If $f$ is any real-valued function, we define its zero set $Z_f = \{ x : f(x) = 0 \}$. Obviously, the zero set of a nice function can be uncountable. e.g., if $f(x) = 0$ on an uncountable domain.

I would like a sufficient condition on functions $f : \mathbb R \to \mathbb R$ for which the following statement holds: $$\mbox{if $Z_f$ is uncountable, then it contains an interval}.$$

If $X_t$ denotes a Brownian motion, then with probability one, the zero set of $X_t$ is homeomorphic to a Cantor set (hence is uncountable but contains no interval). Since $X_t$ is $\tfrac{1}{2}$-Hölder continuous, this is obviously not sufficient.

Edit: Due to Joel David Hamkin's elegant counterexample below, continuous differentiability is not a sufficient condition for the above statement to hold. Is there a natural sufficient condition?

Edit 2: Thanks, all. I've accepted Joel's answer because it doesn't seem like there is a solution to my problem at this level of generality. The motivation for the question comes from stochastic geometry. I take a realization of a random Riemannian metric $g$ on the Euclidean plane, and consider a certain geodesic $\gamma$. Such a curve is (a.s.) smooth but certainly not analytic.

Given the random environment $g$, the path of the geodesic is determined. I then look at the intersection of the geodesic with a given line segment or circular arc. This intersection could be empty, finite, countably infinite, or uncountable. Under the hypotheses in my model, I have already shown that it cannot be an interval. I was hoping that a general argument would reduce other cases of uncountability to that case, proving that the intersection is countable. I may just have to deal with the possibility it can be uncountable, or find a context-specific argument.

  • $\begingroup$ What would make a sufficient condition 'natural'? $\endgroup$ Dec 21, 2011 at 0:30
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    $\begingroup$ $C^\infty$'' is not enough, as you remarked below. Analytic'' is enough, by well-known theorems in complex analysis. $\endgroup$
    – Goldstern
    Dec 21, 2011 at 1:01
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    $\begingroup$ ...and I guess in this case one can replace "uncountable" with "contains a convergent sequence". $\endgroup$ Dec 21, 2011 at 12:25

2 Answers 2


The distance function to a closed set is continuous, even Lipschitz continuous, and is zero exactly on that closed set. A modified version of this function can be made continuously differentiable, by smoothing out the kinks. In the case of the Cantor set, this provides a counterexample to your latter questions.

  • $\begingroup$ Hi Joel, thanks for the very elegant counterexample. $\endgroup$ Dec 20, 2011 at 22:22
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    $\begingroup$ It appears you can make it $C^\infty$ with suitable smoothing---the original distance function looks like a bunch of little triangles with slope $1$ resting on the omitted middle thirds of the Cantor set, and one needs only to smooth out the corners on these triangles. (And hey, I notice were both at NYU; I'm in the Philosophy department there this semester.) $\endgroup$ Dec 20, 2011 at 22:35
  • $\begingroup$ Yeah, as soon as you mentioned the counterexample I saw how it extends to $C^{\infty}$ for this reason. I had no idea you were at NYU this semester! I have a pretty busy week then I'm flying out Friday, but I'll be back at the beginning of next semester. You'll still be in NYC then I presume? Want to meet up for lunch on Wednesday January 25? $\endgroup$ Dec 20, 2011 at 23:05
  • $\begingroup$ Sure, I'll send you an email. $\endgroup$ Dec 20, 2011 at 23:08

I can't imagine how to sell the following as a natural condition; anyway: Let assume: $f:\mathbb{R}\to\mathbb{R}$ is quasi-analytic at every point but (possibly) countably many exceptional points. So, if $Z_f$ is uncountable, it has an accumulation point $a$ where $f$ is locally quasi-analitic, and since all derivatives of $f$ vanish at $a$, $f$ is locally zero there.


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