Find all $x$ in the Cantor set $C$ such that $2x$ is also in $C$. Is this correct?

Question

This is an exercise in my real analysis book; "Find all $x$ such that both $x$ and $2x$ belong to $C$, the Cantor set." I tried tackling this problem in the following way. Let the ternary expansion of $x\in C$ be $x=0.a_1 a_2 a_3 \ldots$. We divide into three cases:

Case 1. $x$ has a 'tail' of 0's

In this case, unless $x=0$, there is the last '2', appearing in $x$'s expansion. Let that digit be $a_N=2$, and write $x=0.a_1 a_2 \ldots a_N$. We now consider $x$'s expansion as consecutive 0's and consecutive 2's taking turns. Unless $x=0.22\ldots 2$ (all 2), there is the last appearance of '0', say at $a_k$, where $1\leq k\leq N-1$. Adding $x$ to itself now shows that the $k$th digit of $2x$ must be 1, so that $2x$ is not in $C$. Hence the only possible $x$ in this case is $x=0$, since $x=0.22\ldots 2$ is clearly impossible by size.

Case 2. $x$ has a 'tail' of 2's

Similarly, unless $x=1$, there is the last '0', appearing in $x$'s expansion. Let that digit be $a_N=0$, and write $x=0.a_1 a_2 \ldots a_N22\ldots=a_1 a_2 \ldots a_{N-1}1$. If '2' appears among $a_1, \ldots, a_{N-1}$, similar reasoning as in case 1 shows that we must have something like $x=0.22\ldots200\ldots01$. But considering the ternary expansion of $2x$, the only possible $x$'s are of the form $x=0.00\ldots01=\frac{1}{3^k}$.

Case 4. $x$ has infinitely many 0's and 2's

Considering $x$'s expansion as consecutive 0's and consecutive 2's taking turns, this case forces the ternary expansion of $2x$ to have infinite 1's. Thus $2x$ cannot be in $C$.

In conclusion, the only possible $x$'s are : 0 or $\frac{1}{3^k}$ for some $k$.

So.. my questions are

• Is the above argument correct?
• If instead of $2x$ we require $2x-1$ to be in $C$, a similar argument will show that the answer in this case must be $x=1$ or $x=0.22\ldots2=1-\frac{1}{3^k}$. Is this also correct?

Thank you in advance.

Answer 1

Your argument is correct, but for some it may be a little easier to start at the other end, with $2x$. If $2x\in C$, then $2x$ has a ternary expansion $2x=0.a_1a_2a_3\ldots$ such that each $a_k$ is $0$ or $2$, so $x$ has a ternary expansion $x=0.b_1b_2b_3\ldots$ such that each $b_k$ is $0$ or $1$. This is fine if $x=0$ and each $b_k=0$. It’s also fine if there is a unique $k\in\Bbb Z^+$ such that $b_k=1$, since then

$$x=\frac1{3^k}=0.\underbrace{00\ldots00}_k222\ldots\in C\;.$$

If there is more than one $k$ such that $b_k=1$, let $m$ be minimal such that $b_m=1$. Then

$$\frac1{3^m}=0.\underbrace{00\ldots00}_m222\ldots=0.\underbrace{00\ldots00}_{m-1}1

and $C\cap\left(\frac1{3^m},\frac2{3^m}\right)=\varnothing$, so $x\notin C$. ($\frac1{3^m}$ and $\frac2{3^m}$ are endpoints of one of the open intervals removed in the construction of $C$.)

The second part is already nicely covered by zaq’s answer.

Answer 2

To be in $C$, a number must have a $0,2$ ternary expansion with integer part being $0$ (so, $1 = 0.2222\cdots$).

Your argument is correct but the different cases could be combined into:

• After the first appearance of $2$, all subsequent digits must be $2$.

Indeed, the ternary expansion begins with $0$ (integer part), and when $02$ is multiplied by $2$, the result is $11$. To fit in the Cantor set, these digits must become $20$ due to carry-over from lower terms. The required amount of carry-over is at least $02$, which means that the tail following $02$ had to have the value of at least $01$ in it, so it must be the tail of $2$s.

The above yields $0$ or $1/3^n$ for $n\in\mathbb N$; note that $1 = 1/3^0$ is excluded for a different reason: when doubled, it falls outside of the interval $[0,1]$.

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The case $2x-1$ is reduced to the previous one by substitution $y=1-x$. Indeed, $x\in C \iff y\in C$, and $2x-1\in C\iff (1-(2x-1))\in C$ where $1-(2x-1) = 2y$. So, $y$ must be as $x$ in the first part.