Characterizing open countable subsets of the Cantor set

Question

The Cantor-Bendixson theorem implies that any closed subset of the Cantor set $\mathcal{C}$ can be described as a disjoint union of a set $\mathcal{C}_c$ that is homeomorphic to the original Cantor set, and a countable open set $\mathcal{C}_o$.

The following answer, and the referenced work by Schoenberg & Grunhage therein, implies that all noncompact open subsets of the Cantor set are homeomorphic to the Cantor set minus a point—say, $\mathcal{C}/ \{ 0\}$. But this would mean that $\mathcal{C}_o$ is homeomorphic to $\mathcal{C}/ \{ 0\}$, which would imply the Cantor set minus a point is countable, which seems strange.

Is this true, or am I missing something?

Answer 1

The "countable open set $\mathcal C_o$" is not open as a subset of the cantor set $\mathcal C$, it is open as a subset of the set that can be written as $\mathcal C_c \sqcup \mathcal C_o$. Non-empty open subsets of Cantor space $\mathcal C$ are never countable.

Answer 2

The Cantor-Bendixsson theorem says that $C$ is a union of it's scattered part (open) and its perfect part (closed). The scattered part here is empty (as $C$ has no isolated points, and $C'=\emptyset$, and there is no scattering part) and it's already perfect. The theorem is valid but void here.

If $O \subseteq C$ is an open non-compact subset, it's locally compact (being open in compact) so it has a one-point compactification $\alpha O = O \cup \{\infty\}$. This is compact metric totally disconnected and has no isolated points so is homeomorphic to $C$ by Brouwer's theorem and so $O$ is homeomorphic to $C$ minus (any) point (by homogeneity).