I refer to the following alternative definition of the Cantor function from Wikipedia:
See figure. To formally define the Cantor function $c:[0,1]\to[0,1]$, let $x\in[0,1]$ and obtain $c(x)$ by the following steps:
Express $x$ in base $3$.
If $x$ contains a $1$, replace every digit after the first $1$ by $0$.
Replace all $2$'s with $1$'s.
Interpret the result as a binary number. The result is $c(x)$.
I find this definition nice and explicit. Sure, it works when I plug in some numbers, but I have no intuition on why it works? Thanks for any help.
Basically, I am trying to see why this definition is equivalent to the other one in my textbook:
I don't think this is an easy problem, because the first definition is an explicit function $f$, while the second is a limit of a sequence of functions. See here and observe that the limit function $F$ is not even found explicitly either; rather, it arises from a Cauchy sequence, using completeness of $[0,1]$.
But for example, if we look at $F_{n}(x)=\frac{3^{n}}{2^{n}}x;\ x\in [0,\frac{1}{3^{n}}]$, then
$F_n(\frac{1}{3^{n}})=\frac{1}{2^{n}}=\underbrace{.00\cdots 0 }_{\text{n-1}}100\cdots $ in its binary expansion.
On the other hand, if we write $\frac{1}{3^{n}}=\underbrace{.00\cdots 0 }_{\text{n}}2222\cdots $, in its ternary expansion then using the explicit formula, we also get $\underbrace{.00\cdots 0 }_{\text{n}}1111\cdots = \underbrace{.00\cdots 0 }_{\text{n-1}}1000\cdots$ in the binary expansion.
It's easy to show that if $m>n$ then $F_m$ and $F_n$ agree at $\frac{1}{3^{n}}$,so the limit $F$ is equal to $f$ at these points.
Obviously, this isn't a proof, but it indicates how you might proceed. In fact, if you look at the definition of the $F_n$, I think you can show, without loss of generality, that it suffices to consider the intervals $[0,1/3^n]$ when doing the analysis because, at the $n^{th}$ stage, there are $2^{n-1}$ intervals of equal length, all obtained by trisecting the intervals from the $(n-1)^{th}$ stage.