Three omitted values force exterior extension

Let $v_1,v_2,v_3$ be the three omitted sphere values. By [F1] choose a Möbius transformation $M$ with $M(v_1)=0$, $M(v_2)=1$, $M(v_3)=\mathrm{\infty }$, and put $h(z):=M(g(1\mathrm{/}z))$ for $0<\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}{R}_1$. Since $\mathrm{\mid }1\mathrm{/}z\mathrm{\mid }>R_1$ there, $g(1\mathrm{/}z)$ omits $v_1,v_2,v_3$; hence $h$ attains neither $\mathrm{\infty }$ nor $0,1$, so $h$ is holomorphic on $0<\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}{R}_1$ and omits $0$ and $1$.

(Local setup) Let $h$ be holomorphic on $0<\mathrm{\mid }z\mathrm{\mid }<r_0$ omitting $0$ and $1$. Set $\rho :=r_0\mathrm{/}8$, ${\rho }_n:=\rho 2^{-n}$ for $n\in \mathbb{N}$, $A:=\{1\mathrm{/}2<\mathrm{\mid }\zeta \mathrm{\mid }<2\}$ and $h_n(\zeta ):=h({\rho }_n\zeta )$ for $\zeta \in A$. Since $0<{\rho }_n\mathrm{\mid }\zeta \mathrm{\mid }<2\rho =r_0\mathrm{/}4<r_0$ for $\zeta \in A$, each $h_n$ is holomorphic on $A$ and omits $0$ and $1$.

(Dense set) Put $K:=\{3\mathrm{/}4\le \mathrm{\mid }\zeta \mathrm{\mid }\le 5\mathrm{/}4\}$ and let $Q:=(\mathbb{Q}+i\mathbb{Q})\cap K$. Since ${\mathbb{Q}}^2$ is countable and dense in ${\mathbb{R}}^2$ and $K$ is the closure of its nonempty interior $\{3\mathrm{/}4<\mathrm{\mid }\zeta \mathrm{\mid }<5\mathrm{/}4\}$, $Q$ is countable and dense in $K$; fix an enumeration $q_0,q_1,q_2,\dots$ of $Q$.

(Dyadic target boxes) Identify the sphere with $S^2\subset [-1,1{]}^3$ through the stereographic homeomorphism of [F5]. For each $m\ge 0$ the $8^m$ half-open dyadic boxes of side $2^{1-m}$ partition $[-1,1{]}^3$; order each level lexicographically. Their diameters are $\sqrt{3}\cdot 2^{1-m}\to 0$.

It suffices to prove the following local claim: a function holomorphic on a punctured disc $0<\mathrm{\mid }z\mathrm{\mid }<r_0$ that omits $0$ and $1$ extends meromorphically at $0$. Indeed, applying the claim to $h$ produces a meromorphic extension of $h$ at $0$; by [F2] the composition $M^{-1}\circ h$ is then meromorphic at $0$, and on the punctured disc it equals $g(1\mathrm{/}z)$, so $z\mapsto g(1\mathrm{/}z)$ is meromorphic near $0$.

(Fixed band) Put $\delta :=1\mathrm{/}16$. The set $K$ of step 1.3 is closed and bounded, hence compact by [F7]. If $a\in K$ and $\mathrm{\mid }\zeta -a\mathrm{\mid }\le 2\delta =1\mathrm{/}8$, then $5\mathrm{/}8\le \mathrm{\mid }\zeta \mathrm{\mid }\le 11\mathrm{/}8$, so $\zeta \in A$; thus $\bar{D}(a,2\delta )\subset A$ for every $a\in K$.

(Center normalization) Fix $a\in K$ and $n\in \mathbb{N}$. If $\mathrm{\mid }{h}_n(a)\mathrm{\mid }\le 2$ put $t:=h_n$; otherwise put $t:=1\mathrm{/}{h}_n$. In both cases $t$ is holomorphic on $A$, omits $0$ and $1$, and $\mathrm{\mid }t(a)\mathrm{\mid }\le 2$: in the first case this is step 1.2; in the second, $h_n$ has no zeros, $t=0$ would force $h_n=\mathrm{\infty }$, and $t=1$ would force $h_n=1$, while $\mathrm{\mid }t(a)\mathrm{\mid }<1\mathrm{/}2$.

