Csdrhrt

Consider the Cantor Set $C={0,1}^{omega}$, that is, the space of all sequences $(b_1,b_2,...)$ with each $b_iin{0,1}$. Define $g:Crightarrow[0,1]$ by $$g(b_1,b_2,...)=sumlimits_{i=1}^{infty}dfrac{b_i}{2^i}$$ 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 $(a_n)_{ninmathbb{N}}$ and $(b_n)_{ninmathbb{N}}$ be elements of the Cantor Set $C.$ For fixed $ninmathbb{N}$ 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 $A_nneq B_n$, then $exists m=min{kin{1,2,...,n}:a_kneq b_k}$ and the following holds:
$$begin{array}{c}
left|sumlimits_{k=1}^ndfrac{a_k}{2^k}-sumlimits_{k=1}^ndfrac{b_k}{2^k}right|=left|sumlimits_{k=1}^ndfrac{a_k}{2^k}-dfrac{b_k}{2^k}right|geqdfrac{|a_m-b_m|}{2^m}-left|sumlimits_{k=m+1}^ndfrac{a_k}{2^k}-dfrac{b_k}{2^k}right|geq\ geqdfrac{|a_m-b_m|}{2^m}-sumlimits_{k=m+1}^ndfrac{|a_k-b_k|}{2^k} \
text{where $|a_m-b_m|$=1 and}\
sumlimits_{k=m+1}^ndfrac{|a_k-b_k|}{2^k}leq sumlimits_{k=m+1}^ndfrac{1}{2^k}=dfrac{1}{2^m}\
text{Now,}\
left|sumlimits_{k=n+1}^{infty}dfrac{a_k}{2^k}-sumlimits_{k=n+1}^{infty}dfrac{b_k}{2^k}right|=left|sumlimits_{k=n+1}^{infty}dfrac{a_k-b_k}{2^k}right|leq\
leqsumlimits_{k=n+1}^{infty}dfrac{|a_k-b_k|}{2^k}leqsumlimits_{k=n+1}^{infty}dfrac{1}{2^k}=dfrac{1}{2^n}\
text{Therefore, if $A_nneq B_n$, then}\
left|f(a)-f(b)right|=left|sumlimits_{k=1}^{infty}dfrac{a_k-b_k}{2^k}right|=left|sumlimits_{k=1}^{n}dfrac{a_k-b_k}{2^k}+sumlimits_{k=n+1}^{infty}dfrac{a_k-b_k}{2^k}right|geq\
geqleft|sumlimits_{k=1}^{n}dfrac{a_k-b_k}{2^k}right|-left|sumlimits_{k=n+1}^{infty}dfrac{a_k-b_k}{2^k}right|geqdfrac{1}{2^n}-dfrac{1}{2^n}=0
end{array}$$

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?

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 $delta_k=2epsilon_k$.

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, $|g(x)-g(y)|<epsilon.$

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.