# Smallest open set in $\mathbb{R}$ that is supset of $\mathbb{Q}$

My problem is that intuitively I would think the whole $\mathbb{R}$ is the only open supset of $\mathbb{Q}$. However this is not true since I can take out for example $\pi$ an I still have an open subset. Now I have two questions, first how to construct a minimal open supset of $\mathbb{Q}$. And the other one is this set countable?

The reason I come up with this is, that I shall prove that the smallest open supset of any given subset of $[0,1]$ has the same Lebesgue measure as the original set.

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There is no "smallest open supset" of a given set in general. What could be the smallest open supset of $\{0\}$? – Xabier Domínguez May 3 '12 at 12:40
You can show in fact that given any open subset of $\mathbb{R}$ containing $\mathbb{Q}$, you can remove a non-rational point and still have an open subset, so there is an infinite descending chain. – Matthew Pressland May 3 '12 at 12:40

There are infinitely many open supsets of $\mathbb Q$ which are strict subsets of $\mathbb R$. To build one of these, fix $x\gt0$ and consider an enumeration $(q_n)_{n\geqslant1}$ of $\mathbb Q$. Then, $$U_x=\bigcup_{n\geqslant1}(q_n-2^{-n}x,q_n+2^{-n}x)$$ is an open subset of $\mathbb R$ such that $\mathbb Q\subset U_x$. Furthermore, the Lebesgue measure of $U_x$ is at most $\sum\limits_n2^{-n}x=x$ which is finite, hence $U_x\ne\mathbb R$.

Furthermore, the family $(U_x)_{x\gt0}$ is nondecreasing and the Lebesgue measure of $U_x$ is at least $x/2$ hence $U_x\ne\mathbb Q$ for every $x$ and $U_x\ne U_y$ for infinitely many $x\ne y$.

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You might be thinking along the lines of the closure operator: $\bar A$ is the smallest closed set containing $A$. Here "smallest" is in the sense of the partial order of inclusion. The reason such a smallest set exists is because the intersection of all closed sets containing $A$ is still closed, so it represents a minimal element. In general however, there is no minimal open set containing a given set, because the intersection of all open sets containing it is no longer in general open. That's why there's no open version of the closure. Similarly, there is no largest closed set contained in a given set, even though there is a largest open set (called the interior).

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Thanks, your answer helped as well, but I can't accept two answers. – Haatschii May 3 '12 at 17:09

Every non-empty open set contains an open interval and is therefore uncountable.

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