# Two related questions on unions of Cantor sets

I need to ask this before I forget about it, although it might be trivial to answer; I'm not sure. I'm asking it out of curiosity. Somehow I am fascinated by Cantor sets.

Definition: The Cantor Set is $$\subset \mathbb{R}$$ the only totally disconnected, perfect, compact metric space up to a homeomorphism (Willard 1970). I think this definition applies not just to the ternary Cantor set, but to all Cantor sets. Am I right about this?

Question a) Is the countable union of disjoint Cantor sets necessarily a Cantor set?

Question b) Can $$[0,1] \cap (\mathbb{R}$$ \ $$\mathbb{Q}$$) be written as the countable union of Cantor sets?

• Applies to any Cantor set, since any Cantor set (by usual definitions) are totally disconnected, ... (need to assume nonempty also, by the way). (a) A countable union of disjoint Cantor sets can be dense --- Just put copies of Cantor sets in every contiguous interval of a given Cantor set, do this again for all the newer contiguous intervals, etc. You'll want to carefully check that this is an appropriate construction, and not hand waving stuff that makes no sense under close examination. (b) No --- Baire category (irrationals are 2nd category, Cantor sets are nowhere dense). Sep 24, 2020 at 15:41
• Is the empty set a Cantor set? Sep 24, 2020 at 15:49
• And I do think that avoiding learning Baire's Category Thm won't do me any good going forward, so it's one of the next things I'm going to learn. Sep 24, 2020 at 15:58
• The empty set is usually not considered to be a Cantor set. Also, some authors assume perfect sets are not empty --- some by saying so explicitly, others implicitly assume this, and probably a few just don't think about it. In this respect the usage of "perfect set" is like the usage of "interval" for a connected subset of the reals. Many beginning real analysis texts allow the empty set and singletons to be intervals, and books heavy with Riemann integration stuff tend to use "interval" only for connected sets of reals having more than one point. Sep 24, 2020 at 15:59
• But then why did you mention I need to assume non-emptiness in your first comment? lol. Anyway, these are minor quibbles. But the main point is: the answer to a) is no; the answer to b) is no. Sep 24, 2020 at 16:03

Let $$C$$ be a Cantor set. The union of countably many pairwise disjoint Cantor sets is homeomorphic to $$C\times\Bbb N$$, where $$\Bbb N$$ has the discrete topology, and is therefore not compact and not a Cantor set. It is also homeomorphic to $$C\setminus\{x\}$$ for any $$x\in C$$.
$$[0,1]\setminus\Bbb Q$$ is not $$\sigma$$-compact, so in particular it is not the union of countably many Cantor sets.