For finding counter examples. That does not sound convincing enough, at least not always. Why as a object in its own right the study of Cantor Set has merit ?

  • 3
    $\begingroup$ Why are you not convinced that counterexamples are an important thing to study? (Re: "at least not always", the beauty of a counterexample is that only needs to be used once, in a particular key spot, to change our understanding forever.) $\endgroup$ Sep 30 '18 at 1:07
  1. in beginning real analysis: to counter the naive notion that a "closed set" is a union of closed intervals, plus a few single points.

  2. In beginning Lebesgue measure: the easiest example of an uncountable set of measure zero

  3. In general topology: sets homeomorphic to the Cantor set are useful in proofs

  4. Fractal geometry: many fractals are homeomorphic to the Cantor set. Mandelbrot calls such a thing a Cantor dust to suggest its appearance.

  • $\begingroup$ Can one claim and substantiate that the study of cantor set is something fundamental and not just a source of ready counterexample ? $\endgroup$
    – Vagabond
    Sep 29 '18 at 15:21
  • $\begingroup$ Point (4) seems to address this. The Cantor set is one of the simplest examples of a fractal, and is therefore studied in detail by students who are first learning fractal geometry or analysis (to the point that a colleague of mine, after writing a masters thesis on Cantor-like sets, started her PhD by stating that she never wanted to see the Cantor set again). $\endgroup$ Sep 30 '18 at 14:51

The Cantor set is quite useful in its own right. My preferred way to think of the Cantor set is as "the most general compact metrizable space." That it is the most general such space means that it is often good for counterexamples because it lacks the particulars. At the same time, one can construct things by specialization.

The formal expression of the Cantor set being the most general compact metrizable space is that every compact metrizable space is the image of the Cantor set under a continuous function. One can use this as a method of proof. For example, to show that a continuous surjection from $[0,1]$ onto $[0,1]^2$ exists, one can use the fact that $[0,1]^2$ is the image of the Cantor set under a continuous function and then extend this continuous function by linear interpolation (the complement of the Cantor set consists of open intervals) to all of $[0,1]$.

A lot more can be proven this way, including a number of very surprising results. A wonderful source for such arguments is the following paper, which (deservedly) won an award for mathematical exposition:

Benyamini, Yoav. "Applications of the universal surjectivity of the Cantor set." The American mathematical monthly 105.9 (1998): 832-839.

Among the results proven there is the entirely classical result of Banach and Mazur that every separable Banach space is linearly isometric to a subspace of $C[0,1]$; a result as far from being a counterexample as can be.


I think that you may be selling short the value of a counterexample! They are quite useful for making sure that you have not proven too much. i.e. When you have a plausible but only semi-formal argument, how do you tell if it is worth the effort in making it rigorous? Checking against counterexamples is often a useful step.

Still, there are relations to unbounded paths in an infinite binary tree without leaves. That is, suppose that you start at the root and write $0.$, thought of as a ternary number. As we travel to the left or right, write a $0$ or $2$, respectively, for the following digit. Continuing on we get a ternary expansion defining a real number in the Cantor set. Correspondingly, the ternary expansion of an element of the Cantor set gives rise to a path from the root in the tree.


Spaces that are homeomorpic to the Cantor set arise naturally in many mathematical settings, particularly in dynamical systems.

For one dynamical example, the Cantor set is homeomorphic to the phase space of any infinite Bernoulli process.

For another, the "nonescaping set" of many simple dynamical systems in the real line (or the complex plane) is homeomorphic to the Cantor set (this is a "Cantor dust" example as in the answer of @GeraldEdgar). Consider for example the dynamical system $$z_n = (z_{n-1})^2 + 10 $$ (You can replace $10$ by any real or complex number of magnitude $>2$). One can prove that there is a subset $C \subset \mathbb C$ homeomorphic to the Cantor set such that if $z_0 \in C$ then the sequence $(z_n)$ is bounded (in fact it stays in $C$), whereas if $z_0 \not\in C$ then $\lim_{n \to \infty} |z_n| =\infty$. In short, points not in $C$ escape to infinity, points in $C$ do not.

Also, there are important theoretical descriptions/properties of the Cantor set, for example:

  • A topological space is homeomorphic to the Cantor set if and only if it is compact, metrizable, has no isolated points, and every component is a point.
  • Any compact zero-dimensional metrizable topological space is homeomorphic to a subspace of the Cantor set.

Cantor sets even occur naturally in number theory! The $p$-adic integers $\mathbb Z_p$ are homeomorphic to the Cantor set.

  • $\begingroup$ Do you mean some subset in $\mathbb{C}$? I don't get the example; for any $z_{-1}$ real one has $z_n \geq 10^{2^n} \to +\infty$. $\endgroup$ Oct 7 '18 at 16:01
  • $\begingroup$ Oops, you're right, that was careless of me. Fixed. $\endgroup$
    – Lee Mosher
    Oct 7 '18 at 21:26

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.