Proving the continuity of the Cantor Function Consider the Cantor Set $C=\{0,1\}^{\omega}$, that is, the space of all sequences $(b_1,b_2,...)$ with each $b_i\in\{0,1\}$. Define $g:C\rightarrow[0,1]$ by $$g(b_1,b_2,...)=\sum\limits_{i=1}^{\infty}\dfrac{b_i}{2^i}$$ In other words, $g(b_1,b_2,...)$ is the real number whose digits in base 2 are $0.b_1b_2...$ 
Prove that $g$ is continuous.
Here is my attempt:
Let sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ be elements of the Cantor Set $C.$ For fixed $n\in\mathbb{N}$ let $A_n=(a_1,a_2,...,a_n)$ and $B_n=(b_1,b_2,...,b_n)$ be the first $n$ terms in each of those sequences. If $A_n\neq B_n$, then $\exists m=min\{k\in\{1,2,...,n\}:a_k\neq b_k\}$ and the following holds:
$$\begin{array}{c}
\left|\sum\limits_{k=1}^n\dfrac{a_k}{2^k}-\sum\limits_{k=1}^n\dfrac{b_k}{2^k}\right|=\left|\sum\limits_{k=1}^n\dfrac{a_k}{2^k}-\dfrac{b_k}{2^k}\right|\geq\dfrac{|a_m-b_m|}{2^m}-\left|\sum\limits_{k=m+1}^n\dfrac{a_k}{2^k}-\dfrac{b_k}{2^k}\right|\geq\\ \geq\dfrac{|a_m-b_m|}{2^m}-\sum\limits_{k=m+1}^n\dfrac{|a_k-b_k|}{2^k}    \\
\text{where $|a_m-b_m|$=1 and}\\
\sum\limits_{k=m+1}^n\dfrac{|a_k-b_k|}{2^k}\leq \sum\limits_{k=m+1}^n\dfrac{1}{2^k}=\dfrac{1}{2^m}\\
\text{Now,}\\
\left|\sum\limits_{k=n+1}^{\infty}\dfrac{a_k}{2^k}-\sum\limits_{k=n+1}^{\infty}\dfrac{b_k}{2^k}\right|=\left|\sum\limits_{k=n+1}^{\infty}\dfrac{a_k-b_k}{2^k}\right|\leq\\
\leq\sum\limits_{k=n+1}^{\infty}\dfrac{|a_k-b_k|}{2^k}\leq\sum\limits_{k=n+1}^{\infty}\dfrac{1}{2^k}=\dfrac{1}{2^n}\\
\text{Therefore, if $A_n\neq B_n$, then}\\
\left|f(a)-f(b)\right|=\left|\sum\limits_{k=1}^{\infty}\dfrac{a_k-b_k}{2^k}\right|=\left|\sum\limits_{k=1}^{n}\dfrac{a_k-b_k}{2^k}+\sum\limits_{k=n+1}^{\infty}\dfrac{a_k-b_k}{2^k}\right|\geq\\
\geq\left|\sum\limits_{k=1}^{n}\dfrac{a_k-b_k}{2^k}\right|-\left|\sum\limits_{k=n+1}^{\infty}\dfrac{a_k-b_k}{2^k}\right|\geq\dfrac{1}{2^n}-\dfrac{1}{2^n}=0
\end{array}$$
I think I must have done something wrong because I wanted $|f(a)-f(b)|$ to be greater than or equal to something in terms of $m$ or $n$ so that I can say choose $n$ (or $m$) such that (something like) $\dfrac{1}{2^n}<\varepsilon$, but I just got 0 instead. How can I improve my proof so that I can achieve this?
 A: Your method of proof will work. Taking your idea, I think we can streamline it, in the following way: 
Let $\epsilon>0$ be given and let $(\epsilon_k)$ be the binary sequence representing $\epsilon.$ Take the ternary sequence for the $\delta$ (that we will show to work) to be $\delta_k=2\epsilon_k$. 
Now, let $N$ be the first non-zero digit of $\delta$ and $\epsilon.$ Then, if $|x-y|<\delta$, it must be the case that $x$ and $y$ agree on the first $N-1$ digits. Hence, the first $N-1$ digits of $g(x)$ and $g(y)$ are the same, as well, by definition of $g$. But then, $|g(x)-g(y)|<\epsilon.$
A: You can prove a homeomorphism from $[0,1]^\omega$ with the product topology to the same set with the dictionary topology, and then use the fact that the functions from $[0,1]^\omega$ to $[0,1]$ are continuous as open sets in $[0,1]$ will give an open preimage in the dictionary order.
A: Seems to me, the argument the poster put down is necessary to prove $|x-y| < 1/3^N$ implies $x,y$ agree in the first $N$ digits.
The correction needed to that calculation is the Cantor set everything is $0$'s or $2$'s so $|a_m-b_m| = 2/3^m \ge 2/3^N$ (numerator = 2 not 1) ... so we get $|x-y| \ge 2/3^N - 1/3^N = 1/3^N$.
