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I am reading about meromorphic functions and it stated that a function that is meromorphic has a set $\{a_0, a_1, a_2, ...\}$ with poles at those points which have no limit point. My question is, if this set is infinite, is it true that $|a_n| \rightarrow \infty$ as $n \rightarrow \infty?$

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    $\begingroup$ If $|a_n| \not\rightarrow \infty$ then the given sequence would contain an infinite bounded subsequence, which in turn would have at least one limit point. $\endgroup$ – Mirko Jun 10 '17 at 21:52
  • $\begingroup$ The question doesn't make a lot of sense the way it is now... you seem to be assuming that the set of poles comes with/is an an enumeration. But it is not - it is a mere set without any natural kind of ordering, and not a sequence. $\endgroup$ – polynomial_donut Jun 10 '17 at 21:53
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Of course. By the Bolzano-Weierstrass theorem, any bounded sequence has a converging subsequence. For a nice combinatorial proof of this classical result, see here.

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The wording is a little confusing, at least for me. I suppose you meant to say that since a meromorphic function only has isolated poles, it can't be that if it has infinite poles which all of them are bounded, since then this infinite set of poles has a limit point, which cannot be.

Thus the answer is yes: $\;|a_n|\xrightarrow[n\to\infty]{}\infty\;$

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