I don' t understand why the ratio of two meromorphic functions is meromorphic Let $D\subseteq\mathbb C $ be  a connected set and consider $f,g\in\mathcal M(D)$.The ratio $r=\frac{f}{g}$ has the pole set $P(r)\subseteq P(f)\cup Z(g)$ (where $Z()$ is the set of all zeros). Why is $P(r) $ discrete in $D$ ? I think that disjoint union of discrete sets isn't discrete, consider for example $$\{0\}\cup\big\{\frac{1}{n} : n\in\mathbb N\big\}$$
 A: The pole set or zero set of a nonzero meromorphic function is not only discrete, it is also closed (meromorphic functions are continuous !), which eliminates your example. 
The closed and discrete subsets of $\mathbb C$ can also be described by the property that their intersection with any compact subset of $\mathbb C$ is finite.
Then it is obvious that the reunion of two such subsets is again closed and discrete.

Suppose $S$ is closed and discrete, and let $K$ be a compact subset of $\mathbb C$.
Since $S$ is discrete, for each element $x \in S \cap K$, there is an open $U_x$ of $K$ containing $x$ such that $U_x \cap S = \{ x \}$. Since $S$ is closed, $K \setminus S$ is an open of $K$. Then, we have that $K \setminus S$ together with all the $U_x$ form an open cover of $K$. Since $K$ is compact, we can extract a finite subcover. But for each $x \in S \cap K$, $U_x$ was the only open containing $x$ so this finite cover still has to contain every $U_x$, thus there were finitely many $x \in S \cap K$ in the first place.
Suppose that for every compact $K$, $S \cap K$ is finite. Then $S$ is closed : if $x \notin S$, then there are finitely many elements of $S$ in the disk of radius $1$ centered at $x$, so their distances to $x$ has a minimum $d>0$, and there is no elements of $S$ in the open disk of radius $d$ centered at $x$. $S$ is discrete : If $x \in S$, repeat the  argument where $S$ is replaced with $S \setminus \{ x \}$
A: Look at this question (and the answers). There, it is proved that any meromorphic function is the ratio of two holomorphic functions, and vice-versa. Therefore, the ratio of two meromorphic functions is again meromorphic, since the ratio of two ratios of holomorphic functions is again a ratio of holomorphic functions. In symbols:
Write $f=\dfrac{f_1}{f_2}$ and $g=\dfrac{g_1}{g_2}$. Then
$$r=\dfrac{f}{g}=\dfrac{\frac{f_1}{f_2}}{\frac{g_1}{g_2}}=\dfrac{f_1g_2}{f_2g_1}.$$
