Let
$F_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$
$F_2 = [0,\frac{1}{3^2}] \cup [\frac{2}{3^2},\frac{3}{3^2}] \cup [\frac{6}{3^2},\frac{7}{3^2}] \cup [\frac{8}{3^2},1]$
and so on.
The Cantor set is then $C=\bigcap^{\infty}_{k=1} F_k$.
Each $x \in C$ can be written in the form $ x = \frac{a_0}{3} + \frac{a_1}{3^2} + \frac{a_2}{3^3} + ...+ \frac{a_n}{3^{n+1}}+ ...$
By Construction, if $x\in F_k$ and $0 \leq j < k$, then $3^jx=a+y$ where $a \in \mathbb{N} \cup \{0\}$ (*) and $y \in F_{k-j}$.
Noting this pattern suppose $x\in C$ therefore $x \in F_k$ for all $k \in \mathbb{N}$. Now assume that for some $m\in \mathbb{N}$ we have $a_m = 1$ Then we have $3^mx=a+\frac{1}{3}+\frac{a_{m+1}}{3^2}+\frac{a_{m+2}}{3^3}+...\ \ \ \Leftrightarrow$ $ \ \ \ \ y=3^mx-a \notin F_1 \Leftrightarrow x \notin F_{m+1} \Leftrightarrow x \notin C$ contradiction.
To see the pattern more clearly, consider $F_3$,
$F_3= [0,\frac{1}{27}] \cup [\frac{2}{27},\frac{3}{27}] \cup [\frac{6}{27},\frac{7}{27}]\cup [\frac{8}{27},\frac{9}{27}]\cup [\frac{18}{27},\frac{19}{27}]\cup [\frac{20}{27},\frac{21}{27}] \cup [\frac{24}{27},\frac{25}{27}] \cup [\frac{26}{27},1]$
Define the "multiplication" of an interval by an scalar, naturally by multiplying the endpoints by that scalar, for example $3F_2= [0,\frac{3}{9}] \cup [\frac{6}{9},1] \cup [\frac{18}{9},\frac{21}{9}] \cup [\frac{24}{9},3]=[0,\frac{1}{3}] \cup [\frac{2}{3},1] \cup [2,\frac{7}{3}] \cup [\frac{8}{3},3]$
This multiplication is important, since it gives us 2 copies of $F_1$, (the 2nd copy is a translation of $F_1$ by an integer, look at (*) )
More generally, $3^jF_k$ gives us $2^j$ copies of $F_{k-j}$
let $x \in F_3$
For $j=0 \ \ \ $, the statement is clear, i.e. $3^0x \in F_{3-0} \Rightarrow x \in F_3$ Trivial !
For $j=1 \ \ \ $Obviously $3^1x \in 3F_3$, i.e. $3^1x$ lies in a translated (by a number $a \in \mathbb{N} \cup \{0\}$) copy of $F_{3-1}=F_2$ Therefore $3x=a+y$ where a is the natural number of translation, and $y\in F_2$
For $j=2 \ \ \ $ Similarly $3^2x \in 3^2F_3 \Rightarrow 3^2x $ lies in a translated copy of $F_{3-2}=F_1$
Yes, it is. You can see this very neatly with an explicit map. Let $K$ denote the Cantor set. Given an element $$x=\sum_{n=1}^\infty\frac{2a_n}{3^n}\in K$$ where $a_n=0$ or $1$ for each $n$, let $$f(x)=\sum_{n=1}^\infty\frac{a_n}{2^n}.$$ That is, $f$ takes the ternary expansion of $x$ using $0$s and $2$s, replaces each $2$ with a $1$, and considers it as a binary expansion. Then $f:K\to [0,1]$ is a surjection that is easily checked to be continuous. Since $K$ is compact and $[0,1]$ is Hausdorff, it follows that $f$ is a quotient map. Finally, the corresponding equivalence relation is exactly the one you describe, since the pairs you are identifying are exactly the pairs whose ternary expansions correspond to the two different binary expansions of some dyadic rational.
Best Answer
It's called the middle-thirds Cantor set because in general you can construct a class of sets with similar properties using a similar but scaled construction. For example, you can start with $[0,1]$, remove an interval of length $\frac{1}{2}$ from the center. Then you have $2$ intervals of length $\frac{1}{4}$, remove an interval of length $\frac{1}{8}$ from the center. Inductively, at each step you have a disjoint union of intervals of length $l$ remaining in your set, and you remove an interval of length $l/2$ from each interval. This set, which we might call the Cantor "middle-halves" set, has many of the same properties of the Cantor middle-thirds set. From this you can imagine constructing the "middle-fourths" set and many other Cantor-type sets. The middle-thirds set is sort of standard because it's the easiest Cantor-type set to construct (mainly it's easy to figure out the lengths of the intervals at each step).
The reason for the ternary expansion is precisely as you stated, the middle-thirds construction removes any numbers with absolutely necessary $1$-s in the ternary expansion. You should check this for yourself for a few cases: for example, check if $0.1abcd...$ (ternary) can be in the middle-thirds set, then $0.01abcd...$, $0.21abcd...$, and so on. Then you'll see why the Cantor set construction removes them. This excludes cases like $1/3$, which lies in the Cantor set and has ternary expansion $0.1000...$, because it can also be written $0.0222...$ .