TITLE: Cantor set minus endpoints homeomorphic to irrationals?
QUESTION [17 upvotes]: $C=$ 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.

REPLY [5 votes]: 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 $C_2\to\mathbb{R}\setminus\mathbb{Q}$.