I'm trying to prove that the following function $ f $ is continuous on $ [0,1] $. The function $ f:[0,1]\rightarrow [0,1] $ is defined as follows.
Let $ x\in [0,1] $. Then $ x= \sum\limits_{n = 1}^\infty \frac{a_n}{3^n} $ ; where $ a_{n}\in \{0,1,2\}$ for each $ n\in \mathbb{N} $.
If $ x\in C $ (the Cantor set) then $ x $ can be written as of the form $ \sum\limits_{n = 1}^\infty \frac{a_n}{3^n} $ ; where $a_{n}\in \{0,2\}$ for each $ n\in \mathbb{N} $ and then we define $ f(x)=\sum\limits_{n = 1}^\infty \frac{a_n}{2^{n+1}} $;(in this case we use the ternary expansion of $x$ which does not contain 1s to define the function $f$ ).
If $ x\notin C $ then there exists $ n_{0}\in \mathbb{N} $ such that $ a_{n_{0}}=1 $.
Put $ N=\min \{n_{0}\in \mathbb{N}:a_{n_{0}}=1\} $.
If $ N=1 $ then define $ f(x)=\dfrac{1}{2} $ and otherwise we define $ f(x)=f(\sum\limits_{n = 1}^\infty \frac{a_n}{3^n})=\sum\limits_{n = 1}^{N-1} \frac{a_n}{2^{n+1}}+\dfrac{1}{2^{N}} $.
My attempt: Let $ x\in [0,1] $ be arbitrary. Then either $ x\in C $ or $ x\notin C $. If $ x\notin C $ then $x$ belongs to some open middle third (say $I$) which has removed in the construction of the Cantor set C. Since $f$ is constant on the open interval $I$, $f$ is obviously continuous at $x$.
But I'm stuck on the Proving that $f$ is continuous at $x$ if $ x\in C $.
Could anyone give me some help ?
Any hints/ideas are much appreciated.
Thanks in advance for any replies.