TITLE: Open superset of $\mathbb{Q}$
QUESTION [5 upvotes]: Let $S$ be an open set such that $\mathbb{Q}\subset S$. We can also define a set $T=\mathbb{R}\setminus S$. I have been trying to prove or disprove whether $T$ could be uncountable. I suspect $T$ has to be at most countable, is my intuition correct?

REPLY [3 votes]: Expressed differently, the condition on $T$ is that it can have no rational elements and no rational limit points.
But even without measure theory, we can easily specify such a $T$ that is uncountable -- for example the set of all numbers in $(0,1)$ in which the $n$th digit of the decimal representation is $3$ when $n$ is a perfect square, and either $4$ or $5$ otherwise.