Let $C$ be a fat Cantor set and $\mathbb{Q}$ is set of rationals.
Q) Is it true that closure of intersection of $C$ and $\mathbb{Q}$ is $C$ that is
$$\overline{ C\cap \mathbb{Q}}=C?$$
Q) If the above is not true then can we atleast say that $\overline{ C\cap \mathbb{Q}}$ is a set of positive Lebesgue measure?
Best Answer
For David's comment: Start with $\big[{-\sqrt2},\sqrt2\,\big]$. Enumerate the rationals in $\big({-\sqrt2},\sqrt2\,\big)$. At the $n$th stage: remove a short interval (with irrational endpoints) containing the $n$th rational (if that rational remains). Result: $C\cap \mathbb Q = \varnothing$.
On the other hand, if you always remove intervals with rational endpoints, then the set of those endpoints remains as a subset of $C$ and is dense in $C$. So in that case $\overline{C \cap \mathbb Q} = C$.