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Does anyone know of any interesting subsets of the Cantor set? When I first started thinking about this, I thought that since the Cantor set $C$ is the intersection of disjoint unions of closed sets, those particular closed sets would be subsets of $C$. Those sets, however, happen to be intervals! The Cantor set contains no isolated points nor (open) intervals, so then what subsets does it contain besides $\varnothing$ and itself?

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Perhaps you should read the corresponding Wikipedia page (at least). Cantor set is homeomorphic to $2^{\mathbb N}$, the space of sequences of 0 and 1. – Grigory M Aug 13 '11 at 18:04
up vote 10 down vote accepted

Some could find the Cantor set minus a point an interesting subspace of the Cantor set. Regardless of which point is deleted, they are homeomorphic to each other. For concreteness denote $C_0=C\setminus\{0\}$. This space is characterized by the following

Theorem: A topological space is homeomorphic to $C_0$ if and only if it is separable, metrisable, zero-dimensional, locally compact, noncompact and has no isolated points.

Now that $C_0$ has been introduced, the nonempty open subsets of the Cantor set have the following nice characterization:

Theorem: Any nonempty compact open subset of $C$ is homeomorphic to $C$. Any noncompact open subset of $C$ is homeomorphic to $C_0$.

Furthermore, this property of having exactly two kinds of open subsets, of which the others are compact and the others noncompact, characterizes the Cantor set among compact metrisable spaces, see Schoenfeld A.H. and Gruenhage G., An Alternate Characterization of the Cantor Set. Proceedings of the American Mathematical Society 53 (1975), 235-236 (available online for free).

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The Cantor set is a very interesting subset of the Cantor set.

It is a metric space which is:

  1. Compact;
  2. Nowhere dense;
  3. It has no isolated points;
  4. Polish (completely metrizable with a dense countable subset);
  5. Zero dimensional.

Not only that, it is also universal with respect to that every Hausdorff, compact, second countable and zero dimensional space is homeomorphic to the Cantor set.

Furthermore, every separable metric space which is zero dimensional is homeomorphic to a subset of the Cantor space. This includes, in particular, the irrationals (with the usual metric, also known as the Baire space).

From this fact we have that the uncountable closed subsets of the Cantor space are themselves Cantor spaces, we also have that any countable product of Cantor spaces is homeomorphic to the Cantor space, and we can partition every Cantor set into continuum many disjoint Cantor sets.

This to show you that "most" interesting subspaces of the Cantor set are themselves Cantor sets...

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...somehow I think the OP did not have an improper subset in mind... ;) – J. M. Aug 13 '11 at 18:19
@J.M.: I also added that proper makes improper ;-) – Asaf Karagila Aug 13 '11 at 18:20
@J.M.: besides, the Cantor set is self similar. – Willie Wong Aug 13 '11 at 19:14
@Willie: right, I forgot about the comb... :) – J. M. Aug 13 '11 at 19:45
@Asaf: But there are other notions of dimension, of course, and I guess that is the source of Gortaur's question. E.g., the Hausdorff dimension of the Cantor set is $\log_3(2)$ (but this is not a topological invariant unlike the dimension you refer to). – Jonas Meyer Aug 13 '11 at 23:12

The Cantor set has a subset that is Lebesgue measurable but not Borel. This is in fact the elementary proof that $Leb(\mathbb{R}^{n})\neq Bor(\mathbb{R}^{n})$ for $n=1$.

Take two cantor sets, $C_{1}$ and $C_{2}$, where $C_{1}$ has zero Lebesgue measure (e.g. the middle third Cantor set) and the second $C_{2}$ as one that has positive Lebesgue measure (you can choose the sequence when constructing the cantor set so that its measure gets any desired value from $[0,1[$). Since all Cantor sets are homeomorphic, there exists a homeomorphism $f:C_{2}\to C_{1}$. Construct a Vitali type of non Lebesgue measurable set $V\subset C_{2}$. Now $f(V)$ has zero Lebesgue measure and is thus Lebesgue measurable, but it is not a Borel set. We argue by contradiction that it is. Being that $C_{2}$ is compact, then in particular $C_{2}\in Bor(\mathbb{R})$. Since $f$ was a homeomorphism, then it is in particular a Borel function (which in this case implies that the preimage of any Borel set is also a Borel set), and thus $V=f^{-1}(f(V))\in Bor(\mathbb{R})$, which is a contradiction since $V$ was not even Lebesgue measurable. Hence $f(V)\notin Bor(\mathbb{R})$, so $Leb(\mathbb{R})\neq Bor(\mathbb{R})$.

If you found the above interesting, I am willing to shed light on any of the steps.

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The question of which rational numbers belong to the Cantor set might be considered interesting. The endpoints of deleted middles thirds, of course, but also $1/4$ and $3/10$ and lots of others. See

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The endpoints of the removed intervals are the numbers that can be represented by ternary expansions using only $0$s and $2$s and that end with repeating $0$s or repeating $2$s. The rational numbers in the Cantor set are all numbers that can be represented by eventually repeating ternary expansions using only $0$s and $2$s. – Jonas Meyer Aug 14 '11 at 0:13
Correct: and just which ones are those? The sequence of denominators doesn't seem to follow a really simple pattern. – Michael Hardy Aug 14 '11 at 19:17

The cantor set has measure zero, so it can eat other sets freely and never get chubby ;)

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1. Perhaps you could elaborate on your answer. 2. I've never see a username match an answer so well. – Austin Mohr Dec 11 '12 at 21:43

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