An element is in the cantor set iff it can be written in ternary form with $a_n \neq 1$, for all $n \in \Bbb N$

Question

An element of $[0,1]$ is in the cantor set iff it can be written in ternary form (base $3$) $(0.a_1 a_2 ... a_n ...)$ with $a_n \neq 1$, for all $n \in \Bbb N$.

How is this possible? The book I'm reading says it should be obvious and does not provide any proof. If $(0.a_1 a_2 ... a_n ...) = (0.1)$ in base $3$, then in base $10$ it's $\frac{1}{3}$, which is an element of the cantor set, which seems to be a contradiction.

Answer 1

But $\frac 13$ can also be written in ternary as $0.02222222\ldots$, which does not have a $1$ anywhere.

The standard "middle third" construction of the Cantor set may be interpreted the following way: Iteration $n$ removes, from the remaining set, all numbers that in the $n$-th place of any of its ternary expansions must have a $1$ (remember that numbers that has a terminating ternary representation also has a ternary representation ending in repeated $2$'s). Thus, the ones that are left are the ones that may be written using only $0$ and $2$.

So, for instance, in the first iteration, $0.1$ is not removed because it may be written as $0.022222\ldots$, and $0.1222222\ldots$ is not removed because it may be written as $0.2$. Any number between those two, however, must have a $1$ in its $a_1$-place, and is therefore removed.

Answer 2

Consider the open intervals removed from $[0,1]$ to get the Cantor set. By induction on $n,$ these are all those,and only those intervals of the form $(a 3^{-n}, (a+1)3^{-n})$ where $3^n>a\in N$ and $a\equiv 1 \pmod 3.$

1. Show that any member of one of these intervals cannot be represented in base $3$ without using the digit $1.$

2. If $x\in [0.1]$ and $x$ cannot be represented in base $3$ without using the digit $1,$ take a representation of $x$ and consider the least $n$ such that the $n$th digit is $1.$ Now observe:

....... The subsequent digits cannot all be $0$'s, else we have another representation of $x$ by replacing the $n$th digit with $0$, and all subsequent digits with $2$'s, but that would be a representation of $x$ with no $1$'s.

.......The subsequent digits cannot be all $2$'s, else we have another representation of $x$ by replacing the $n$th digit with $2$, and all subsequent digits with $0$'s,but that would be a representation for $x$ with no $1$'s.

....... So $x=3^{-n}a +y$ where $3^n>a\in N$ with $a\equiv 1 \pmod 3,$ and $0<y<3^{-n},$ so x is in one of the open intervals that were removed.

Please note the comment to your Q from GEdgar.