Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A function on a subset of Rm is continuous at x iff its oscillation there is 0, and every oscillation superlevel set is closed

Statement

For f:A→R, f is continuous at c∈A if and only if ωf(c)=0. If f is bounded, then for every ε>0, the relative superlevel set {c∈A:ωf(c)≥ε} is closed in A.

Proof

technique · direct
1.1L1L2

If f is continuous at c, choose a ball on which ∣f(x)−f(c)∣<ε/3; pairwise differences are then below 2ε/3, so the ball oscillation is at most 2ε/3<ε and ωf(c)=0.

1.2L1L2

If ωf(c)=0, choose r with ball oscillation below ε. Holding one point at c gives ∣f(x)−f(c)∣<ε, proving continuity.

1.3L1L2given

If ωf(c)<ε, choose r with ωf(A∩B(c,r))<ε. Every d∈A∩B(c,r/2) has a sufficiently small ball contained in B(c,r), so ωf(d)<ε. Thus the sublevel set is relatively open.

2.1step 1.1step 1.2step 1.3given∎

Steps 1.1 and 1.2 give the equivalence; step 1.3 makes the complementary superlevel set closed.

Depends on

Dependency tree · 2 levels

Sources