Can every compact metric space be realized as the continuous image of a cantor set?
Real Analysis – Cantor Set and Compact Metric Spaces
measure-theoryreal-analysis
Related Solutions
The general outline is to write your given compact metric space $A$ as a decomposition into two sets, then four sets, then eight sets, etc., in the exact same fashion that the Cantor set is decomposed, except that whereas the decomposition elements of the Cantor set are pairwise disjoint, the decomposition elements in $A$ need not be pairwise disjoint. Be careful about how your decomposition elements are indexed. Once similarly indexed decomposition elements of $C$ and of $A$ are put side-by-side, the formula for writing the desired function $C \mapsto A$ is evident.
Here's a few details.
For any compact metric space $A$, you can write it as a union of two compact subsets $$A = A_0 \cup A_1 $$ and then for each $i_1 \in \{0,1\}$ you can write $A_{i_1}$ as a union of two compact subsets $$A_{i_1} = A_{i_1,0} \cup A_{i_1,1} $$ and then for each $i_1,i_2 \in \{0,1\}$ you can write $A_{i_1,i_2}$ as a union of two compact subsets $$A_{i_1,i_2} = A_{i_1,i_2,0} \cup A_{i_1,i_2,1} $$ and so on and so on inductively, so that as the length of the subscript sequence increases, the diameter of the set decreases to zero. It follows for any infinite sequence $\omega = (i_1,i_2,i_3,...)$ of $0$'s and $1$'s the nested decreasing intersection $$A_\omega = A_{i_1} \cap A_{i_1, i_2} \cap A_{i_1, i_2, i_3} \cap \cdots $$ is a single point.
Then you take the usual description of the Cantor set $C$ as the set of real numbers $x$ written in trinary as $$x = .j_1 j_2 j_3 \cdots $$ where $j_n \in \{0,2\}$, let $i_n = j_n/2 \in \{0,1\}$, let $\omega(x) = (i_1,i_2,i_3,...)$, and map $x$ to the point $A_{\omega(x)}$.
Cantor set is defined as $C=\cap_n C_n$ where $C_{n+1}$ is obtained from $C_n$ by dropping 'middle third' of each closed interval in $C_n$
As you have noted, Cantor set is bounded.
Since each $C_n$ is closed and $C$ is an intersection of such sets, $C$ is closed (arbitrary intersection of closed sets is a closed set).
As $C$ is closed and bounded, it is compact by Heine-Borel theorem.
PS: You cannot say that Cantor set is a union of closed intervals. Rudin is giving Cantor set as an example for a perfect set that contains no open interval!
Best Answer
Yes (assuming it's nonempty, of course). Moreover if you google "continuous image of the Cantor set", the first hit takes you to
http://en.wikipedia.org/wiki/Cantor_set
where you can read that this theorem is true and a reference is given to Willard's General Topology. (The article does not say this and perhaps it should, but it is specifically Theorem 30.7 on p. 217 of the Dover edition.)