Split perfect and Cantor sets

Question

Let $C$ be a Cantor ternary set. We can write it as union of $\mathfrak{c}$ many disjoint Cantor sets as follows: $$ C\times C= \bigcup_{x\in C} (\{x\}\times C)$$ and clearly $\{x\}\times C$ is Cantor set. Moreover, $(\{x\}\times C) \bigcap (\{y\}\times C)=\emptyset$. So, we can define a homemorhpic between $C$ and $C\times C$ in the natural way just by sending $x$ to $\{x\}\times C$. We can conclude that $C=\bigcup_{\alpha<\mathfrak{c}} C_{\alpha}$ where $C_{\alpha}$'s are pairwise disjoint Cantor set. I have two questions. Kindly check if what I did is correct and Also I know the same fact is correct for perfect, that is, every perfect set can be written as union of $\mathfrak{c}$-many pairwise disjoint perfect sets.

Answer 1

The function you defined (at least before any edits you might make after I first posted my answer) is NOT a homeomorphism. Indeed, it’s clearly not an injective function. I’ll prove the result for perfect subsets of $\mathbb R$ (similar proofs can be used to prove the result for perfect subsets of ${\mathbb R}^n),$ and I’ll leave it to you or others for generalizations to an arbitrary metric space or beyond.

Theorem: Let $P$ be a nonempty perfect subset of ${\mathbb R}.$ Then $P$ contains $c$-many pairwise disjoint nowhere dense perfect subsets.

Proof 1: First, map $P$ continuously onto $[0,1].$ For intervals this is trivial; for Cantor sets use a monotone increasing function similar to the Cantor function; if both types appear, pick one and work within that subset. Then follow this with a continuous mapping of $[0,1]$ onto the unit square (use a Peano curve). Now observe that the inverse images of the vertical fibers $\{r\} \times [0,1],$ as $r$ varies over $[0,1],$ gives continuum many pairwise disjoint nowhere dense perfect subsets of the given perfect set.

Proof 2: Since $P$ contains a nonempty compact perfect nowhere dense set (simply carry out a Cantor set type construction process relativized to $P),$ without loss of generality we can assume that $P$ is a nonempty compact perfect nowhere dense set. Since $P$ and $P \times P$ are nonempty totally disconnected perfect compact metric spaces, it follows from a classical topological result that $P$ is homeomorphic to $P \times P$ [Willard’s General Topology, Corollary 30.4, p. 217]. Hence, the inverse images of $\{x \} \times P$ for any fixed homeomorphism from $P$ onto $P \times P,$ as $x$ varies over $P,$ gives $c$-many pairwise disjoint perfect subsets of $P.$

The result itself is due to Mahlo (1913) [1] and the first proof above is due to Luzin/Sierpiński (1917) [2].

[1] Friedrich Paul Mahlo, Über teilmengen des kontinuums von dessen mächtigkeit [On subsets of the continuum with its power], Berichte über die Verhandlungen der Königlich Sächsischen Gesellschaft der Wissenschaften zu Leipzig, Mathematisch-Physische Klasse 65 (1913), 283-315.

[2] Nikolai Nikolaevich Luzin [Lusin] and Wacław Franciszek Sierpiński, Sur une propriété du continu [On a property of the continuum], Comptes Rendus Hebdomadaires des Séances de l’Académie des Sciences (Paris) 165 #16 (15 October 1917), 498-500.