Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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.

Basic weak neighborhoods

Statement

For a real or complex normed space X, a weak neighborhood base at xX consists of

U(x;f1,,fm;ε)={yX:fj(yx)<ε (1jm)},ε>0,fjX.

Here m=0 is allowed and gives X. The weak topology is a locally convex vector topology: addition and joint scalar multiplication are continuous, and the displayed zero neighborhoods are convex and balanced.

Facts & Assumptions

[F1]

Weak topology on a normed space defines the weak topology by inverse images of scalar open sets; finite intersections form a basis.

Proof

Given: X, x, a finite list of bounded scalar-linear functionals, and positive radii.

1.1

Every displayed U is a finite intersection of inverse images of open disks centered at fj(x), so is weakly open and contains x. Conversely, a finite subbasic intersection containing x contains inverse images of disks of radii rj>0 about fj(x); take ε=minjrj. With no conditions the intersection is X. Thus these sets form a neighborhood base.

F1given
2.1

Put p(v)=maxjfj(v), taking p=0 for an empty list. Scalar linearity and the triangle inequality give p(v+w)p(v)+p(w) and p(av)=ap(v). Consequently {p<ε} is balanced and real-convex. Moreover p((v+w)(v0+w0))<ε whenever p(vv0),p(ww0)<ε/2. This proves continuity of addition at every pair.

step 1.1algebra
3.1

At (a0,v0) write ava0v0=a(vv0)+(aa0)v0. Require aa0<min(1,ε/(2(p(v0)+1))) and p(vv0)<ε/(2(a0+1)). Then p(ava0v0)<ε. These are product neighborhoods and work also at a0=0 and v0=0. Thus scalar multiplication is jointly continuous, and the convex zero-neighborhood base proves local convexity.

step 2.1algebra

Depends on

Used by

Dependency tree · two levels

3 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