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?
Best Answer
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.