17
$\begingroup$

$C=$ 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.

$\endgroup$
1
  • 3
    $\begingroup$ 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. $\endgroup$ Mar 24, 2016 at 22:45

2 Answers 2

15
$\begingroup$

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 $\mathbf{1.9.8}$.

$\endgroup$
8
  • 1
    $\begingroup$ oh wow, that's a nice characterization of the irrationals $\endgroup$ Mar 24, 2016 at 22:45
  • 3
    $\begingroup$ @ForeverMozart: I like the fact that $C$, $\Bbb Q$, and $\Bbb R\setminus\Bbb Q$ all have really nice topological characterizations. $\endgroup$ Mar 24, 2016 at 22:47
  • $\begingroup$ You need to add separable metrisable here. Or we get other spaces like $\omega^{\omega_1}$ as well, or the Sorgenfrey line. $\endgroup$ Mar 25, 2016 at 8:47
  • 1
    $\begingroup$ Do you have a reference for this theorem? $\endgroup$ Mar 25, 2016 at 9:40
  • $\begingroup$ @Najib: Jan van Mill, The Infinite-Dimensional Topology of Function Spaces (North-Holland Mathematical Library), Theorem $1.9.8$. $\endgroup$ Mar 25, 2016 at 13:41
5
$\begingroup$

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 $C_2\to\mathbb{R}\setminus\mathbb{Q}$.

$\endgroup$
1
  • 2
    $\begingroup$ To be explicit: "nonterminating" here includes "not ending in an infinite stream of 2s (or 1s)". $\endgroup$ Mar 25, 2016 at 5:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .