Showing that meromorphic functions form a field (without using Laurent series) Thinking of meromorphic functions on a Riemann surface $X$ as holomorphic maps $X → \hat{ℂ}$, where $\hat{ℂ}$ is the Riemann sphere, how can one show that the set of meromorphic functions $\mathcal{M}(X)$ forms a field? I want to avoid Laurent series.
I tried to define addition and multiplication in the following way:
\begin{align}
f + g&\colon X → \hat{ℂ},\; z ↦ \lim_{w → z} (f(w) + g(w))\\
f · g&\colon X → \hat{ℂ},\; z ↦ \lim_{w → z} (f(w)·g(w))
\end{align}
I have to show that this is well-defined, then the assertion follows easily.
By the identity theorem, the poles of $f$ and $g$ are discrete and closed in $X$, so one can indeed look at the sums and products of $f$ and $g$ in a punctured neighbourhood of $z$.
So I have to establish that those limits exist.
I also know $\lvert h (w)\rvert → ∞ ⇔ h (w) → ∞$ which I then can apply to $h = f + g$ and $h = f · g$. This is where I get stuck.
I’m also open to different ways of proving that meromorphic functions form a field. For example one can prov it for the unit disk first and then lift it to $X$ by the use of charts. I’m also interested in plausible arguments to convince me, that Laurent series or lifting are the way to go.
 A: One way is to start with a slightly different definition of a meromorphic function on $X$: namely, it is a holomorphic function $f$ on an open subset $Y$ of $X$ such that for each $x_0 \in X \setminus Y$ there is an open coordinate neighborhood* $U$ of $x_0$ such that $U \setminus \{x_0\} \subset Y$ and there is some positive integer $k$ such that 
$x^k f$ extends to be holomorphic on $U$.  This gives a meromorphic function on your sense in an evident way, but it is easier to add, subtract, multiply and invert meromorphic functions as defined above.  
*: Here I mean that there is a biholomorphic map from $U$ to the open unit disk in the  complex plane which carries $x_0$ to the origin.  We write $x$ for the pulback of the identity function $z \mapsto z$ under this map, so that $x$ has a simple zero at $x_0$.  
A: Use another definition of meromorphic:
A function is meromorphic if it is analytical except for an isolated set of points.
This way things are quite easy now.
E.g. $f,g$ meromorphic with exceptional sets $F,G$ which consist of isolated points. $\Rightarrow f\cdot g$ meromorphic, since
$f\cdot g$ is analytical except on the set $F\cup G$ which also consists of isolated points, since we did no infinite union. 
Etc. ... 
So do not care so much about the function values/limits, work with the poles.
