Prove that any open interval $(a, b)$ that intersects C (Cantor's set) contains some subinterval of some $F_m$. Then conclude that $(a, b)$ also intersects the complement from C.
My try:
In other words what the first part is asking me to prove is:
If $(a,b)\cap$ C $\neq \emptyset \Rightarrow$ $[\frac{n}{3^m}, \frac{n+1}{3^m}] \subseteq (a,b)$ for some $n, m\in \mathbb{N}$
Let $\xi\in [\frac{n}{3^m}, \frac{n+1}{3^m}]$ ...
I don't know if it's the right way to go. Any hints would be great!
If $(a,b)\cap\mathbf C\neq\emptyset$, you can take some $c\in(a,b)\cap\mathbf C$. Take $m\in\mathbb Z_+$ so large that $a<c-3^{-m}$ and that $c+3^{-m}<b$. Now, take some $n$ such that $c\in\bigl[n3^{-m},(n+1)3^{-m}\bigr]$. Since the length of this interval is $3^{-m}$, it follows from the choice of $m$ that $\bigl[n3^{-m},(n+1)3^{-m}\bigr]\subset[a,b]$.