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One definition I have seen is that a Meromorphic function has at most a countable number of poles. Another says that a function f is Meromorphic if every point is either a pole or the function is analytic there.

Now first an easy question, but I am unsure of it:

1. That poles are isolated, that is usually taken as a definition of the pole?

Now come the hard question:

2. Can it be proved that if we use definition 2 of a Meromorphic function it can at most have a coutnable number of poles? If the answer to question 1 is yes, I guess we can assume for contradiction that it has more than a countable number of poles, and then show that it must have a limit point, and hence not be isolated? Is this hard to prove?

PS: I think an equivalent question for 2 in terms of $\mathbb{R}^2$ is that lets say that the set A is bigger than just beeing countable(I don't know if this implies if it is uncountable?). Then the set A must have a sequence of distinct points, with a limit point in $\mathbb{R}^2$. But I don't think the limit point must be in A? Is this something we can prove?

user119615
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2 Answers2

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Ad 1. Yes, it is part of the definition or a consequence of the definition that a pole is an isolated singularity of the function.

Ad 2. Yes, since poles are isolated singularities, a meromorphic function can have at most countably many poles. That is not hard to prove: Every compact subset $K$ of the domain of the function can only contain finitely many poles [otherwise the set of poles would have a limit point in $K$], and every open subset of $\mathbb{C}$ (or $\widehat{\mathbb{C}}$) is the union of countably many compact subsets (since it is locally compact and second countable), so the set of poles is a countable union of finite sets.

Daniel Fischer
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  • You are so smart, amazing! – user119615 Oct 07 '14 at 14:43
  • Just curious, what kind of singularity is a branch cut? ( like, e.g., the negative Real axis for Logz ) – gary Jul 13 '17 at 16:28
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    @gary A non-isolated singularity. Some people count non-isolated singularities as essential singularities, others reserve the term "essential singularity" for isolated singularities that are neither removable nor poles. – Daniel Fischer Jul 13 '17 at 17:48
  • Please take a look at my related question: https://math.stackexchange.com/q/3991330/47771. I don't get why a meromorphic function can contain only finitely many poles inside a ball (or why the claim is only true for a "small enough" ball). – 0xbadf00d Jan 20 '21 at 06:13
  • @0xbadf00d An infinite subset $A$ of a compact set $K$ has an accumulation point in $K$, i.e. there is a point $z \in K$ such that every neighbourhood of $z$ contains infinitely many points of $K$. The set of poles of a meromorphic function on $U$ is, by definition, a discrete and closed subset of $U$, hence it has no accumulation points in $U$. Thus every compact subset of $U$ can contain only finitely many poles. – Daniel Fischer Jan 20 '21 at 11:32
  • @DanielFischer Thank you for reminding me on the fact about infinite subsets, but please note that I'm using a different definition of "meromorphic" in that question. The conclusion should still hold. Do you know how we need to argue? – 0xbadf00d Jan 20 '21 at 18:19
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Poles are by definition isolated singularities, so there is an open disk of positive radius around each of them in which no other singularity occurs. This is sufficient to prove there are at most countably many poles in the complex plane (or in some open domain in the complex plane). This can be seen as a corollary to no uncountable sum of positive real numbers (the areas of these disks) can be finite.

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hardmath
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  • Thanks, but even though we only have a countable number, their sum won't necessarily be finite either? – user119615 Oct 07 '14 at 14:46
  • Their sum won't necessarily be finite (we have the whole complex plane to work with), but if you have uncountably many points in the plane, some bounded subset of the plane would already have uncountably many of these points. Consider the nested union of disks of radius $r=1,2,...$. If each contains only countably many of the points, their nested union (the entire complex plane) would only have countably many. – hardmath Oct 07 '14 at 14:50
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    I think I understand. So if we had an uncountable number of poles, then atleast one of the discs you mention would have a an uncountable number of poles. Around each of these poles there is an open disc, since we have an uncountable sum, the area of these discs must be infinity, but that is a contradiction since we are in a big disc with finite area? – user119615 Oct 07 '14 at 14:53
  • Yes, you've got it. – hardmath Oct 07 '14 at 14:56
  • Thanks, that was also a very good solution! – user119615 Oct 07 '14 at 14:58
  • Why exists a big disc with finite area such that all ball are contained in it? I don't see this – eraldcoil Jul 28 '20 at 16:29
  • @eraldcoil: Suppose there are uncountably many poles of a meromorphic function. Even a pole "at infinity" needs to be an isolated singularity. It follows that a bounded disk, complementary to one "at infinity" that isolates a pole there (if any), would contain an uncountable number of these poles. The sum of areas in disjoint subdisks that isolate those remaining poles is finite, so there can be at most countably many (as the partial sums of any uncountable set of positive numbers are unbounded). Does that help? If still unclear I will amend my Answer.. – hardmath Jul 28 '20 at 17:27