Let $G \subseteq \mathbb C$ be a region and $H(G),M(G)$ and $C(G,\mathbb C_{\infty})$ be respectively denote the spaces of holomorphic functions, meromorphic functions and extended complex valued continuous functions on $G$ respectively. Then in Theorem 3.4 (Page no. $156$) in Conway's book on complex analysis it has been proved that $M(G) \cup \{\infty\}$ is a complete metric space. As a corollary it has been claimed that $H(G) \cup \{\infty\}$ is a closed subset of $C(G,\mathbb C_{\infty}).$ It is clear to me that $H(G) \cup \{\infty\}$ is contained in the closure of $H(G)$ when considered as a subset of $C(G, \mathbb C_{\infty}).$ But how to prove that there is nothing outside $H(G) \cup \{\infty\}$ in the closure of $H(G)\ $? Is it due to the fact that $H(G) \subseteq M(G)\ $? If so, how do I show that there won't be any meromorphic function in the closure of $H(G)$ when viewing it as a subset of $C(G,\mathbb C_{\infty})\ $?
Any help in this regard would be warmly appreciated. Thanks for investing your valuable time.
Let $(f_n)$ be a sequence in $H(G)$ which converges to $f \in M(G) \cup \{\infty\}$ with respect to the spherical metric. We want to show that $f$ is holomorphic or identically $\infty$.
If $f(z_0) = \infty$ then there is a small neighborhood $V$ of $z_0$ such that $1/f$ and all $1/f_n$ are holomorphic $V$, and $1/f_n \to 1/f$ uniformly in $V$ with respect to the spherical metric. It follows that also $1/f_n \to 1/f$ uniformly with respect to the Euclidean metric.
All $1/f_n$ are non-zero in $V$. Hurwitz's theorem states that $1/f$ is non-zero in $V$ or identically zero in $V$. But $1/f(z_0) = 0$, so $1/f$ is identically zero in $V$.
So we have shown the following: If $f(z_0) = \infty$ then $f(z) = \infty$ in a neighborhood of $V$. Since poles of meromorphic functions are isolated, we have that either $f$ is identically $\infty$, or $f$ has no poles, i.e. $f \in H(G)$.