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?
 A: Your intuition is reasonable, but incorrect. HINT: the rationals are countable, so we can list them as $\mathbb{Q}=\{q_1, q_2, q_3, . . .\}$ (obviously this listing isn't "in order," in any sense, but that's fine). 
Now, let $U_n=(q_n-2^{-n}, q_n+2^{-n})$. Each $U_n$ is open, so the union $V=\bigcup U_n$ is open, and clearly $V$ contains $\mathbb{Q}$. Do you see how to show that $\mathbb{R}\setminus V$ is uncountable?

If you are unfamiliar with Lebesgue measure, the above hint will be very difficult. An alternate approach is via the Baire category theorem: let $C$ be the Cantor set, and show that there is some real $r$ such that $C+r=\{x+r: x\in C\}$ contains no rational numbers. Then the complement of $C+r$ is open, contains the rationals, and has uncountable complement. HINT: To use BCT, show that for each rational $q$, the set $B_q=\{r: q\in C+r\}$ is nowhere dense. Then, since there are only countably many rationals, BCT implies that $\bigcup_{q\in\mathbb{Q}} B_q$ is meager, and hence $\not=\mathbb{R}$ . . .
A: 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.
A: (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.
A: AS rationals form a countable set, enumerate them, and take an open interval of length $\frac{\epsilon}{2^n}$ around the $n$-th rational number. Union of these open intervals is a candidate for $S$ in your question. The measure for this open set is less than $\epsilon$. So the complement has positive measure. Can it be countable?
