Every locally finite effective divisor on a plane domain is a holomorphic zero divisor

Every point of $\mathrm{\Omega }$ has a closed disc neighbourhood contained in $\mathrm{\Omega }$, so local finiteness makes each point of $S$ isolated and prevents any accumulation point of $S$ in $\mathrm{\Omega }$. By [L6], enumerate all open discs with rational centre coordinates and positive rational radius in a fixed order. For each $a\in S$, let $k(a)$ be the least index of such a disc containing $a$ and meeting $S$ only at $a$; isolation and density of the rationals ensure its existence. Distinct points have distinct indices. Thus $S$ injects into the nonnegative integers. By [L6], encode the pairs $(k(a),r)$ with $1\le r\le m(a)$ by a fixed pairing of integers, and list the resulting codes in increasing order. This gives the points of $S$ repeated exactly $m(a)$ times, without a choice principle. A compact subset contains only finitely many entries of this list. If the list is finite, $F(z)={\prod }_{a\in S}(z-a{)}^{m(a)}$ works, with empty product $1$. Henceforth assume it is infinite.

If $C:=\mathbb{C}\setminus \mathrm{\Omega }$ is empty, put all entries into a list $(c_j)$ and leave the list $(b_j)$ empty. If $C$ is nonempty, set $\delta (z):=\mathrm{dist}(z,C)$ and split the repeated list, preserving its order, according to $D:=\{z\in \mathrm{\Omega }:\delta (z)<1\mathrm{/}(\mathrm{\mid }z\mathrm{\mid }+1)\}$: $(b_j)$ contains the entries in $D$, and $(c_j)$ those outside $D$. Either sublist may be empty, finite, or infinite. The closedness of $C$ and openness of $\mathrm{\Omega }$ imply $\delta (z)>0$ on $\mathrm{\Omega }$, and [L4] makes $\delta$ continuous.

Suppose $(b_j)$ is infinite. For each $\epsilon >0$, an entry $b_j$ with $\delta (b_j)\ge \epsilon$ satisfies $\mathrm{\mid }{b}_j\mathrm{\mid }<1\mathrm{/}\epsilon$. Such entries lie in the closed bounded set $\{z:\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}\epsilon ,\delta (z)\ge \epsilon \}$, contained in $\mathrm{\Omega }$ and compact by [L4]. Step 1.1 makes their number finite. Thus $\delta (b_j)\to 0$. To specify a nearest boundary point without simultaneous choices, consider the compact nonempty set $C\cap \bar{B}(b_j,\delta (b_j)+1)$. The minimum of $\mathrm{\mid }{b}_j-p\mathrm{\mid }$ on it is $\delta (b_j)$: points outside it have distance greater than $\delta (b_j)+1$, and the definition of the infimum supplies points inside it with distance arbitrarily close to $\delta (b_j)$. The nearest-point set is therefore nonempty compact. First minimize $\mathrm{Re}p$ on that set, then minimize $\mathrm{Im}p$ on the minimizers. This specifies a unique $p_j\in C$ with $\mathrm{\mid }{b}_j-p_j\mathrm{\mid }=\delta (b_j)$.

Suppose $(c_j)$ is infinite. If $C=\mathrm{\varnothing }$, every closed disc is compact in $\mathrm{\Omega }=\mathbb{C}$, so step 1.1 shows that only finitely many entries have modulus at most any fixed $R$. If $C\mathrm{\ne }\mathrm{\varnothing }$, entries with $\mathrm{\mid }{c}_j\mathrm{\mid }\le R$ satisfy $\delta (c_j)\ge 1\mathrm{/}(R+1)$; they lie in the compact subset $\{z:\mathrm{\mid }z\mathrm{\mid }\le R,\delta (z)\ge 1\mathrm{/}(R+1)\}$ of $\mathrm{\Omega }$, which again contains only finitely many entries. In either case $\mathrm{\mid }{c}_j\mathrm{\mid }\to \mathrm{\infty }$. Let $m_0$ be the finite number of zero entries and enumerate the nonzero entries as $(d_j)$. Then $\mathrm{\mid }{d}_j\mathrm{\mid }\to \mathrm{\infty }$.

For infinite $(b_j)$ define $Q_j(z):=E_j((b_j-p_j)\mathrm{/}(z-p_j))$. Its denominator is nonzero on $\mathrm{\Omega }$, so [L1] makes it holomorphic. Its only zero is $b_j$, and that zero is simple because $1-(b_j-p_j)\mathrm{/}(z-p_j)=(z-b_j)\mathrm{/}(z-p_j)$ and the exponential factor never vanishes. For a nonempty compact $K\subseteq \mathrm{\Omega }$, [L4] gives $d_K:={\min }_{z\in K}\delta (z)>0$. Since $p_j\in C$, ${\sup }_{z\in K}\mathrm{\mid }(b_j-p_j)\mathrm{/}(z-p_j)\mathrm{\mid }\le \delta (b_j)\mathrm{/}{d}_K\le 1\mathrm{/}2$ for all sufficiently large $j$. These tail factors have no zeros on $K$, and [L2] gives ${\sup }_K\mathrm{\mid }1-Q_j\mathrm{\mid }\le 2^{-j-1}$. Hence [L3] yields a holomorphic function $F_1:={\prod }_{j\ge 1}{Q}_j$ with precisely the listed $b_j$ zeros and their multiplicities. If $(b_j)$ is finite, take its finite polynomial instead, and if it is empty take $F_1=1$.

On each disc $\mathrm{\mid }z\mathrm{\mid }\le R$, eventually $\mathrm{\mid }{d}_j\mathrm{\mid }>2R$, so $E_j(z\mathrm{/}{d}_j)$ is zero-free there and [L2] gives ${\sup }_{\mathrm{\mid }z\mathrm{\mid }\le R}\mathrm{\mid }1-E_j(z\mathrm{/}{d}_j)\mathrm{\mid }\le 2^{-j-1}$. By [L1] and [L3], $H(z):={\prod }_{j\ge 1}{E}_j(z\mathrm{/}{d}_j)$ is entire with exactly the listed nonzero zeros and their multiplicities; $H(0)=1$ since every partial product equals $1$ there. Thus $F_2(z):=z^{m_0}H(z)$ contributes the listed $c_j$ zeros, with $z^0=1$ contributing no zero at $0$. If $(c_j)$ is finite, take its finite polynomial instead, and if it is empty take $F_2=1$.

Set $F:=F_1{F}_2$. It is holomorphic by [L1], and steps 4.1 and 4.2 show that its zero set is exactly $S$, with order $m(a)$ at each $a\in S$: at a given point the orders of the finitely many contributing factors add and the remaining local product is nonzero. Every point of $\mathrm{\Omega }$ has a disc meeting $S$ in finitely many points, and that disc contains a point outside this finite set, so $F$ is not identically zero. It is nonzero at each point of $\mathrm{\Omega }\setminus S$, and [L5] gives order $0$ there, including each $a\in A$ with $m(a)=0$. This proves the stated realization.