34
$\begingroup$

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?

Some 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.

$\endgroup$
32
$\begingroup$

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.

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:

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 $2^{-2^n}$.

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.

$\endgroup$
  • $\begingroup$ If you work through the details, I suspect you may find you need the bumps at step $n$ to have heights decreasing faster than $2^{-n}$. The reason is that the bump has to fit into an interval of length $3^{-n}$, and the resulting squeeze makes the derivatives big (relatively speaking). $\endgroup$ – Harald Hanche-Olsen May 9 '10 at 19:03
  • $\begingroup$ @Harald: Thanks you're absolutely right. I meant to say 2^{-2^n}. That should work. $\endgroup$ – Roland van der Veen May 9 '10 at 19:21
  • 2
    $\begingroup$ Each bump in the bump function has width 1/3^n. If the height is h(n), then the largest value of the kth derivative of of the stage n bumps will be 3^{kn}h(n) times the largest value of the kth derivative of the original bump function. If you want these to converge uniformly, the exact condition you need is that 3^{kn}h(n) converges to 0. So the necessary and sufficient condition on h to make the construction work is that for all a>0, h(n) is eventually less than a^n. So yes, 2^{-2^n} will work. $\endgroup$ – skeptical scientist May 16 '11 at 18:24
176
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Damn! My stupidity is now all the more evident. I wish it was possible to choose two answers! $\endgroup$ – auniket May 9 '10 at 23:34
  • 1
    $\begingroup$ Deserves to be "the answer"! Works in any metric space, smooth ones can be attained by mollifying these. $\endgroup$ – Behnam Esmayli Mar 31 '18 at 19:06
29
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I admit I have never worked through the proof of this result. So I wonder, why partitions of unity? It seems to me that the local result is no easier than the global result. $\endgroup$ – Harald Hanche-Olsen May 9 '10 at 19:06
  • 1
    $\begingroup$ It's true on any (paracompact) smooth manifold - you certainly need partitions of unity for that. $\endgroup$ – Robin Chapman May 9 '10 at 19:10
  • $\begingroup$ Yes, that makes sense. $\endgroup$ – Harald Hanche-Olsen May 9 '10 at 19:21
  • $\begingroup$ One can avoid using the partition of unity for that, just summation of "1-bump" functions with suitable coefficients.. $\endgroup$ – Petya May 10 '10 at 3:22
  • $\begingroup$ Petya, I'm sure you're correct, but it's a bit more fiddly to ensure the uniform convergence of the sum of the derivatives of the bump functions without a convenient global coordinate system. $\endgroup$ – Robin Chapman May 10 '10 at 9:07
19
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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 $E_0=\{x\colon\operatorname{dist}(x,F)\ge1\}$, and for $k=1,2,\ldots$ let $E_k=\{x\colon2^{-k}\le\operatorname{dist}(x,F)\le2^{1-k}\}$. Let $\omega$ be a standard mollifier, and put $$f=\sum_{k=0}^\infty \alpha_k\chi_{E_k}*\omega_k,\qquad\omega_k(y)=2^{nk}\omega(2^ky),$$ where $\alpha_k>0$ decays fast enough so all derivatives converge uniformly ($\alpha_k=2^{-k^2}$ ought to be sufficient).

$\endgroup$
7
$\begingroup$

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, $\{t : B_t = 0 \}$ is almost-surely a topological Cantor set (see, for example, Section 8 of Lalley's lecture notes).

$\endgroup$
4
$\begingroup$

To mention a further point not covered in existing answers: while any closed subset of $\mathbb{R}$ can be the zero set of a smooth function, the zero set of an analytic 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 $\mathbb{R}$.

http://en.wikipedia.org/wiki/Identity_theorem

$\endgroup$
4
$\begingroup$

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 $< x$ in $C$ (and $\alpha=-\infty$ if $C$ contains no elements which are $< x$) and where similarly $\beta$ is the infimum of all elements $> x$ in $C$ (respectively $\beta=\infty$ if $C$ contains no elements $> x$).

$\endgroup$
  • $\begingroup$ This is not C^inf. Example: take the closed set to be {-1,1}, then your function is not C^1 at 0. $\endgroup$ – André Henriques May 10 '10 at 16:47
  • $\begingroup$ Of course. I have fixed this stupid error. $\endgroup$ – Roland Bacher May 10 '10 at 17:54
  • $\begingroup$ Of course, a closed set need not have an upper bouned Also what is $\beta$? $\endgroup$ – Robin Chapman May 10 '10 at 18:30
  • $\begingroup$ Sorry, part of the correction got lost du to a missing space between $<$ and $a$. $\endgroup$ – Roland Bacher May 11 '10 at 7:13
  • $\begingroup$ So we back in one dimension again. $\endgroup$ – Robin Chapman May 11 '10 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.