Given that $S= K \bigcap \{ \dfrac {a+b}{2}~|~|a,b \in K, a \ne b \}$ where $K$ denotes the cantor set. Prove that $S = K \bigcap \{\dfrac {l}{3^n}~|~n \in \mathbb N, l \in \mathbb N_{3^n}, \dfrac {l}{3} \notin \mathbb N \}~$ where $\mathbb N_{3^n}$ means natural numbers upto $3^n$
Attempt: Suppose $a,b \in K, a \ne b$. let $a= \sum_{n=1} ^\infty \dfrac {x_n}{3^n}$ and $b = \sum_{n=1} ^\infty \dfrac {y_n}{3^n} $ where $x_n , y_n$ are all zeroes or twos. Then $ \dfrac {a+b}{2} = \sum_{n=1}^{\infty} \dfrac {z_n}{3^n} $ where $z_n = \dfrac {x_n+y_n}{2}~\forall n \in \mathbb N$. Thus, $z_n$ can take any value from $\{ 0,1,2\}$
Since $a \ne b $, there is atleast one $ s \in \mathbb N~|~ x_s \ne y_s$, so that one of these numbers is $0$ and the other $2$ making $z_s=1$. Now, $\dfrac {a+b}{2} \in K$ if and only if it has a ternary expansion in which no $1$ occurs.
If $z_s=1$ only once and $z_n \in K$, this should occur only if either $z_n=0 ~\forall~ n \ge s $ or $z_n=2~\forall~ n \ge s$ so that the the series converges to the other $(\dfrac {2}{3})$rd tip of the cantor set.
If $z_s=1$ happens more than once ,I don't think in that case $z_n$ could belong to the Cantor Set.
Could you please help me move forward from here. Thanks for reading !
Suppose $a,b \in K, a \ne b$. let $a= \sum_{n=1} ^\infty \dfrac {x_n}{3^n}$ and $b = \sum_{n=1} ^\infty \dfrac {y_n}{3^n} $ where $x_n , y_n$ are all zeroes or twos. Then $ \dfrac {a+b}{2} = \sum_{n=1}^{\infty} \dfrac {z_n}{3^n} $ where $z_n = \dfrac {x_n+y_n}{2}~\forall n \in \mathbb N$. Thus, $z_n$ can take any value from $\{ 0,1,2\}$
Since $a \ne b $, there is atleast one $ s \in \mathbb N~|~ x_s \ne y_s$, so that one of these numbers is $0$ and the other $2$ making $z_s=1$. Now, $\dfrac {a+b}{2} \in K$ if and only if it has a ternary expansion in which no $1$ occurs.
If $z_s=1$ only once and $z_n \in K$, this should occur only if either $z_n=0 ~\forall~ n \ge s $ or $z_n=2~\forall~ n \ge s$ so that the the series converges to the other $(\dfrac {2}{3})$rd tip of the cantor set. [![enter image description here][1]][1]
Suppose $z_s = 1 $ at the $s$th place and this happens only once. Then $\{z_n\}_{n \ge s} = \dfrac {1} {3^s} + \dfrac {2} {3^{s+1}} + \dfrac {2} {3^{s+2}} + \cdots = \dfrac {2}{3^s} $
$\implies z_n = \dfrac {\alpha_1} {3 } + \dfrac {\alpha_2} {3^2 } + \cdots + \dfrac {\alpha_{n-1}} {3^{s-1} } + \dfrac {2}{3^s} $ where $\alpha_i = 0 $ or $2$.
$\implies z_n = \dfrac {3^{s-1} \alpha_1 + 3^{s-2} \alpha_2 +\cdots+ 3 \alpha_n +2}{3^s} \le 3^s -1$ when each $\alpha_i=2$
Also, $2 = z_n \mod 2$
Thus, all the properties are satisfied. Similarily for the case when $z_n=0$
If $z_s=1$ happens more than once , in that case $z_n$ does not belong to the Cantor Set.