Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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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1/sin(1/z) has a nonisolated singularity at 0

Statement refuted

Refuted claim: every singularity at a point of a complex function is automatically isolated.

The witness is

f(z)=1sin(1/z)

at a=0. The function has poles at points tending to 0, so the singularity at 0 is not isolated.

Facts & Assumptions

Given: The function f(z)=1/sin(1/z).

[L1]

The zeros of sine are exactly the integral multiples of π (The zero sets of sine and cosine and the least positive common period 2 pi).

Counterexample

technique · direct
1.1

For every positive integer n, the point zn:=1/(nπ) satisfies sin(1/zn)=sin(nπ)=0 by [L1], so f has a pole at each zn.

L1algebra
2.1

The sequence (zn) tends to 0, so every punctured neighbourhood of 0 contains some pole of f. Hence no punctured neighbourhood of 0 is a region on which f is holomorphic, and the singularity at 0 is not isolated.

step 1.1algebra

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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