(Extraction at one point) Let $I\subset \mathbb{N}$ be infinite and $q\in K$. Define $I_0:=I$ and, recursively, for $m\ge 0$: among the finitely many level-$(m+1)$ dyadic boxes, at least one contains $\mathrm{\Sigma }(h_i(q))$ for infinitely many $i\in I_m$; let $B_m$ be the first such box and $I_{m+1}:=\{i\in I_m:\mathrm{\Sigma }(h_i(q))\in B_m\}$. Let $s(m)$ be the least element of $I_m$ with $s(m)>s(m-1)$, where $s(-1):=-1$. Then every $I_m$ is infinite and nested, $s$ is strictly increasing with range in $I$, and the values $\mathrm{\Sigma }(h_{s(m)}(q))$ are eventually inside boxes of diameter tending to $0$, so they form a Cauchy sequence; being contained in the compact metric space $[-1,1{]}^3$, it converges, and the limit lies in the closed subset $S^2$. Every selection above is a least element in a finite or well-ordered list, so no choice principle is used.

(Schottky bound) The map $F(\zeta ):=t(a+2\delta \zeta )$ is holomorphic on the unit disc and omits $0,1$, with $\mathrm{\mid }F(0)\mathrm{\mid }=\mathrm{\mid }t(a)\mathrm{\mid }\le 2$; the disc $D(a,2\delta )$ lies in $A$ by step 2.2. Applying [F3] with center bound $2$ and inner radius $1\mathrm{/}2$ gives a constant $C_0=C(2,1\mathrm{/}2)$, independent of $a$ and $n$, with $\mathrm{\mid }t(\zeta )\mathrm{\mid }\le C_0$ for $\mathrm{\mid }\zeta -a\mathrm{\mid }\le \delta$.

(Nested refinements) Let $s_0$ be the strictly increasing sequence produced by step 2.4 with $I=\mathbb{N}$ and $q=q_0$. Define states $(j,s_j)$ by recursion [F10], taking $s_{j+1}$ to be the sequence produced by step 2.4 from $I:=\mathrm{range}(s_j)$ and $q:=q_{j+1}$. Then each $s_j$ is strictly increasing, $\mathrm{range}(s_{j+1})\subseteq \mathrm{range}(s_j)$, and, since $s_j$ was extracted at $q_j$, the values $h_{s_j(m)}(q_j)$ converge in the chordal sphere as $m\to \mathrm{\infty }$ for every $j$.

(Lipschitz bound for $t$) By [F4] applied to $t$ on $D(a,2\delta )$ with bound $M=C_0$ on the circle $\mathrm{\mid }\zeta -a\mathrm{\mid }=\delta$ and $r=\delta \mathrm{/}2$, we get $\mathrm{\mid }{t}^{\mathrm{\prime }}(\zeta )\mathrm{\mid }\le \frac{\delta C_0}{(\delta \mathrm{/}2{)}^2}=\frac{4C_0}{\delta }$ for $\mathrm{\mid }\zeta -a\mathrm{\mid }\le \delta \mathrm{/}2$. Hence $\mathrm{\mid }t(z)-t(w)\mathrm{\mid }\le \frac{4C_0}{\delta }\mathrm{\mid }z-w\mathrm{\mid }$ for $z,w\in D(a,\delta \mathrm{/}2)$.

(Diagonal) Put $n_i:=s_i(i)$ for $i\in \mathbb{N}$. Then $n_{i+1}=s_{i+1}(i+1)>s_{i+1}(i)\ge s_i(i)=n_i$, because $\mathrm{range}(s_{i+1})\subseteq \mathrm{range}(s_i)$ forces $s_{i+1}(i)\ge s_i(i)$ by induction on the position in the increasing enumeration. Hence $(n_i)$ is strictly increasing. For fixed $j$ and $i\ge j$ we have $n_i\in \mathrm{range}(s_i)\subseteq \mathrm{range}(s_j)$, so $(n_i{)}_{i\ge j}$ is a strictly increasing sequence of elements of $\mathrm{range}(s_j)$, i.e. a subsequence of $s_j$; by step 3.2 the values $h_{n_i}(q_j)$ converge to a point $H(q_j)$ of the sphere.

