CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Regular values have null complement and are dense

Statement

For a smooth map $F : M \to N$ , the complement of the regular values is a null subset of $N$ . In particular, the regular values are dense in $N$ .

Facts & Assumptions

Given: A smooth map $F : M \to N$ .

[F1]

The complement of the regular values is the critical value set (The critical locus and critical value set).

[L1]

The critical value set is null (Morse-Sard for smooth manifolds).

[L2]

In a positive-dimensional manifold, a null set has dense complement (A null set has dense complement in a positive-dimensional manifold).

Proof

technique · direct

1.1

By [F1] and [L1], the complement of the regular values is null in $N$ .

F1L1given

2.1

If $\dim N > 0$ , [L2] implies that the complement of that null set is dense. If $\dim N = 0$ , then $N$ is discrete and every value is regular because the target tangent spaces are zero, so the regular-value set is all of $N$ and is certainly dense.

L2step 1.1cases

3.1

Therefore regular values have null complement and are dense.

step 1.1step 2.1∎

Depends on

Used by

Regular values form a dense G_δ set Corollary

Dependency tree · two levels

8 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

Encyclopedia of Mathematics, Sard theorem (standard reference, not scraped)