On the uncountability of zero sets

2011-12-20T22:12:20Z

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.

Answer by Joel David Hamkins for On the uncountability of zero sets

2011-12-20T22:17:39Z

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.

Answer by Pietro Majer for On the uncountability of zero sets

2011-12-21T10:04:12Z

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.

On the uncountability of zero sets - MathOverflow

On the uncountability of zero sets

Answer by Joel David Hamkins for On the uncountability of zero sets

Answer by Pietro Majer for On the uncountability of zero sets