Exercise: Let $G$ be a Cantor set.
a) Prove that $G$ has a countable dense subset.
b)If $X$ is a countable subset of $G$ then $G\setminus X$ is dense in $G$
a)
It is know that the Cantor set for example the ternary Cantor set has the same cardinality of $\mathbb{R}$ despite its length is $0$. Since $\mathbb{Q}$ are dense set in $\mathbb{R}$ and $G$ is a subset of $\mathbb{R}$ with uncountable elements hence $G\cap\mathbb{Q}$ must be a countable dense subset of $G$.
b) I have no concrete idea on how to taclkle the question of countability and density in an abstract set $X$.
Questions:
1) Is my proof insofar right? If not. Why not?
2) How is b) solved?
Thanks in advance!
Best Answer
A compact metric space is separable, so the Cantor set is too. For a concrete dense set in the Cantor set: take all the end points of the intervals in the construction: at stage $0$ we have $[0,1]$ and we add $0$ and $1$ to the dense set; at stage $1$ we have $[0,\frac13]$ and $[\frac23,1]$ and we add $\frac13,\frac23$ to the dense set; at the next stage we add $\frac{1}{9}, \frac29, \frac79, \frac89$ etc. In the end we get all "rationals" in the ternary expansion: all those numbers (that have only $0$ and $2$'s in a ternary expansion) that are eventually periodic.
All non-empty open subsets of $C$ are uncountable (they all contain sets homeomorphic to the whole Cantor set, even), so all of them intersect the complement of any countable set.