most recent 30 from http://mathoverflow.net 2013-05-19T16:03:44Z http://mathoverflow.net/feeds/question/24034 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function auniket 2010-05-09T17:57:46Z 2010-05-11T07:11:38Z

<p>More generally, can the zero set $V(f)$ of a continuous function $f : \mathbb{R} \to \mathbb{R}$ be nowhere dense and uncountable? What if $f$ is smooth?</p> <p>About two days ago I discovered that in this proof I am working on, I have implicitly assumed that $V(f)$ has to be countable if it is nowhere dense - hence this question.</p>

http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24036#24036 Robin Chapman 2010-05-09T18:22:36Z 2010-05-09T18:22:36Z

<p>It is a standard result that each closed subset of $\mathbb{R}^n$ is a zero set of some smooth function on $\mathbb{R}^n$. One proves this using smooth bump functions and partitions of unity.</p>

http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24037#24037 Roland van der Veen 2010-05-09T18:33:37Z 2010-05-09T19:20:15Z

<p>Here's a semi-explicit construction for a smooth function f that is zero precisely on the classical Cantor set. By this set I mean the one that is obtained from I_0 = [0,1] by repeatedly removing the middle third of any ensuing interval. So let's denote by I_n the n-th set in this process.</p> <p>Now let's make a smooth function f_n on [0,1] such that its zero set is exactly I_n. Starting with f_0 = 0 we obtain f_{n+1} from f_n as follows:</p> <p>set f_{n+1} = f_n on I_{n+1} and on an interval that is removed from I_n make f_{n+1} equal to a bump function that is 0 only at the boundary of the interval. We can choose the bump function to be of height 2^{-2^n}.</p> <p>This choice of heights of the bump functions will ensure that the derivatives of f all converge uniformly to their pointwise limits. Hence the limit function f_n is again smooth. By construction its zero set is exactly the Cantor set. </p>

http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24040#24040 Harald Hanche-Olsen 2010-05-09T19:20:42Z 2010-05-09T19:20:42Z

<p>I can't resist trying my hand at sketching a proof of the general result given in Robin Chapman's answer. Let $F\subset\mathbb{R}^n$ be any closed set. Let <code>$E_0=\{x\colon\operatorname{dist}(x,F)\ge1\}$</code>, and for $k=1,2,\ldots$ let <code>$E_k=\{x\colon2^{-k}\le\operatorname{dist}(x,F)\le2^{1-k}\}$</code>. Let $\omega$ be a standard mollifier, and put <code>$$f=\sum_{k=0}^\infty \alpha_k\chi_{E_k}*\omega_k,\qquad\omega_k(y)=2^{nk}\omega(2^ky),$$</code> where <code>$\alpha_k&gt;0$</code> decays fast enough so all derivatives converge uniformly ($\alpha_k=2^{-k^2}$ ought to be sufficient).</p>

http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24041#24041 André Henriques 2010-05-09T19:34:29Z 2010-05-09T19:34:29Z

<p>The continuous function is very easy to construct: it's the distance to the closed set.</p>

http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24044#24044 Henno Brandsma 2010-05-09T20:06:05Z 2010-05-09T20:06:05Z

<p>In a normal topological space, the zero-sets of continuous functions are precisely the closed $G_{\delta}$ sets. Hence in any metric space all closed sets are, including the Cantor set.</p>

http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24046#24046 Ian Morris 2010-05-09T20:22:42Z 2010-05-09T20:22:42Z

<p>To mention a further point not covered in existing answers: while any closed subset of $\mathbb{R}$ can be the zero set of a <em>smooth</em> function, the zero set of an <em>analytic</em> function either consists entirely of isolated points, or is all of $\mathbb{R}$. To see this we note that if the zero set of an analytic function $f$ contains an accumulation point, then by taking a power series expansion of $f$ at the accumulation point we may extend $f$ locally to a small complex disc around that point, and apply the Identity Theorem from complex analysis to show that $f$ is everywhere zero within that disc. In particular the zero set contains an open neighbourhood in $\mathbb{R}$ of the accumulation point, and using connectedness we can repeat this argument to show that the zero set must be all of $\mathbb{R}$.</p> <p><a href="http://en.wikipedia.org/wiki/Identity_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Identity_theorem</a></p>

http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24065#24065 Tom LaGatta 2010-05-10T06:18:25Z 2010-05-10T06:18:25Z

<p>Here's an answer from probability: a Brownian motion $B_t$ is a random, continuous function whose zero set is closed, nowhere dense, and has no isolated points. That is, <code>$\{t : B_t = 0 \}$</code> is almost-surely a topological Cantor set (see, for example, Section 8 of <a href="http://galton.uchicago.edu/~lalley/Courses/390/Lecture5.pdf" rel="nofollow">Lalley's lecture notes</a>).</p>

http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function/24119#24119 Roland Bacher 2010-05-10T15:41:25Z 2010-05-11T07:11:38Z

<p>André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where $\alpha$ is the supremum of all elements $&lt; x$ in $C$ (and $\alpha=-\infty$ if $C$ contains no elements which are $&lt; x$) and where similarly $\beta$ is the infimum of all elements $> x$ in $C$ (respectively $\beta=\infty$ if $C$ contains no elements $> x$). </p>