Assume that $C$ is a Cantor Set defined in the following way: $$C = \left\{\sum_{n = 1}^{\infty}\frac{a_n}{3^n} : a_i \in \{0 ,2\}\right\} \subset [0, 1] =: I$$ with induced topology.
I want to show that the following function is continuous: $$ f: C \to I \times I $$ $$ \sum_{n = 1}^{\infty}\frac{a_{n}}{3^n} \overset{f}{\mapsto} \left(\sum_{n = 0}^{\infty}\frac{a_{2n + 1}/2}{2^n}, \sum_{n = 1}^{\infty}\frac{a_{2n}/2}{2^n}\right) \iff 0.a_1a_2\dots_3 \overset{f}{\mapsto} \left( 0.\frac{a_1}{2} \frac{a_3}{2} \dots_2; 0.\frac{a_2}{2}\frac{a_4}{2} \dots_2 \right) $$ To do it by definition i need to show, that for every open subset $(a , b) \times (c, d) \subset I \times I$ the inverse image $f^{-1}[(a , b) \times (c, d)]$ is open subset in $C$. But can I find the inverse image of this subset directly ?
UPDATED: Here is my attempt: $$(0.x_1x_2\dots_2, 0.y_1y_2\dots_2) \in (a , b) \times (c, d) $$ Then the following inequalities hold: $$ \sum_{n = 1}^{\infty}\frac{a_n}{2^n} < \sum_{n = 1}^{\infty}\frac{x_n}{2^n} < \sum_{n = 1}^{\infty}\frac{c_n}{2^n} $$ $$ \sum_{n = 1}^{\infty}\frac{b_n}{2^n} < \sum_{n = 1}^{\infty}\frac{y_n}{2^n} < \sum_{n = 1}^{\infty}\frac{d_n}{2^n} $$ But then: $$ \sum_{n = 1}^{\infty}\frac{a_n}{3^{2n - 1}} < \sum_{n = 1}^{\infty}\frac{x_n}{3^{2n - 1}} < \sum_{n = 1}^{\infty}\frac{c_n}{3^{2n - 1}} $$ And also: $$ \sum_{n = 1}^{\infty}\frac{b_n}{3^{2n}} < \sum_{n = 1}^{\infty}\frac{y_n}{3^{2n}} < \sum_{n = 1}^{\infty}\frac{d_n}{3^{2n}} $$ So we can conclude: $$ \sum_{n = 1}^{\infty}\frac{a_n}{3^{2n - 1}} + \sum_{n = 1}^{\infty}\frac{b_n}{3^{2n}} < \sum_{n = 1}^{\infty}\frac{x_n}{3^{2n - 1}} + \sum_{n = 1}^{\infty}\frac{y_n}{3^{2n}} < \sum_{n = 1}^{\infty}\frac{c_n}{3^{2n - 1}} + \sum_{n = 1}^{\infty}\frac{d_n}{3^{2n}} $$ $$ \Rightarrow \sum_{n = 1}^{\infty}\frac{a_n}{3^{2n - 1}} + \sum_{n = 1}^{\infty}\frac{b_n}{3^{2n}} <f^{-1}(x) < \sum_{n = 1}^{\infty}\frac{c_n}{3^{2n - 1}} + \sum_{n = 1}^{\infty}\frac{d_n}{3^{2n}} $$ So $f^{-1}[ (a , b) \times (c, d)] = C \cap \left( \sum_{n = 1}^{\infty}\frac{a_n}{3^{2n - 1}} + \sum_{n = 1}^{\infty}\frac{b_n}{3^{2n}};\sum_{n = 1}^{\infty}\frac{c_n}{3^{2n - 1}} + \sum_{n = 1}^{\infty}\frac{d_n}{3^{2n}} \right)$ - open subset of Cantor Set in induced topology.