Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$C=$ Cantor set

$C_1=$ set of points in $C$ that are adjacent to removed intervals

$C_2=C\setminus C_1$ (all of the "non-endpoints")

QUESTION: Is $C_2$ homeomorphic to $\overline {\mathbb Q}$, the set of irrationals?

I see no obvious reason why they would not be homeomorphic. Both are zero dimensional, nowhere locally compact, cardinality $2^\omega$, etc.

share|cite|improve this question
A theorem of Alexandroff & Urysohn from 1928 characterizes the irrationals as the unique separable, completely metrizable, zero-dimension al space for which every compact subset has empty interior. – Grumpy Parsnip Mar 24 at 22:45
up vote 12 down vote accepted

Yes. $\Bbb R\setminus\Bbb Q$ is the unique non-empty, separable, completely metrizable, nowhere locally compact, zero-dimensional space. $C_2$ is clearly a $G_\delta$ in $C$, so it’s topologically complete, and you’ve already observed that it’s nowhere locally compact and zero-dimensional.

Added: One reference for this theorem is Jan van Mill, The Infinite-Dimensional Topology of Function Spaces (North-Holland Mathematics Library), Theorem $\mathbf{1.9.8}$.

share|cite|improve this answer
oh wow, that's a nice characterization of the irrationals – Forever Mozart Mar 24 at 22:45
@ForeverMozart: I like the fact that $C$, $\Bbb Q$, and $\Bbb R\setminus\Bbb Q$ all have really nice topological characterizations. – Brian M. Scott Mar 24 at 22:47
You need to add separable metrisable here. Or we get other spaces like $\omega^{\omega_1}$ as well, or the Sorgenfrey line. – Henno Brandsma Mar 25 at 8:47
Do you have a reference for this theorem? – Najib Idrissi Mar 25 at 9:40

Here is a more direct proof. Note that $C_2$ consists of those numbers between $0$ and $2$ whose base $3$ expansion is nonterminating and consists of $0$s and $2$s. We can take any such expansion, replace the $2$s with $1$s, and consider it as a binary expansion. We then get a number between $0$ and $1$ which has nonterminating binary expansion, i.e. a number which is not a dyadic rational. Writing $D$ for the dyadic rationals in $(0,1)$, we now have a bijection $C_2\to (0,1)\setminus D$, and it is not too hard to check directly that this bijection is a homeomorphism. (It is actually the restriction to $C_2$ of the Cantor function.)

Now $D$ is a countable dense linear order without endpoints, so by a standard back-and-forth argument it is order-isomorphic to $\mathbb{Q}$. This isomorphism extends to an isomorphism between the Dedekind-completions of $D$ and $\mathbb{Q}$, which are just $(0,1)$ and $\mathbb{R}$. So we have an order-isomorphism (and hence homeomorphism) $(0,1)\to \mathbb{R}$ which sends $D$ to $\mathbb{Q}$. It thus also sends $(0,1)\setminus D$ to $\mathbb{R}\setminus\mathbb{Q}$. We thus get a homeomorphism $(0,1)\setminus D\to \mathbb{R}\setminus\mathbb{Q}$, which we can compose with our earlier homeomorphism to get a homeomorphism $C_2\to\mathbb{R}\setminus\mathbb{Q}$.

share|cite|improve this answer
To be explicit: "nonterminating" here includes "not ending in an infinite stream of 2s (or 1s)". – Ben Millwood Mar 25 at 5:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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