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In complex analysis, meromorphic functions are simply functions defined on an open subset $U\subseteq\mathbb{C}$ that are holomorphic in all of $U$, except for a set of isolated points, or poles.

I want to understand more rigorously this set of poles.

I mean, Lang, in his book Complex Analysis, says that such a set of poles has to be "discrete".

I get the intuitive meaning of that, but I want to know exactly what is meant.

Discreteness here is in the topological sense? i.e., such that every point of our set of poles can be separated by disjoint open neighbourhoods? (Much like the Hausdorff condition.)

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  • $\begingroup$ In this case it's equivalent to the Hausdorff condition. The term comes from topology: The condition is that the induced (subspace) topology on the set of poles is the discrete topology. $\endgroup$ – Travis Willse Apr 14 '16 at 12:42
  • $\begingroup$ Oh thanks. This makes so much sense now. I always thought that it was a bad thing in topology that we had the word 'discrete' for these two diferent notions, but now I see how connected they are $\endgroup$ – Shoutre Apr 14 '16 at 12:44
  • $\begingroup$ You're welcome, I'm glad you found the answer useful. Since it basically contains the answer to your question, I'll submit my comment as an answer now. $\endgroup$ – Travis Willse Apr 14 '16 at 12:48
  • $\begingroup$ By the way, your definition of meromorphic is not quite correct. For example, $z \mapsto \exp\left(\frac{1}{z}\right)$ is a defined on all of $\Bbb C$ except for the single point $z = 0$, but this point is an essential singularity, not a pole. $\endgroup$ – Travis Willse Apr 14 '16 at 12:50
  • $\begingroup$ Right. I'll read more about it. Thanks. $\endgroup$ – Shoutre Apr 14 '16 at 12:51
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In this setting discrete is equivalent to the Hausdorff condition. The term comes from topology: The condition is that the induced (subspace) topology on the set of poles is the discrete topology.

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