Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Every closed subset of a manifold is the zero set of a smooth nonnegative function

Statement

Every closed subset A of a smooth manifold M is the zero set of some smooth nonnegative function g:M[0,).

Facts & Assumptions

Given: A closed subset A of a smooth manifold M.

[L1]

Every open cover of a manifold has a countable cover by relatively compact coordinate balls subordinate to it (Every open cover of a manifold has a countable relatively compact coordinate-ball subcover).

[L2]

A countable cover by coordinate balls with compact closures has a countable locally finite shrinking WkVk (A countable coordinate-ball cover has a countable locally finite shrinking).

[L3]

For every compact set inside an open set there is a smooth manifold bump equal to 1 near that compact set and supported in the open set (A manifold bump for a compact set inside an open set).

[L4]

A locally finite sum of smooth functions is smooth (A locally finite sum of smooth functions is smooth).

Proof

technique · direct
1.1

Apply [L1] to the one-set open cover {MA} of the open manifold MA to obtain a countable cover by coordinate balls with compact closures contained in MA. Then apply [L2] to obtain a countable locally finite shrinking WkVk of that cover. For each k, apply [L3] to WkVk to obtain a smooth function bk:M[0,1] that is positive on Wk and supported in MA.

L1L2L3givenchoose
2.1

The family (bk) is locally finite, so g:=k12kbk is smooth and nonnegative by [L4]. One has g=0 on A because every bk vanishes there, and g>0 on MA because each point there lies in some Wk.

L4step 1.1
3.1

Therefore A=g1(0).

step 2.1

Depends on

Used by

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