Why does Mittag-Leffler's theorem work over the entire complex plane? I have a minor factual clarification about the statement of Mittag-Leffler's theorem, and also a more substantive question about its proof.
Q1. The Wikipedia article on Mittag-Leffler's theorem states that given any closed, discrete set $E$ in an open subset $D \subset \mathbb{C}$, and a desired polynomial in $1/(z-a)$ for each point $a \in E$, there exists a meromorphic function on $D$ with poles at these points, with the corresponding principal parts at each pole. Is there always a meromorphic function on $D$ with poles only at the points in $E$? Or are there sometimes necessarily additional poles in $D$ as well?
Q2. The proof outline in the next paragraph starts by taking a straightforward sum of the $p_a(z)$ over finite subsets $F \subset E$. Then it says "While the [partial sums] may not converge as $F$ approaches $E$, one may subtract well-chosen rational functions with poles outside of $D$ (provided by Runge's theorem) without changing the principal parts of [the partial sums] and in such a way that convergence is guaranteed."
Runge's theorem applies to compact subsets of $\mathbb{C}$, so this argument makes sense to me when the region $D$ is bounded, and so contained within a compact subset of $\mathbb{C}$. But what if $D$ is the entire complex plane $C$? In that case, if the answer to Q1 is "the former" (as I suspect is the case), then you can't put any poles outside of $D$ because there is nowhere to put them. Let's say you want to put a simple pole at each integer $n$ with residues that blow up ridiculously fast at large $n$, like as the Busy Beaver sequence or something. Then how can you keep the partial sums from diverging?
The only loophole that I can think of is that there is one place "outside of $D = \mathbb{C}$" to put a pole, which is at $\infty$. But the only entire functions with poles at infinity are the polynomials. And I don't see how a polynomial could possibly grow fast enough at large $z$ to subtract off a divergence (and thereby ensure convergence) for a sequence of principal parts that grows really fast.
 A: The answer to Q1 is no, there is no need no add more poles than required. The reason is because this is always possible by construction, the proof of the theorem is by contructing directly a function satisfying the requirements. Actually the resultant function is defined everywhere, except by the singularities in $E$, so you can actually assume that $D=\mathbb{C}\setminus E$ (if for some reason one is working in a proper domain of $\mathbb{C}$, then one needs just to restrict the original function).
The answer to Q2 is as follows. Suppose $\{h_j(\frac{1}{z-z_j})\}$ are the desired singular parts of the desired meromorphic function, where $h_j$ is a polinomial and $z_j\neq 0$. Assume $|z_j|$ is monotone increasing in $j$. Let $D_j = D(0,|z_j|/2)$ and choose a polinomial $f_j$ (Runge theorem allows to assume this when the complement of the compact set is connected, if you take a rational function it may add extra singularities outside $E$, so this is in fact important) such that
$$|h_j(\frac{1}{z-z_j})-f_j(z)|\leq  \frac{1}{j^2},$$
for $z\in D(0,|z_j|/2)$. Fix an $r>0$ arbitrary and let $z\in D(0,r)\setminus \{z_j\}$, then eventually the series $\sum_{j\geq 1} h_j(\frac{1}{z-z_j})-f_j(z)$ is dominated by a series of the form $\sum \frac{1}{j^2}$, and therefore we have absolute and uniform convergence. Hence $f(z) = \sum_{j\geq 1} h_j(\frac{1}{z-z_j})-f_j(z)$ is the desired meromorphic function (it converges in $D(0,r)\setminus \{z_j\}$ for any $r>0$). However, we assumed $z_j\neq 0$, so if originally you wanted also a singularity at the origin, it suffices just to add the desired singular part at this point to the constructed $f$.
