Give an example or argue that an example is impossible. A nonempty open set that contains all rational numbers.
This was a question on my undergrad Real Analysis exam. There is a similar question that has been asked on the site but I don't understand the answer given. We have not discussed topology at length. I want to say that it is impossible since I think $\mathbb{Q}$ is neither open nor closed. But I'm not sure how to explain this. Any help would be appreciated, thanks
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If you want strict subset then take the union of intervals $(\sqrt{2}+n-1,\sqrt{2}+n)$ for every integer $n$. Note that only points missing are the points of the form $\sqrt{2}+n$ are irrationals since $n$ is integer. So the union contains all the rational numbers and being a union of open sets is open.
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If you want an open subset containing only irrationnal numbers, it doesn't exist. Since it has to be opened, then for any $x$ it contains, it also contains a neighbourhood of $x$. This neighbourhood can be taken without loss of generality as $U=(x-\epsilon,x+\epsilon)$. Now, since both the rationnals and the irrationnals are dense in $\mathbb{R}$, then by definition there is at least a rationnal number in $U$.
It's kind of hard to talk about open sets without some conception of topology, so I'll try to give a crash course. A set is open iff there is an open ball around every element in the set. An open ball of radius $r$ centered at $x_0$ is the set of all $x$ st $\lvert x_0 - x \rvert < r$. In particular, all the open balls in $\mathbb R$ are the open intervals. For instance, the open interval (a, b) is an open ball centered around $\frac{a + b}{2}$ with radius $\frac{b-a}{2}$.
Now to answer your question, we are trying to find an open set- a set such that every point in the set is enclosed by an open interval- that contains all the rational numbers. That's why some of the answers are suggesting the real number line, which itself is open. Or, as pre-kidney suggested, take $\mathbb R$ and remove a point in $\mathbb Q$\ $\mathbb R$, like $\mathbb R$ \ $\sqrt2$. Then every point on $\mathbb R$ has an open interval around it except for $\sqrt{2}$, which isn't in the set anyways, so the set is open and contains all the rational numbers.