Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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.

Sard on the nonflat critical strata

Statement

Let n1, let URm be open, let f:URn be Cr, and for j1 define

Cj:={xU:Dαf(x)=0 for every multi-index 1αj}.

If 1j<r, KCjCj+1 is compact, and the Morse-Sard conclusion is already known for Crj maps from open subsets of Rm1 to Rn, then f(K) is null.

Facts & Assumptions

Given: An integer n1, a Cr map f:URn, an integer 1j<r, and a compact set KCjCj+1.

[L1]

If a Euclidean map has invertible derivative at a point, it becomes a coordinate there after shrinking (The Euclidean inverse function theorem).

Proof

technique · direct
1.1

Fix xK. [L1, given, choose] Because xCj+1, some partial derivative of order j+1 of some component of f is nonzero at x. After reordering coordinates and components, choose a multi-index α with α=j and a component fμ such that, for

g:=Dαfμ,

one has g/x1(x)0. Since g is Crj, [L1] applied to

Φx(y):=(g(y),y2,,ym)

gives a neighbourhood Wx of x and a Crj diffeomorphism from Wx onto an open set Ix×ΩxR×Rm1.

L1givenchoose
2.1

Every point of KWx lies in Cj, so g vanishes there by definition of Cj. [step 1.1, algebra] Hence

Φx(KWx){0}×Ωx.

Define

hx:ΩxRn,hx(u):=f ⁣(Φx1(0,u)).

This map is Crj. If q=Φx1(0,u)KWx, then qCj and j1, so Dfq=0. Therefore

Dhx(u)=DfqD(Φx1{0}×Ωx)u=0,

so u is a critical point of hx. The induction hypothesis therefore gives that

f(KWx)=hx(Φx(KWx))

is a null subset of Rn.

step 1.1algebra
3.1

Finitely many neighbourhoods Wx cover the compact set K, so f(K) is a finite union of null sets and therefore null.

step 2.1givenchoose

Depends on

Used by

Dependency tree · two levels

12 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