Question: Show that there exists a Locally Compact, Perfect set in $[0,1]$ such that ternary expansion of each of its points consists of $0$ and $1$ only.
I know that the Cantor Set has almost all the above properties except the last one.
$1.$ Cantor set is compact and hence locally compact.
$2.$ Cantor set is a perfect set.
$3.$ Cantor set consists of members whose ternary expansion consists of $0$ and $2$ only. But here we required $0$ and $1$ only. How should I proceed? Thanks in advance!
Best Answer
Apply $f:[0,1] \to [0,1]$ defined by $f(x)=\frac{x}{2}$ to the middle third Cantor set $C$. $f[C]$ is clearly homeomorphic to $C$ and fits your requirements:
If $x\in C$ is written as $x=\sum_{n \ge 1} \frac{a_n}{3^n}$ with all $a_n \in \{0,2\}$, then $f(x) = \sum_{n \ge 1} \frac{b_n}{3^n}$ with $b_n = \frac{a_n}{2} \in \{0,1\}$.