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 infinitely flat critical stratum

Statement

Let n1 and r1, let URm be open, and let f:URn be Cr with rnm. Let

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

If KCr is compact, then f(K) is null in Rn.

Facts & Assumptions

Given: Integers n,r1, a Cr map f:URn with rnm, and a compact set KCr.

[L1]

The multivariable Taylor formula with Lagrange remainder expresses the order-r remainder using the order-r derivatives at points of the joining segment (Multivariable Taylor formula with a Lagrange remainder along a line segment).

[L2]

A continuous map on a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[F1]

Euclidean nullity is proved by box covers of arbitrarily small total volume (Measure zero and content zero in Rm by countable and finite cube covers).

Proof

technique · direct
1.1

If m=0, then K has at most one point and f(K) is finite, hence null in Rn by cubes of arbitrarily small side. Assume henceforth that m1. For each xK, choose nested closed cubes QxQ^xU with xint(Qx) and Q^x having Qx in its interior. Compactness gives finitely many inner cubes Q1,,Qs whose interiors cover K. It suffices to prove that each f(KQi) is null, because finite unions of Euclidean null sets are null directly from the cube-cover definition [F1].

F1givenchoosecases
2.1

Fix one pair QQ^, let KQ:=KQ, and let λ be the side length of Q. The finitely many order-r partial derivatives of the components of f are uniformly continuous on the compact cube Q^ by [L2]. Since they vanish at every xKQ, applying [L1] componentwise with degree r1 shows that for every η>0 there is a single δ>0 such that f(y)f(x)ηyxr whenever xKQ, yQ, and yx<δ.

L1L2step 1.1algebra
3.1

Fix ε>0. If rn=m, choose η>0 so small that (2η)nmrn/2λrn<ε; if rn>m, choose any η>0. Let δ be furnished by step 2.1, and choose N so large that the congruent subcubes in the subdivision of Q into Nm cubes have diameter below δ. Use the following target cubes. [step 2.1, choose, cases] For each subcube Qν meeting KQ, choose a point xνKQQν. If yKQQν, then yxνmλ/N, so step 2.1 gives f(y)f(xν)η(mλN)r. Hence f(KQQν) lies in an n-cube of side length at most 2η(mλN)r.

step 2.1choosealgebra
4.1

The union of those target cubes covers f(KQ), and its total n-volume has the following bound. [F1, step 3.1, cases, algebra] It is at most Nm(2η(mλN)r)n=(2η)nmrn/2λrnNmrn. If rn>m, increase N until this quantity is below ε; if rn=m, the choice of η in step 3.1 already makes it smaller than ε. In either case the total covering volume is below ε. Therefore [F1] implies that f(KQ) is null.

F1step 3.1casesalgebra
5.1

Applying step 4.1 to the finite cover from step 1.1 shows that f(K) is null. Thus the infinitely flat critical stratum has null image.

F1step 1.1step 4.1

Depends on

Used by

Dependency tree · two levels

24 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