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Sard's famous theorem asserts that

Theorem. The set of critical values of a smooth function from a manifold to another has Lebesgue measure $0$.

I am asking for the curiosity that is it possible to find such a function whose set of critical values is

  1. Cantor set or
  2. Any other uncountably infinite set?
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    $\begingroup$ You might have looked at the first version of Sard's Theorem, which is more heavily cited. He published a follow-up paper where he gave an upper bound on the Hausdorff dimension. $\endgroup$
    – Ryan Budney
    Mar 24, 2020 at 17:19

3 Answers 3

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It is not hard to construct a smooth function $f$ on $\mathbb R$ such that $f \ge 0$ with $f(x) = 0$ if and only if $x$ is in the Cantor set $E$. If $F$ is an antiderivative of $f$, the critical values of $F$ will be an uncountably infinite perfect set.

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By a theorem of Whitney (easy in this $1$-dimensional case), any compact subset $K$ in the interval $I$ is the set of zeroes of a smooth ($C^\infty$) nonnegative function $f$.
As Robert said, take a primitive $F$. Provided that $K$ has no interior point, the critical values $F(K)$ of $F$ are homeomorphic with $K$.

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The paper "Hausdoff Measures and the Morse-Sard Theorem" by Moreira gives in section 4 a general construction of near-counterexamples involving generalized Cantor sets.

Reference

Carlos Gustavo T. de A. Moreira, "Hausdorff measures and the Morse-Sard theorem" (English), Publicacions Matemàtiques 45, No. 1, 149-162 (2001), MR1829581, Zbl 0995.58007.

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