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The locally zero locus of a holomorphic function is clopen
Statement
For a holomorphic function on an open set , the set of points having a neighbourhood on which vanishes is both open and closed in .
More precisely, put Then and are open in (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).
Facts & Assumptions
Given: An open set , a holomorphic function , and the locally zero locus defined above.
If a holomorphic function has finite order at , then near it has the form with holomorphic and ; moreover, its order at is exactly when it vanishes on a neighbourhood of (The order of a zero is the exponent in its local holomorphic factorization).
A function complex differentiable at a point is continuous at that point (Complex differentiability at a point implies continuity there).
Proof
If , one of the neighbourhoods appearing in the definition of is contained in , so is open in ; this also covers .
Let . If , [L2] gives a neighbourhood on which is nonzero. If , then [L1] and make the order finite, so near with ; after shrinking by [L2], is nowhere zero there, and is the only zero. In either case a neighbourhood of contains no point of , so is open.
Thus is open and its complement in is open, so is both open and closed in .
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)