$K$ : Cantor set
$\phi$ : $K \to [0,1]$, given by $\sum_{k=1}^{\infty} \dfrac{2\epsilon_k}{3^k} \mapsto \sum_{k=1}^{\infty} \dfrac{\epsilon_k}{2^k} \ (\epsilon_k=0,1)$
I want to prove that $\phi$ is continuous on $K$.
Let $\epsilon >0.$
I have to prove
$\exists \delta>0, |x-a|<\delta \ (a\in K)\Rightarrow |\phi (x)-\phi (a)|< \epsilon.$
I have to decide $\delta= \cdots$,
and I can write $x=\sum_{k=1}^{\infty} \dfrac{2\epsilon_k}{3^k}, a=\sum_{k=1}^{\infty} \dfrac{2\epsilon'_k}{3^k}$ and when $|x-a|<\delta$,
\begin{align} |\phi (x)-\phi (a)| &=\left| \sum_{k=1}^{\infty} \dfrac{\epsilon_k}{2^k} - \sum_{k=1}^{\infty} \dfrac{\epsilon'_k}{2^k} \right| \\ &= \left| \sum_{k=1}^{\infty} \dfrac{\epsilon_k-\epsilon'_k}{2^k} \right|\\ &\leqq \sum_{k=1}^{\infty} \left| \dfrac{\epsilon_k-\epsilon'_k}{2^k} \right| \end{align}
and I want to use $|x-a|=\left|\sum_{k=1}^{\infty} \dfrac{\epsilon_k-\epsilon'_k}{3^k}\right|<\delta.$
I have no idea about how I have to determine $\delta$ and how I have to evaluate $|\phi (x)-\phi (a)|$.
I would like you to give me some ideas.
Hint:
If $k$ is the first position at which $\epsilon_k$ and $\epsilon'_k$ differ. You can prove that $|x-y| \ge \frac{1}{3^k}$ and that $\left| \sum_{k\ge 1} \frac{\epsilon_k}{2^k} - \sum_{k\ge 1} \frac{\epsilon'_k}{2^k} \right| \le 2^{-k+1}$