Are these Cantor spaces and what numbers does $\sum_{n=0}^{\infty}\dfrac{a_n}{4^{n+1}}$ contain?
The Cantor set in the real numbers can be defined as the set of infinite sums:
$\displaystyle\sum_{n=0}^{\infty}\dfrac{2a_n}{3^{n+1}}$
where $a_n$ is drawn from $\{0,1\}$
If instead we take the set :
$\displaystyle\sum_{n=0}^{\infty}\dfrac{3a_n}{4^{n+1}}$
Then from the definitions I've read, being generated by $2^\omega$ I see no reason why this would not be a Cantor space.
I'm not clear what numbers it contains. These seem to be numbers in the interval $[0,1]$ again, but those which can be written in base $4$, using only $0$s and $3$s.
Is this built by repeatedly removing the centre halves of every segment?
Then finally what I'm really interested in is the set:
$\displaystyle\sum_{n=0}^{\infty}\dfrac{a_n}{4^{n+1}}$
which I think is the same set as the last, this time divided throughout by $3$ so scaled down to the interval $[0,\frac13]$
Is it correct that is still a Cantor space, and what numbers does it contain? It seems to be the numbers written in base $4$ using only $1$s and $0$s and is the interval $[0,\frac13]$ with the centre halves of every interval removed.