False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)

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The critical-value set need not be closed

Statement

False claim: the critical value set of a smooth map is always closed.

Facts & Assumptions

Given: A smooth bump $β : R \to R$ supported in $[- 1, 1]$ with $β (0) = 1$ , $β^{'} (0) = 0$ , and $0 < β (t) < 1$ for $0 < ∣ t ∣ < 1$ , together with the smooth map $f (x) : = \sum_{k = 1}^{\infty} (1 - \frac{1}{k}) β (2^{k} (x - k)) .$

Refutation

technique · direct

1.1givenalgebra

The supports of the summands are pairwise disjoint, so the series defines a smooth function. At each center $x = k$ , the derivative is zero and $f (k) = 1 - \frac{1}{k} .$ Thus every value $1 - \frac{1}{k}$ is a critical value.

2.1step 1.1algebra

The sequence $1 - \frac{1}{k}$ converges to $1$ . But $f (x) \neq 1$ for every $x$ , because every summand has height strictly below $1$ and outside the supports the function is $0$ . Hence $1$ is a regular value with empty fibre, not a critical value.

3.1step 2.1∎

Therefore the critical value set contains ${1 - \frac{1}{k} : k \geq 1}$ but not its limit $1$ , so it is not closed.

Used by

A smooth map with a nonclosed critical-value set Example

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Marco Gualtieri, Topology I: Smooth Manifolds, cumulative notes (standard reference, not scraped)