Prove that there exists $a \in \mathbb{R}$, such that set $\{a+x: x\in K\}$ is included in set of irrational numbers. Let $K$ be closed subset of $\mathbb{R}$ with empty interior.
Prove that there exists $a \in \mathbb{R}$, such that set $\{a+x: x\in K\}$ is included in set of irrational numbers.
Well, I know that Baire theorem is what should solve that problem. My problems stem from that I am confused how $K$ can look like. Surely we cannot have all irrational numbers included in it. But can we have countably many of them? I suspect we cannot, probably because it would contradict with the fact (no ontological obligations here) that it has empty interior. But what are implications, if any, of that?
 A: Your intuition is incorrect. $K$ can contain infinitely many - even uncountably many! - irrationals and still be a closed set with empty interior (and I'm not sure what "ontological obligations" means in this context). 
For an example of a closed set with empty interior containing infinitely many irrationals, consider $\{z+\pi: z\in\mathbb{Z}\}$. This can be modified to be compact, in addition, without much effort. 
For an example of a closed set with empty interior containing uncountably many irrationals, think of the Cantor set - although note that it also contains lots of rationals. Coming up with a closed set with empty interior containing uncountably many irrationals and not containing any rationals is a harder task, which I leave as an exercise.

For solving the problem, here's a hint: for $q\in\mathbb{Q}$, let $A_q=\{r\in\mathbb{R}: q\not\in K+r\}$. Then:


*

*How many sets of the form $A_q$ are there?

*What kind of set, topologically, is $A_q$ (open? closed? neither?)? Use the fact that $K$ is closed.

*What kind of set, distribution-wise, is $A_q$ (dense? not dense? nowhere dense?)? Use the fact that $K$ has empty interior.

*Finally, what does the Baire category theorem say about $\bigcap A_q$, and how does that solve your problem?
