6
$\begingroup$

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.

  • 0
    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.2011-09-11

1 Answers 1

8

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.

  • 1
    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.2011-09-02