Open superset of $\mathbb{Q}$ 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?

(I assume you are working in the topological space $\mathbb{R}$)
There's a nice trick to this with measure theory. By using the fact that the rationals are countable — and thus enumerable — you can construct $S$ by picking a positive real number $m$ and defining $S$ to be the union of


*

*An open set of length $m/2$ containing on the first rational 

*An open set of length $m/4$ containing on the second rational

*An open set of length $m/8$ containing on the third rational

*An open set of length $m/16$ containing on the fourth rational

*$\vdots$


Then the total measure of $S$ is clearly no greater than the sum of the lengths of these intervals — i.e. $m$. In particular, it is finite.
Therefore, the measure of the corresponding $T$ must be infinite, and thus uncountable.
