# Classification of a kind of compact sets

We define a Cantor set in the real line as a set that is:

1. compact
2. perfect
3. with empty interior

Is it true that if we have 2 sets in the real line with this properties (and no countable because the singleton also has this) then they are homeomorphic?

Please don't give me a solution, I want to do this problem, but i need some advice. Clearly if I have a continuous bijection then it'll be a homeomorphism.

Because the closed sets are compact and $f$ continuous map preserve this property, and so the image is also closed, so this map is also homeomorphism. But I don't know how to use the property of empty interior in this problem.

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A continuous bijection between two spaces is not always a homeomorphism, unless, I think, the sets are finite; not even a continuous bijection from a space to itself is necessarily a homeomorphism. – gary Sep 2 '11 at 0:05
@gary: A continuous bijection even between two finite spaces does not have to be a homeomorphism. Consider the identity mapping on a finite set into a strictly coarser topology. However, a continuous bijection from a compact space onto a Hausdorff space is always closed and thus a homeomorphism. – LostInMath Sep 2 '11 at 0:27
@LostInMath: I meant a bijection $f:(X,T)\rightarrow (X,T)$ , where the two topologies are equal; I should have been more clear about that , tho. – gary Sep 2 '11 at 0:30
@gary: Yes, a continuous bijection to the same space is a homeomorphism if the space is finite. Actually finiteness of the topology suffices. – LostInMath Sep 2 '11 at 10:38
Daniel: In its current form the title does not reflect the question very much, since some compact sets are closed intervals which are not Cantor sets, and others are not perfect sets, and some are a combination of a Cantor set with a disjoint interval, etc etc. – Asaf Karagila Sep 11 '11 at 18:27

Theorem: Suppose $X$ is a nonempty, compact, totally disconnected, perfect and metrizable space. Then $X$ is homeomorphic to the Cantor set.

The proof is not very hard, but nontrivial. This gives an answer much stronger than just subspaces of the real line.

To prove this theorem note that you can represent the Cantor set as infinite binary sequences, and that given a countable clopen basis $\{U_n\mid n\in\mathbb N\}$ for $X$ the map $x\mapsto \langle \chi_{U_n}(x)\mid n\in\mathbb N\rangle$ is an interesting map (where $\chi_A(x)=1$ if and only if $x\in A$)

In the specific case of the real line, you might want to use the fact that empty interior implies the set is totally disconnected. Otherwise you can take some connected subspace and show it contains an open interval.

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my hypothesis are enough? The property to be totally unconnected, it follows from what I said? I guess it has to do with nonempty interior, and perhaps something else, or am I wrong? – Daniel Sep 1 '11 at 23:10
Kechris (7.4), p.35. It's not so hard to give a hint. The Cantor scheme is sufficient. – t.b. Sep 1 '11 at 23:21
In the real line, totally disconnected is the same as having empty interior, since the connected subsets of the real line are precisely the intervals. – gary Sep 2 '11 at 0:03