact and disjoint from $K$, so their distance $d$ is positive. Partition the path into finitely many steps shorter than $d/2$.

Suppose $1/(b-z)\in \mathcal{A}$ and $|a-b|<d/2$, with $b$ on the path. Thenconverges uniformly on $K$, since $|b-z|\ge d$. Every partial sum belongs to $\mathcal{A}$, by its algebra property, and so does the limit. Working backwards along the partition transfers the pole from $W$ to $w$. Thus $1/(w-z)\in \mathcal{A}$: a polynomial $P$ exists withThis pole-moving polynomial approximation is the elementary mechanism behind the polynomial Runge theorem.

The prescribed-pole form of Runge theorem is this: let $K\subset \mathbb{C}$ be compact, and let $E\subset \hat{\mathbb{C}}\setminus K$ meet every connected component of the complement in the Riemann sphere. If $f$ is holomorphic on a neighbourhood of $K$, then for every $ϵ>0$ there is a rational function with all its poles in $E$ whose uniform distance from $f$ on $K$ is less than $ϵ$. A polynomial is regarded as having its only possible pole at infinity. The empty compact set is trivial, so assume $K\ne \varnothing$.

First suppose $\infty \in E$. Let $A\subset C(K)$ be the uniform rational approximation algebra with prescribed poles, the uniform closure of rational functions whose finite poles lie in $E$. It is a commutative unital Banach algebra, and contains $u(z)=z$. Setwhere the inverse in this definition is the pointwise continuous function on $K$. The set $S$ is relatively open: if the inverse at $\lambda$ belongs to $A$, the Neumann series gives inverses at nearby points. It is relatively closed: when ${\lambda }_n\to \lambda \notin K$, the corresponding scalar functions converge uniformly on $K$, and $A$ is closed.

Each bounded complementary component meets $E$ in a finite point $a$, and $(u-a1{)}^{-1}$ is an allowed rational function. The unbounded component meets $S$ because for $|\lambda |>\Vert u\Vert$ the geometric series for $(u-\lambda 1{)}^{-1}$ belongs to $A$. Being both open and closed in the complement, $S$ therefore contains every component, so it is the entire complement. Conversely, evaluation at any $z\in K$ prevents $u-z1$ from being invertible. HenceApply the holomorphic functional calculus in $A$ to $f$ near $K$. For every $z\in K$, the evaluation character of an algebra and the contour formula from (b) give $f(u)(z)=f(z)$. Thus $f{|}_K\in A$. By the definition of this uniform closure, the required rational approximations exist.

If $\infty \notin E$, choose a finite point $a\in E$ in the component containing infinity. The Möbius transformation $w=(z-a{)}^{-1}$ sends $K$ to a compact subset $K^{\prime }$ of the plane and sends $E$ to a set $E^{\prime }$ containing infinity. It preserves complementary components, so $E^{\prime }$ meets every component of the complement of $K^{\prime }$. The transformed function $F(w)=f(a+1/w)$ is holomorphic near $K^{\prime }$, since $0\notin K^{\prime }$. The case already proved approximates $F$ by rational functions with poles in $E^{\prime }$. Pulling them back gives rational approximations to $f$ on $K$ with poles only in $E$: a pole at infinity in the $w$-plane becomes a pole at $a$, and every other pole has its prescribed preimage. In particular the pullbacks do not acquire a pole at the original infinity, since $0\notin E^{\prime }$.

This proves the full prescribed-pole theorem, including the case where a pole at infinity is forbidden. If $\mathbb{C}\setminus K$ is connected, take $E=\{\infty \}$ to obtain the polynomial Runge theorem as a corollary.

The set $K$ is compact, and its complement is connected: two disjoint closed disks do not enclose any bounded complementary component. Since $f$ is holomorphic on the open neighbourhood $\Omega$ of $K$, the polynomial Runge theorem gives, for each $n$, a polynomial $p_n$ such thatHence

Define a function on the disconnected open set $\Omega$ byIt is holomorphic on $\Omega$. Applying the first part produces polynomials $P_n$ with

For an obstruction, takeThese 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 giveThey therefore cannot also satisfy $Q_n(0)\to 1$. Thus this pair $K_1,K_2$ has the requested non-approximation property.

The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.

Statement (ii) is true. LetEach $K_n$ is compact and has connected complement. By the polynomial Runge theorem, choose a polynomial $P_n$ withEvery fixed $z\in E$ belongs to $K_n$ for all sufficiently large $n$, so $P_n(z)\to f(z)$. Thus the convergence is pointwise on $E$.

There is a sequence of complex polynomials converging pointwise on the whole plane to the sign of the imaginary part, with value zero on the real axis. For each $n$, take the compact union of the upper rectangle $[-n,n]+i[1/n,n]$, its lower reflection and the real segment $[-n,n]$. Its complement is connected. The function equal to one, minus one and zero on disjoint neighborhoods of the three components is holomorphic. The polynomial Runge theorem provides a polynomial within $1/n$ of it on the union. Every fixed point lies in its corresponding component for all large $n$, proving pointwise convergence. The limit is discontinuous, so the convergence cannot be locally uniform near the real axis. This illustrates how pointwise approximation by holomorphic functions differs from locally uniform approximation.

For the unit disk with the nonpositive real radius removed, compact sets $K_m=\{r{e}^{i\theta }:1/m\le r\le 1-1/m,|\theta |\le \pi -1/m\}$, $m\ge 3$, exhaust the domain. Each has connected complement: its inner and outer complementary regions join through the missing angular sector. The polynomial Runge theorem gives a polynomial within $1/m$ of a given holomorphic function on $K_m$. Every compact subset lies in some $K_m$, so these polynomials converge uniformly on compact subsets.