Proof of Runge's theorem Reference-Functions of one complex variable by John B.Conway(Graduate text in Mathematics)
The proof starts with some  important introductory information-
I've  already  shown that $ C(K,\mathbb C)$ is a complete metric space,but i'm not getting how to show $\rho (f_n,f)\rightarrow $ iff $f=u-lim f_n$
Now the main proof begins-The proof comprises of 3 lemmas-





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*I got stuck in the very first lemmma(1.8).Please tell me how to show if $f,g\in B(E)$ then $f+g\in B(E)$?

*What is in the square block in lemma(1.9)?


PROOF OF LEMMA-1.8
Since,$f_n\implies f$ & $g_n\implies g$($\implies$ represents uniform convergence),so for given $\epsilon >0$ we can find $N(\epsilon) \in \mathbb N$ such that $\vert f_n(z)-f(z)\vert<\frac{\epsilon}{2}    \forall z\in K$ whenever $n\geq N(\epsilon)/2$.
Similarly, for given $\epsilon >0$ we can find $M(\epsilon)\in\mathbb N$ such that $\vert g_n(z)-g(z)\vert<\frac{\epsilon}{2}  \forall z\in K$ whenever $n\geq M(\epsilon)/2$.
Now consider,$\vert (f_n+g_n)(z)-(f+g)(z) \vert \leq \vert f_n(z)-f(z)\vert +\vert g_n(z)-g(z)\vert <\epsilon /2 +\epsilon/2 =\epsilon \forall z\in K$ and$ n\geq L(\epsilon):=\max{M,N}$.
so,$f+g\in B(E)$
Is it correct?How to show that $f+g$  has poles in $E$?
PROOF OF LEMMA 1.9

How does $\partial G \subset \partial V$ happen??
 A: $f+g\in B(E)$:
You have proved  that if $f_n\Rightarrow f$ and $g_n\Rightarrow g$ on $K$ then $f_n+g_n\Rightarrow f+g$ on $K$. And it will be better if you state that the pole of $f_n+g_n$ lies in poles of $f_n$ and $g_n$.
There is no need to prove that $f+g$ has a pole in $E$ since $f+g$ is defined on $K$.
$\partial G\subset \partial V$:
Lemma 1. Suppose $X$ is a topological space and $A$ is a component of $X$, then $A$ is closed in $X$.
Lemma 2. Suppose $X$ is an open subset of $\mathbb{C}$ and $A$ is a component of $X$, then $A$ open.
Back to your question, $G$ is open by lemma 2. Let $E=V\backslash G$, then $E$ is open in $V$ by lemma 1 and thus $E$ is open in $\mathbb{C}$.
\begin{align*}
\partial V & =\overline{V^c}\cap \overline{V}\\
& =\overline{(G\cup E)^c}\cap \overline{G\cup E}\\
& =(G^c\cap E^c)\cap (\overline{G}\cup\overline{E})\\
& =(G^c\cap E^c\cap \overline{G} )\cup (G^c\cap E^c\cap \overline{E})\\
& =(G^c\cap \overline{G})\cup (E^c\cap \overline{E})\\
& =\partial G\cup \partial E.
\end{align*}
 The penultimate equation is because $\overline{G}\subset E^c$ and $\overline{E}\subset G^c$.
So the following conclusion is obatained:

Conclusion. Suppose $X$ is an open subset of $\mathbb{C}$ and $A$ is a component of $X$, then $\partial X=\partial A\cup\partial (X\backslash A)$.

