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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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    $\begingroup$ 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. $\endgroup$
    – Grigory M
    Aug 13, 2011 at 18:04
  • $\begingroup$ Homeomorphic copies of $C$ also show up inside many other spaces. Foe example a non-empty completely metrizable space with no isolated points has a sub-space homeomorphic to $C$. $\endgroup$ May 10, 2018 at 15:04

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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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  • $\begingroup$ An amusing consequence for your first theorem is that $(C+\Bbb Z)\setminus \Bbb Z=$ $\{x+z: x\in C, z\in \Bbb Z\}\setminus \Bbb Z$ is homeomorphic to $C_0.$.... Applying your first paragraph, your first theorem could be stated as :The one-point (Alexandroff) compactification of a separable metrizable zero-dimensional locally-compact, non-compact space with no isolated points is homeomorphic to C. $\endgroup$ May 10, 2018 at 15:20
  • $\begingroup$ @LostInMath These are interesting characterizations of $C_0$. Do you also know whether all clopen subsets of $C_0$ are homeomorphic to $C_0$? (The same question for $C$ - are all clopen subsets of $C$ homeomorphic to $C$?) $\endgroup$ Jan 4, 2022 at 11:18
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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 as a subset of $[0,1]$;
  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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  • $\begingroup$ ...somehow I think the OP did not have an improper subset in mind... ;) $\endgroup$ Aug 13, 2011 at 18:19
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    $\begingroup$ @J.M.: besides, the Cantor set is self similar. $\endgroup$ Aug 13, 2011 at 19:14
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    $\begingroup$ @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). $\endgroup$ Aug 13, 2011 at 23:12
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    $\begingroup$ The second point - the claim that it is nowhere dense - seems suspicious. It would make more sense to say it is nowhere dense if it is embedded in another space. But if I have any non-empty metric space $X$, then $\operatorname{Int} \overline X=X$. So $X$ is not nowhere dense in $X$. $\endgroup$ May 10, 2018 at 13:27
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    $\begingroup$ "the uncountable closed subsets of the Cantor space are themselves Cantor spaces" - but couldn't we have (by the Cantor-Bendixson theorem) an uncountable closed subset of the Cantor space that was a disjoint union of a Cantor space and a countable space? So that the subset would be uncountable and closed, but not itself a Cantor space? (Since it would have isolated points.) $\endgroup$ May 16, 2018 at 18:46
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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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  • $\begingroup$ The cardinal of the set of all Borel subsets of $\Bbb R$ is $c=2^{\aleph_0}$ but the cardinal of the set of all subsets of the Cantor set is $2^c.$ And every subset of the Cantor set $C$ is Lebesgue-measurable because $C$ is Lebesgue-null (measure $0$). $\endgroup$ May 10, 2018 at 14:56
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The cantor set has measure zero, so it can eat other sets freely and never get chubby ;)

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    $\begingroup$ 1. Perhaps you could elaborate on your answer. 2. I've never see a username match an answer so well. $\endgroup$ Dec 11, 2012 at 21:43
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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 http://oeis.org/A054591

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  • $\begingroup$ 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. $\endgroup$ Aug 14, 2011 at 0:13
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    $\begingroup$ Correct: and just which ones are those? The sequence of denominators doesn't seem to follow a really simple pattern. $\endgroup$ Aug 14, 2011 at 19:17
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Let $C\subset[0,1]$ be the Cantor set and $f:[0,1]\to[0,1]$ the Cantor function. Define $$D=\{x\in C: f'(x){\rm\ exists}\}.$$ This subset has the following two interesting properties:

  1. $\lambda[f(D)]=1$ where $\lambda$ is Lebesgue measure, and

  2. $C\setminus D$ is uncountable.

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