LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The locally zero locus of a holomorphic function is clopen

Statement

For a holomorphic function $h$ on an open set $U$ , the set of points having a neighbourhood on which $h$ vanishes is both open and closed in $U$ .

More precisely, put $L (h) : = {a \in U : there is an open neighbourhood V \subseteq U of a such that h ∣_{V} = 0} .$ Then $L (h)$ and $U ∖ L (h)$ are open in $U$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

Facts & Assumptions

Given: An open set $U \subseteq C$ , a holomorphic function $h : U \to C$ , and the locally zero locus $L (h)$ defined above.

[L1]

If a holomorphic function has finite order $m$ at $b$ , then near $b$ it has the form $(z - b)^{m} g (z)$ with $g$ holomorphic and $g (b) \neq 0$ ; moreover, its order at $b$ is $+ \infty$ exactly when it vanishes on a neighbourhood of $b$ (The order of a zero is the exponent in its local holomorphic factorization).

[L2]

A function complex differentiable at a point is continuous at that point (Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1given

If $a \in L (h)$ , one of the neighbourhoods appearing in the definition of $L (h)$ is contained in $L (h)$ , so $L (h)$ is open in $U$ ; this also covers $L (h) = \emptyset$ .

1.2L1L2algebra

Let $b \in U ∖ L (h)$ . If $h (b) \neq 0$ , [L2] gives a neighbourhood on which $h$ is nonzero. If $h (b) = 0$ , then [L1] and $b \notin L (h)$ make the order finite, so $h (z) = (z - b)^{m} g (z)$ near $b$ with $g (b) \neq 0$ ; after shrinking by [L2], $g$ is nowhere zero there, and $b$ is the only zero. In either case a neighbourhood of $b$ contains no point of $L (h)$ , so $U ∖ L (h)$ is open.

2.1step 1.1step 1.2∎

Thus $L (h)$ is open and its complement in $U$ is open, so $L (h)$ is both open and closed in $U$ .

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

B. V. Shabat, Introduction to Complex Analysis, Theorem 2.28 (standard reference, not scraped)