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What would be your example of an open proper subset $S \subsetneq \mathbb R$ such that $\mathbb{Q} \subsetneq S$?

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up vote 12 down vote accepted

An example of such a set is $$( - \infty , \sqrt{2} ) \cup ( \sqrt{2} , + \infty ) = \mathbb{R} \setminus \{ \sqrt{2} \}.$$ Of course, there is nothing special about $\sqrt{2}$, and choosing simply removing your favourite irrational number from $\mathbb{R}$ will result in another such set. Also, since any finite subset of $\mathbb{R}$ is closed, you can similarly remove any finite number of irrational numbers from $\mathbb{R}$ to obtain another example of such a set.

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a) $A_k:=\bigcup_{j=1}^{+\infty}(r_j-2^{-j-k},r_j+2^{-j-k})$ where $\{r_j\}$ is an enumeration of rationals. This show that such that set can be taken as small (in measure) as we want.

b) Take $O_n:=(-1/n,1-1/n)$; then $\{O_n,n\in\Bbb N^*\}$ is an open cover of $[0,1)$ which doesn't have any finite subcover.

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