TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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Poles of a meromorphic function form a closed discrete set and are at most countable

Statement

Let $f$ be meromorphic on a plane domain $Ω$ , and let $P \subseteq Ω$ be its pole set. Then:

every $a \in P$ has a neighbourhood in $Ω$ containing no other pole, so $P$ is discrete in $Ω$ ;
$Ω ∖ P$ is open, so $P$ is closed in $Ω$ ;
$P$ is at most countable.

Facts & Assumptions

Given: A meromorphic function $f : Ω ∖ P \to C$ on a nonempty connected open set $Ω$ .

[L1]

By definition, every point of $P$ is a pole, and $f$ is holomorphic on $Ω ∖ P$ (Meromorphic functions on a plane domain, Isolated singularities: removable, poles, and essential singularities).

[L2]

The rationals are countable, the product of two at most countable sets is at most countable, a subset of an at most countable set is at most countable, and every nonempty at most countable set admits a surjection from $N$ whose least-hit map gives an injection into $N$ ( $Q$ is countably infinite, A product of two at most countable sets is at most countable, Every subset of an at most countable set is at most countable, A nonempty set is at most countable iff it is a surjective image of $N$ ).

[L3]

Between any two real numbers lies a rational (The rationals embed densely in the reals).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

Proof

technique · direct

1.1L1

Fix $a \in P$ . By [L1], $a$ is a pole, so some radius $r_{a} > 0$ has $∣ z - a ∣ < r_{a}$ contained in $Ω$ and $f$ holomorphic on $0 < ∣ z - a ∣ < r_{a}$ . If $b \in P$ and $0 < ∣ b - a ∣ < r_{a}$ , then $b$ lies in a region where $f$ is holomorphic, contradicting $b \in P$ . Thus $B (a, r_{a})$ contains no pole other than $a$ .

1.2L2

Let $D$ be the family of discs $B (p + i q, s)$ with $p, q, s \in Q$ and $s > 0$ . By [L2], $Q^{3}$ is at most countable, so $D$ is at most countable and admits an injection $e : D \to N$ .

2.1step 1.1L1

Step 1.1 proves that $P$ is discrete in $Ω$ . If $c \in Ω ∖ P$ , then [L1] says $f$ is holomorphic on a neighbourhood of $c$ , and that neighbourhood contains no point of $P$ ; hence $Ω ∖ P$ is open and $P$ is closed in $Ω$ .

2.2step 1.1L3choose

For each $a \in P$ , step 1.1 gives $r_{a} > 0$ . Write $a = x + i y$ . By [L3], choose rationals $p, q$ with $∣ x - p ∣ < r_{a} / 8$ and $∣ y - q ∣ < r_{a} / 8$ , so $∣ a - (p + i q) ∣ < r_{a} / 4$ ; choose a rational $s$ with $∣ a - (p + i q) ∣ < s < r_{a} / 2$ , again by [L3]. Then $a \in B (p + i q, s) \subseteq B (a, r_{a})$ , so the set of discs in $D$ containing $a$ and contained in $B (a, r_{a})$ is nonempty.

3.1step 1.1step 1.2step 2.2L4

For each $a \in P$ , the set $E_{a} : = {e (D) : D \in D, a \in D \subseteq B (a, r_{a})}$ is nonempty by step 2.2, so [L4] gives its least element; call it $j (a)$ . If $j (a) = j (b)$ , then injectivity of $e$ makes the corresponding discs equal, and that disc lies inside $B (a, r_{a})$ and contains both $a$ and $b$ , so step 1.1 forces $a = b$ . Therefore $a \mapsto j (a)$ is injective from $P$ into $N$ .

4.1step 2.1step 1.2step 3.1L2∎

The injection of step 3.1 makes $P$ at most countable, completing the proof.

Remarks

The pole set need not be closed in all of $C$ when $Ω \neq C$ : it may accumulate at boundary points of the domain. The theorem says precisely that no accumulation can happen inside $Ω$ .

Poles of a meromorphic function form a closed discrete set and are at most countable

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources