On the uncountability of zero sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:16:24Z http://mathoverflow.net/feeds/question/83970 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83970/on-the-uncountability-of-zero-sets On the uncountability of zero sets Tom LaGatta 2011-12-20T22:12:20Z 2011-12-21T15:20:56Z <p>If $f$ is any real-valued function, we define its zero set <code>$Z_f = \{ x : f(x) = 0 \}$</code>. Obviously, the zero set of a nice function can be uncountable. e.g., if $f(x) = 0$ on an uncountable domain. </p> <p>I would like a sufficient condition on functions $f : \mathbb R \to \mathbb R$ for which the following statement holds: $$\mbox{if Z_f is uncountable, then it contains an interval}.$$</p> <p>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 <code>$\tfrac{1}{2}$</code>-Hölder continuous, this is obviously not sufficient.</p> <p><b>Edit:</b> 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?</p> <p><b>Edit 2:</b> 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.</p> <p>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.</p> http://mathoverflow.net/questions/83970/on-the-uncountability-of-zero-sets/83971#83971 Answer by Joel David Hamkins for On the uncountability of zero sets Joel David Hamkins 2011-12-20T22:17:39Z 2011-12-20T22:17:39Z <p>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. </p> http://mathoverflow.net/questions/83970/on-the-uncountability-of-zero-sets/83996#83996 Answer by Pietro Majer for On the uncountability of zero sets Pietro Majer 2011-12-21T10:04:12Z 2011-12-21T10:04:12Z <p>I can't imagine how to sell the following as a natural condition; anyway: Let assume: $f:\mathbb{R}\to\mathbb{R}$ is <a href="http://en.wikipedia.org/wiki/Quasi-analytic_function" rel="nofollow">quasi-analytic</a> 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.</p>