8
$\begingroup$

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?

$\endgroup$

2 Answers 2

Reset to default
11
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ You are so smart, amazing! $\endgroup$
    – user119615
    Commented Oct 7, 2014 at 14:43
  • $\begingroup$ Just curious, what kind of singularity is a branch cut? ( like, e.g., the negative Real axis for Logz ) $\endgroup$
    – gary
    Commented Jul 13, 2017 at 16:28
  • 1
    $\begingroup$ @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. $\endgroup$
    – Daniel Fischer
    Commented Jul 13, 2017 at 17:48
  • $\begingroup$ Please take a look at my related question: 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). $\endgroup$
    – 0xbadf00d
    Commented Jan 20, 2021 at 6:13
  • $\begingroup$ @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. $\endgroup$
    – Daniel Fischer
    Commented Jan 20, 2021 at 11:32
3
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Thanks, but even though we only have a countable number, their sum won't necessarily be finite either? $\endgroup$
    – user119615
    Commented Oct 7, 2014 at 14:46
  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Commented Oct 7, 2014 at 14:50
  • 1
    $\begingroup$ 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? $\endgroup$
    – user119615
    Commented Oct 7, 2014 at 14:53
  • $\begingroup$ Yes, you've got it. $\endgroup$
    – hardmath
    Commented Oct 7, 2014 at 14:56
  • $\begingroup$ Thanks, that was also a very good solution! $\endgroup$
    – user119615
    Commented Oct 7, 2014 at 14:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .