Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A divergent probability sum does not force infinitely many occurrences without independence

Statement refuted

If (An)nN is a sequence of events with n=0P(An)=+, then P(An i.o.)=1.

Facts & Assumptions

Given: An event A with 0<P(A)<1, and define An:=A for every nN.

[L1]

The event An i.o. is the event that infinitely many of the An occur. (Limsup and the infinitely often event)

[L2]

Probability measures respect complements and monotone set identities. (Basic identities for a probability measure)

Counterexample

technique · constructive
1.1

Since every An equals A, one has n=0P(An)=n=0P(A)=+.

givenalgebraconstruct
1.2

For every outcome ω, either ωA and then ωAn for all n, or ωA and then ωAn for all n. Hence the infinitely-often event is exactly A itself: An i.o.=A.

L1
2.1

Therefore P(An i.o.)=P(A), and the chosen hypothesis 0<P(A)<1 makes this probability strictly between 0 and 1. So the displayed implication is false without an independence hypothesis.

step 1.1step 1.2L2discharge-construct

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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