I know that this question has been asked a few times before, but somehow I cannot find or come up with a satisfying answer. Consider the residue theorem:

Let $D \subseteq \mathbb{C}$ be open and $f$ be holomorphic except isolated singularities. Then for any cycle $\Gamma$ which is homologous to zero holds $$\int_\Gamma f(\zeta)d\zeta = 2\pi i \sum_{z \in D}n(\Gamma,z)\mathrm{res}_zf$$

The set where $n(\Gamma,z) \neq 0$ is relatively compact as the remark in the book I use states. By this, we have that only a finite number of singularities occur on the right hand side.

I do not quite see why the set should be relatively compact and why then only a finite number of singularities lies in this set.

  • $\begingroup$ The usual answer is something like the singularities of a meromorphic function form a discrete subspace (which can be infinite), and the intersection of a discrete subspace with a compact subspace is finite. $\endgroup$
    – sharding4
    Jun 19, 2017 at 13:27
  • $\begingroup$ The point is that because if you had even countably infinite singularities, because you are in a bounded sunset of $\mathbb{C}$, it will have an accumulation point. This will lead to a contradiction of your statement $\endgroup$
    – user456218
    Jun 24, 2017 at 15:56

1 Answer 1


The trace $\operatorname{Tr} \Gamma$ of $\Gamma$ is a compact subset of $D$. The winding number $n(\Gamma,z)$ is constant on every component of $U := \mathbb{C}\setminus \operatorname{Tr} \Gamma$, and it is zero on the unbounded component of $U$. Thus

$$A := \{ z \in U : n(\Gamma,z) \neq 0\}$$

is a bounded set, hence relatively compact in $\mathbb{C}$. And $\overline{A} \subset A \cup \operatorname{Tr} \Gamma$. That is true for all cycles.

Now we use the assumption that $\Gamma$ is homologous to zero, which is another way to say $A \subset D$. But then $K := A \cup \operatorname{Tr} \Gamma$ is a compact subset of $D$, so $A$ is relatively compact in $D$.

And a compact $C \subset D$ can contain only finitely many singularities of $f$, since $f$ has only isolated singularities in $D$. By compactness, every infinite subset of $C$ has an accumulation point in $C$. If $C$ contained infinitely many singularities of $f$, then those would accumulate at a point $p \in C$. But then $p$ would be a non-isolated singularity of $f$, contradicting the assumption that all singularities of $f$ in $D$ are isolated.

  • $\begingroup$ @ Daniel Fisher If I understand well the assumption in the theorem is $B'\cap D=\emptyset$, where $B$ is the set of singularities of $f$ ($B'$ is just the set of accumulation points). I have seen versions of the theorem indicating that $B$ is a discrete set (i.e., $B$ has only isolated points). But, in this case, your proof must be changed. I am missing something? $\endgroup$
    – aly
    Dec 22, 2018 at 23:30
  • $\begingroup$ @aly I'm not quite sure how exactly such a version would look like. If we denote the set of non-removable singularities of $f$ in $D$ by $N$, then it is the same to say that $N' \cap D = \varnothing$ or that $N$ is a discrete set. For if $z \in D$ were an accumulation point of $N$, then $z$ would itself be a non-removable singularity. $\endgroup$ Oct 3, 2019 at 0:06

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