(Chordal form) For finite $u,v$ the stereographic coordinates give $\chi (u,v)=\frac{2\mathrm{\mid }u-v\mathrm{\mid }}{\sqrt{(1+\mathrm{\mid }u{\mathrm{\mid }}^2)(1+\mathrm{\mid }v{\mathrm{\mid }}^2)}}\le 2\mathrm{\mid }u-v\mathrm{\mid }$, and $\chi (1\mathrm{/}u,1\mathrm{/}v)=\chi (u,v)$ for $u,v\mathrm{\ne }0$: substituting $1\mathrm{/}u,1\mathrm{/}v$ in the formula clears the factors $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }$. Therefore for $z,w\in D(a,\delta \mathrm{/}2)$, in the notation of step 2.3, $\chi (h_n(z),h_n(w))=\chi (t(z),t(w))\le 2\mathrm{\mid }t(z)-t(w)\mathrm{\mid }\le \frac{8C_0}{\delta }\mathrm{\mid }z-w\mathrm{\mid }$, uniformly in $n$ and $a$.

(Equicontinuity on $K$) For all $z,w\in K$ with $\mathrm{\mid }z-w\mathrm{\mid }<\delta \mathrm{/}2$, apply step 5.1 with $a:=z\in K$ and $w\in D(z,\delta \mathrm{/}2)$: $\chi (h_n(z),h_n(w))\le \frac{8C_0}{\delta }\mathrm{\mid }z-w\mathrm{\mid }$ for every $n$. Thus the family $\{h_n\}$ has the uniform chordal modulus of continuity $\omega (s):=\min \{2,\frac{8C_0}{\delta }s\}$ on $K$, where the constant $2$ bounds the chordal distance because $\chi$ is the chordal length on the unit sphere.

(Uniform convergence on $K$) Let $\epsilon >0$ and choose $\eta >0$ with $\omega (\eta )<\epsilon \mathrm{/}3$. The discs $D(q,\eta )$, $q\in Q$, cover $K$; by compactness [F7] a finite subcover exists, and we take the least one in a fixed enumeration of the finite subsets of $Q$. For each of its finitely many centers $q_{\mathrm{\ell }}$, step 4.2 gives $N_{\mathrm{\ell }}$ with $\chi (h_{n_i}(q_{\mathrm{\ell }}),h_{n_m}(q_{\mathrm{\ell }}))<\epsilon \mathrm{/}3$ for $i,m\ge N_{\mathrm{\ell }}$. Let $N:={\max }_{\mathrm{\ell }}{N}_{\mathrm{\ell }}$. For $z\in K$ choose $\mathrm{\ell }$ with $\mathrm{\mid }z-q_{\mathrm{\ell }}\mathrm{\mid }<\eta$; then for $i,m\ge N$, $\chi (h_{n_i}(z),h_{n_m}(z))\le \chi (h_{n_i}(z),h_{n_i}(q_{\mathrm{\ell }}))+\chi (h_{n_i}(q_{\mathrm{\ell }}),h_{n_m}(q_{\mathrm{\ell }}))+\chi (h_{n_m}(q_{\mathrm{\ell }}),h_{n_m}(z))<\epsilon$. Thus $(h_{n_i})$ is uniformly Cauchy on $K$, and since the chordal sphere is compact, hence complete [F8], it converges uniformly on $K$ to a map $H:K\to \widehat{\mathbb{C}}$.

(Limit on the annulus) The chosen functions $h_{n_i}$ are holomorphic on the annulus $U:=\{3\mathrm{/}4<\mathrm{\mid }\zeta \mathrm{\mid }<5\mathrm{/}4\}\subset A$ and converge chordally uniformly on $U$ by step 7.1. By [F11] the limit $H:U\to \widehat{\mathbb{C}}$ is meromorphic or identically $\mathrm{\infty }$; since all approximants are holomorphic, $H$ is holomorphic on $U$ or identically $\mathrm{\infty }$.

