Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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  • 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):={aU: there is an open neighbourhood VU of a such that hV=0}. Then L(h) and UL(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 UC, a holomorphic function h:UC, 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 (zb)mg(z) with g holomorphic and g(b)0; moreover, its order at b is + 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
1.1

If aL(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)=.

given
1.2

Let bUL(h). If h(b)0, [L2] gives a neighbourhood on which h is nonzero. If h(b)=0, then [L1] and bL(h) make the order finite, so h(z)=(zb)mg(z) near b with g(b)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 UL(h) is open.

L1L2algebra
2.1

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

step 1.1step 1.2

Depends on

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