Perfect set with empty interior

Question

Let $G\subset (-1,1)$ be a perfect, compact set with an empty interior. Since $(-1,1)\setminus G$ is an open set we can write it as follows $(-1,1)\setminus G=\cup (a_i,b_i)$, es decir $$G=(-1,1)\setminus\bigcup_{i=1}^\infty (a_i,b_i)$$ Let $Y=\{a_1,b_1,a_2,b_2,\dots\}$. I have to prove that $$ s\in G\setminus Y\Leftrightarrow\forall\epsilon>0: G\cap (s-\epsilon,s)\neq\emptyset\text{ and }G\cap (s,s+\epsilon)\neq\emptyset $$ When I see the set $G$, with these characteristics, the first thing I imagine is the Cantor set and I think it is possible to obtain a homeomorphism between them under these hypotheses, what I do not know is if that is enough to conclude what I need. Any ideas?

Answer 1

If $s \in G$ and for some $\epsilon \in (0,1-s)$, the interval $(s, s+\epsilon)$ is disjoint from $G$, then it is contained in the disjoint union $\bigcup_i (a_i,b_i)$. Since $s\in G$ this forces the existence of $i$ such that $s=a_i$.

The analysis of the case $(s-\epsilon,s) \cap G=\emptyset$ is similar.