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?
  • 5
    $\begingroup$ 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 '20 at 17:19

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.


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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