As we know, every real number has a binary representation, so there must be a binary representation of Cantor set?
I am familiar with ternary representation of cantor set. But I didn't saw anywhere (in my text books and unable to do it by myself) binary representation of cantor set! Is there is such representation? Please help me.
Edit(for those who are unclear what I am asking): as we can represent cantor set using ternary representation as
$F=\{x∈ [0,1] | x= ∑ a_n/3^n, a_n = 0 or 2\}$
Is in the same manner can we represent cantor set using binary representation? If yes! Then how?
The Cantor set consists of two self-similar copies of itself; the first is shrunk by a factor of 3 and not displaced, and the second is shrunk by a factor of 3 and is displaced by 2/3. In other words, the transformations that generate the Cantor set are the following:
$$f_1(x)=x/3$$ $$f_2(x)=x/3+2/3$$
As you can see, if you have a representation of numbers that allows you to easily divide things by 3, then you can construct a natural definition of the Cantor set under that representation. For ternary numbers, division by 3 just moves the decimal point by one spot (i.e. it's the representation for which division by 3 is the easiest), so we usually use ternary numbers for the definition of the Cantor set: it's the set containing all numbers in [0,1] whose ternary expansion never contains the digit 1.
There are other representations for which dividing by 3 is also easy; for example, in base-9, division by 3 is equivalent to multiplication by 0.3, and so the definition of the Cantor set again comes naturally: it's the set containing all numbers in [0,1] whose base-9 expansion never contains the digits 4, 5, or 6. You can easily generalize this to base-$3^n$ systems.
The trick I've been using for the above examples relies on the fact that division by 3 in certain representations amounts to an operation that only acts on a finite number of digits while moving the decimal point to a point just beyond the affected digits. By investigating what happens to those digits under the two transformations, I can make conclusions about the entire decimal expansion by iterating the operation for every digit in the expansion. Since, for a finite number of digits, there are a finite number of cases, this is easy. This fact is not true for all or even most representations. For base-2, the fraction 1/3 is represented as .0101010101..., so dividing by 3 means multiplying by .01010101..., which affects an infinite number of digits, so it's hard to make a nice definition out of it.
However, there are other "Cantoresque" fractals that do admit a nice definition in binary. Consider the fractal generated by the following two transformations:
$$f_1(x)=x/4$$ $$f_2(x)=x/4+3/4$$
You can see that this will generate something similar, but not equivalent to, the Cantor set (you can get this one by iteratively deleting the middle two fourths of any segment, rather than the middle third as in the normal Cantor set). In binary, division by 4 is equivalent to multiplying by .01. After the first iteration of the two transformations, the only numbers left will be of the form .00xxx... and .11xxx.... As such, the definition of this "Cantoresque" set in binary is: the set of all numbers in [0,1] whose binary expansions are made entirely from concatenations of the pairs 00 and 11. In base-4, the definition is even easier: the set of all numbers in [0,1] whose base-4 expansions do not contain the digits 2 or 3. In contrast, in base-3, 1/4 = .0202020202..., so the ternary numbers are not a natural representation for this fractal.
So asking the question: "Which base-$n$ systems admit natural representations of Cantoresque fractals?" is equivalent to asking the question: "for which base-$n$ systems does division by an integer $m$ only affect a finite number of digits?" This latter question is more tractable. If division by $m$ only affects a finite number of digits, then $1/m$ must truncate in base-$n$. This means that
$$\frac{1}{m}=\frac{a}{n^k}$$
for some integers $a<n^k$ and $k>0$. Equivalently, $m$ must divide $n^k$ for some $k$. If all of the prime factors of $m$ are also present in $n$ (maybe to differing powers), then $m$ will always divide $n^k$ for some $k$, because you can set $k$ to be the product of the least common multiples of the powers on each of the prime factors. If $m$ contains a prime factor that is not present in $n$, on the other hand, then $m$ can never divide $n^k$ for any $k$. As such, the following result holds:
A Cantoresque fractal involving division by $m$ only has a natural representation in base-$n$ for any $n$ whose prime factorization includes all prime factors of $m$.