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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 $$\{0\}\cup\big\{\frac{1}{n} : n\in\mathbb N\big\}$$

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up vote 5 down vote accepted

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 $S \setminus \{ x \}$

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How can i prove that $D$ is closed and discrete iff $D$ intersects any compact in a finite set? –  Galoisfan Aug 8 '12 at 8:41
    
@ Galoisfan : I edited. There are other ways of showing that the reunion of two closed discrete sets is closed discrete, this was the first thing that popped in my mind. –  mercio Aug 8 '12 at 11:33
    
A perfect explanation, mercio, +1. Unfortunately many books and courses write discrete instead of the correct discrete and closed, for example in the context of the Mittag-Leffler or Weierstrass theorems. –  Georges Elencwajg Aug 8 '12 at 11:39
    
I think it depends on how you define discrete. I've always had it given as a countable set without limit points, which consequently makes it closed as well. –  Greg Zitelli Aug 8 '12 at 11:44
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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 $$r=\dfrac{f}{g}=\dfrac{\frac{f_1}{f_2}}{\frac{g_1}{g_2}}=\dfrac{f_1g_2}{f_2g_1}.$$

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Yes, this is true, but I'd like (if it is possible) an answer not involving the result that $\mathcal M (D)$ is the quotient field of $O(D)$. –  Galoisfan Aug 7 '12 at 19:58
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