a variant of the Cantor function

Question

The classical Cantor function $\phi$ maps the space $\Omega$ consisting of sequences $({\omega}_{k})_{k\geq 1}$ where for every integer $k\geq 1$, ${\omega}_{k}\in \{0,1\}$, into the unit interval $[0,1]$ via the formula $${\phi}(\omega) = {{\sum}_{k=1}^{\infty}}{\frac{2{\omega}_{k}}{3^{k}}}.$$ The image is the usual "middle-thirds" Cantor set. Is there a similar formula to map the space ${\Omega}$ into the set $[0,1/9]\cup[8/9,1]$? The image must be the intersection of these intervals with the middle thirds Cantor set. Thanks!

Answer 1

This seems to be a generalization of the classical Cantor function.

Fix a positive integer $m$. Consider the mapping ${{\phi}_{m}}:{\Omega}\rightarrow {[0,1]}$ defined by $$\phi _{m}\left( \omega \right) =\sum_{k=1}^{\infty }\frac{\left(3^{m}-1\right) \omega _{k}}{3^{mk}}.$$

Then the image $ \phi _{m}\left( \Omega \right)$ is homeomorphic to the Cantor set and lies in $[0,{\frac{1}{3^m}}]\cup{[1-{\frac{1}{3^m}},1]}.$ The case $m =1$ is the usual middle-thirds Cantor set, and the case $m=2$ is what I was trying to construct. The map ${\phi}_{m}$ seems to be an order-preserving homeomorphism of $\Omega$ onto $ \phi _{m}\left( \Omega \right)$ if we impose the lexicographic order on $\Omega$, and topologize $\Omega$ it by giving it the topology of pointwise convergence.