(Finite case: a bound on the circle) Suppose $H$ is holomorphic on $U$. By [F9] applied to $\mathrm{\mid }H\mathrm{\mid }$ on the compact circle $\{\mathrm{\mid }\zeta \mathrm{\mid }=1\}\subset U$ there is $M$ with $\mathrm{\mid }H(\zeta )\mathrm{\mid }\le M$ there, and the continuous positive function $\zeta \mapsto \chi (H(\zeta ),\mathrm{\infty })$ attains a positive minimum. So the image of the circle is a compact subset of $\mathbb{C}$ and there is $c_0>0$ with $\chi (H(\zeta ),\mathrm{\infty })\ge c_0$ on it. Uniform chordal convergence (step 7.1) then gives, for all large $i$, $\chi (h_{n_i}(\zeta ),H(\zeta ))<c_0\mathrm{/}2$ on $\mathrm{\mid }\zeta \mathrm{\mid }=1$, so $\chi (h_{n_i}(\zeta ),\mathrm{\infty })\ge c_0\mathrm{/}2$ there; writing $\chi (u,\mathrm{\infty })=2\mathrm{/}\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}$, this means $\mathrm{\mid }{h}_{n_i}(\zeta )\mathrm{\mid }\le M^{\mathrm{\prime }}$ on $\mathrm{\mid }\zeta \mathrm{\mid }=1$ for a constant $M^{\mathrm{\prime }}$ independent of $i$. Since $h_{n_i}(\zeta )=h({\rho }_{n_i}\zeta )$, the function $h$ satisfies $\mathrm{\mid }h\mathrm{\mid }\le M^{\mathrm{\prime }}$ on each circle $\mathrm{\mid }z\mathrm{\mid }={\rho }_{n_i}$ with $i$ large.

(Infinite case: a bound for the reciprocal) Suppose $H\equiv \mathrm{\infty }$ on $U$. Uniform chordal convergence to $\mathrm{\infty }$ gives, for all large $i$, $\chi (h_{n_i}(\zeta ),\mathrm{\infty })<\sqrt{2}$ on $\mathrm{\mid }\zeta \mathrm{\mid }=1$. Since $\chi (u,\mathrm{\infty })=2\mathrm{/}\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}$, this is exactly $\mathrm{\mid }{h}_{n_i}(\zeta )\mathrm{\mid }>1$, that is $\mathrm{\mid }1\mathrm{/}h({\rho }_{n_i}\zeta )\mathrm{\mid }<1$, on the circle $\mathrm{\mid }\zeta \mathrm{\mid }=1$. As $h$ omits $0$, the reciprocal $1\mathrm{/}h$ is holomorphic on the punctured disc, and $\mathrm{\mid }1\mathrm{/}h\mathrm{\mid }\le 1$ on each circle $\mathrm{\mid }z\mathrm{\mid }={\rho }_{n_i}$ with $i$ large.

(Propagation, finite case) For each large $i$, the function $h$ is continuous on the closed annulus ${\rho }_{n_{i+1}}\le \mathrm{\mid }z\mathrm{\mid }\le {\rho }_{n_i}$ and holomorphic in its interior, with $\mathrm{\mid }h\mathrm{\mid }\le M^{\mathrm{\prime }}$ on both boundary circles by step 8.2; [F12] gives $\mathrm{\mid }h\mathrm{\mid }\le M^{\mathrm{\prime }}$ throughout that annulus. Consecutive retained annuli cover $\{0<\mathrm{\mid }z\mathrm{\mid }\le {\rho }_{n_{i_0}}\}$ for a suitable $i_0$, so $h$ is bounded on a punctured neighbourhood of $0$.

(Propagation, infinite case) Likewise, in the setting of step 8.3 the reciprocal $1\mathrm{/}h$ is holomorphic on each such annulus and bounded by $1$ on both boundary circles, so [F12] gives $\mathrm{\mid }1\mathrm{/}h\mathrm{\mid }\le 1$ throughout the annulus; hence $1\mathrm{/}h$ is bounded on a punctured neighbourhood of $0$.

(Meromorphic extension at the centre) If $h$ is bounded near $0$, then [F13] makes $0$ a removable singularity of $h$ and $h$ has a holomorphic extension across $0$. If instead $1\mathrm{/}h$ is bounded near $0$, then [F13] extends $1\mathrm{/}h$ holomorphically to a function $\phi$ with $\phi (0)=c$: if $c\mathrm{\ne }0$ then $h=1\mathrm{/}\phi$ is holomorphic near $0$, and if $c=0$ then $\phi$ has a zero of finite order $m\ge 1$ at $0$, and the pole criterion of [F14], applied to $f:=h$ whose reciprocal $\phi =1\mathrm{/}h$ extends holomorphically across $0$ and vanishes there, shows that $h$ has a pole of order $m$ at $0$. In every case $h$ extends meromorphically at $0$.

The local claim of step 2.1 is proved, so by step 2.1 the function $z\mapsto g(1\mathrm{/}z)$ is meromorphic in a neighbourhood of $0$; equivalently, $g$ extends meromorphically across $w=\mathrm{\infty }$. Every selection above was the least element of a finite or well-ordered explicitly enumerated list, so no choice principle is used.