Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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.

Evaluation of characters is jointly continuous

Facts & Assumptions

[F1]

G is a topological group whose translations and inversion are homeomorphisms, and G is locally compact: every point has a compact neighbourhood. (Left and right translations and inversion in a topological group are homeomorphisms, Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space)

[F3]

The dual consists of the continuous homomorphisms γ:G→T, with the compact-open subbasis S(K,V)={γ:γ[K]⊆V} for compact K⊆G and open V⊆T; every such set is open in the subspace topology and contains every character mapping K into V. (The Pontryagin dual with the compact-open topology, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace)

[F5]

For all z,w∈T: ∣z+w∣≤∣z∣+∣w∣, ∣zw∣=∣z∣∣w∣, z−1 has modulus 1, and ∣z−1−w−1∣=∣z−w∣; in particular every γ(x) and every γ0(x0) has modulus 1. (The multiplicative unit circle is a compact metrizable topological abelian group, Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive)

Proof

Given: A locally compact Hausdorff abelian group G, a character γ0∈G^, a point x0∈G, and ε>0.

1.1F1F2F3

Choose an open neighbourhood U of x0 with γ0[U]⊆B(γ0(x0),ε/2), possible because γ0 is continuous at x0 by [F3]; by [F2] choose an open V with x0∈V⊆V‾⊆U and K:=V‾ compact. Then K is a compact neighbourhood of x0 with γ0[K]⊆B(γ0(x0),ε/2).

2.1step 1.1F1F3F4F5

The translate K−x0={x−x0:x∈K} is compact and is a neighbourhood of 0: it is the image of K under the homeomorphism x↦x−x0 of [F1, F4], and it contains the open translate V−x0 of V, which contains 0. Moreover γ0[K−x0]⊆B(1,ε/2), because for x∈K⊆U one has γ0(x−x0)=γ0(x)γ0(x0)−1 and ∣γ0(x)γ0(x0)−1−1∣=∣γ0(x)−γ0(x0)∣<ε/2 by [F5].

3.1step 2.1F3F5

Let γ∈S({x0},B(γ0(x0),ε/2))∩S(K−x0,B(1,ε/2)) and x∈K. Since γ is a homomorphism, γ(x)=γ(x0)γ(x−x0), so γ(x)−γ0(x0)=(γ(x0)−γ0(x0))γ(x−x0)+γ0(x0)(γ(x−x0)−1) and hence ∣γ(x)−γ0(x0)∣≤∣γ(x0)−γ0(x0)∣+∣γ(x−x0)−1∣<ε/2+ε/2=ε by [F5] and the choice of γ.

4.1step 1.1step 2.1step 3.1F2F3∎

The set W:=(S({x0},B(γ0(x0),ε/2))∩S(K−x0,B(1,ε/2)))×V is a neighbourhood of (γ0,x0) in G^×G: it is a product of an open set containing γ0 and an open set containing x0, the first because γ0(x0)∈B(γ0(x0),ε/2) and γ0[K−x0]⊆B(1,ε/2) by step 2.1, the second because V is open and x0∈V⊆K. By step 3.1 the pairing maps W into B(γ0(x0),ε); since open balls form a neighbourhood base at γ0(x0), the pairing is continuous at the arbitrary point (γ0,x0), hence continuous.

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Sources