I have the following problem:
Consider the set $M=\{0,2\}^\mathbb{N}$ and the map $f:M\rightarrow C$ where C is the Cantor set such that $f(a)=\sum_{n=1}^\infty \frac{a_n}{3^n}$ and show that f is continuous.
I somehow have no idea how to show this because I can't work with open sets, do I?
Thank you for your help.
On
It followed from the discussion in the comments that $\{ 0, 2 \}^{\mathbb{N}}$ is to be equipped with the standard product topology.
Sketch: first show that sets of the form $\left[ \frac{k}{3^n}, \frac{k+1}{3^n} \right] \cap C$, where $k = \sum \limits_{i=1}^n \frac{a_i}{3^i}$ for some $a_i \in \{ 0, 2 \}$, form a basis of the Cantor set. So it suffices to check that the preimages of these sets through $f$ are open in $\{ 0, 2 \}^{\mathbb{N}}$. Then prove that if $k = \sum \limits_{i=1}^n \frac{a_i}{3^i}$, then
$$f^{-1} \left[ \left[ \tfrac{k}{3^n}, \tfrac{k+1}{3^n} \right] \cap C \right] = \left\{ (x_n)_{n \in \mathbb{N}} \in \{ 0, 2 \}^{\mathbb{N}} : \bigwedge_{i=1}^n x_i = a_i \right\} $$
which by definition is open in $\{ 0, 2 \}^{\mathbb{N}}$, so it will finish the proof.
Together with @Louis Pan I remarked that on M we have the discrete topology, i.e. every set is open. So we can say that the preimage of every set in $C$ is open, more precisly this holds for all open sets in $C$, therefore $f$ is continuous.