How to view the middle third Cantor set as a fixed point via it's self-similarity

Question

The definition of middle third cantor set is given in this link" https://en.wikipedia.org/wiki/Cantor_set and we need to use Hausdorff metric, the definition of the Hausdorff metric is given in this link:https://en.wikipedia.org/wiki/Hausdorff_distance The idea is that, first, if we denote the middle third Cantor set by C, then we need to find a function f from the set of all nonempty compact subsets of R(denote this set by K) to itself such that f(C)=C, then we need to show that f is a contraction and then use Banach fixed point theorem(notice that the Hausdorff metric is complete), so how can we achieve this?

Answer 1

We are working on the metric space $(K, d_{H})$ where $K$ is the set of all nonempty compact subsets of $\mathbb{R}$ and $d_{H}$ is the Hausdorff metric. Consider the map $f: K \to K$ given by $$f(S) = \frac{S}{3} \cup \frac{S+2}{3}$$ where we are using the notation $S+c = \{x + c \, : \, x \in S\}$ and $cS = \{cx \, : \, x \in S\}$. It remains to show $f$ is a contraction in the Hausdorff metric.

Suppose $x \in f(A)$. Then either $3x \in A$ or $3x-2 \in A$. By the definition of Hausdorff metric $\delta := d_{H} (A,B)$, for every $a \in A$ there exists $b \in B$ with $d(a, b) \le \delta$; here we are using compactness of $A$ and $B$ to produce distance minimizers.

Now we check both cases:

• In the first case ($3x \in A$), we deduce the existence of some $y \in B$ such that $d(3x,y) \le \delta$, and then $d(x,y/3) \le \delta/3$. Since $y/3 \in f(B)$, it follows that $d(x, f(B)) \le \delta/3$.

• The second case ($3x-2 \in A$) gives some $y \in B$ such that $d(3x-2,y) \le \delta$ so $d(x,(y+2)/3) \le \delta/3$ and again, since $(y+2)/3 \in f(B)$ we have $d(x, f(B)) \le \delta/3$.

We have shown that $d(x, f(B)) \le \delta/3$ for all $x\in f(A)$, so in fact $\sup_{x\in f(A)} d(x,f(B)) \le \delta/3$. An entirely analogous argument shows $\sup_{y \in f(B)} d(y, f(A)) \le \delta/3$. We conclude $$d_{H} (f(A), f(B)) \le \frac{1}{3} d_{H}(A,B)$$ so $f$ is indeed a contraction. Banach gives a unique fixed point, which proves the existence of the Cantor set (if one defines the Cantor set as the unique fixed point of the map $f$).