Where does the following reasoning that $\mathbb{Q}$ is supposedly a $G_\delta$ set fail?

"Proof": $\mathbb Q$ may be covered by selecting open sets $O_n$ such that $m(O_n)<\frac{1}{n}$ for all $n \in \mathbb{N}$ since $\mathbb{Q}$ is a (Lebesgue) measurable set of measure zero. Intersect these sets to get $\mathbb{Q}$. Therefore $\mathbb{Q}$ is $G_\delta$. $\qed$

I know that this reasoning is not correct, since any such intersection will include irrationals (c.f. the Wikipedia article on $G_\delta$ sets). However, I do not understand why it is the case that any such intersection would include the irrationals. Why wouldn't the intersection of the above construction containing the rationals be of measure zero, since the limit infimum of the measures of all open covers of the rationals is zero?

  • 1
    $\begingroup$ The intersection is of measure zero. But that doesn't mean it is countable. It doesn't include "the irrationals", only some irrationals (uncountably many, but a set of measure $0$). $\endgroup$ – Daniel Fischer Nov 7 '14 at 15:30
  • $\begingroup$ Why would the intersection be equal to ${\bf Q}$??? It contains it, sure, but there's no reason to believe that there's nothing else. $\endgroup$ – tomasz Nov 7 '14 at 15:30

$\forall_n \mathbb{Q} \subseteq O_n$, so indeed $\mathbb{Q} \subseteq \cap_n O_n$. The reverse is shown nowhere. The measure is indeed $0$, by the "continuity from above" property of any measure. But sets of measure $0$ can be uncountable (think of the Cantor middle third set, e.g.). So that also does not settle that.

In fact, the argument why it cannot be equal to the rationals, is essentially topological: any $G_\delta$ subset of a complete metric space (Or locally compact Hausdorff space, if you proper; the reals are both) is Baire, where a topological space $X$ is Baire iff every countable intersection of open and dense subsets of $X$ is dense in $X$. And $\mathbb{Q}$ is not Baire: for every rational $q$, the set $O_q = \mathbb{Q}\setminus\{q\}$ is open and dense in $\mathbb{Q}$ while their intersection is empty (so not dense).

So why there are irrationals in $\cap_n O_n$: if not, $\mathbb{Q} = \cap O_n$, and then $\mathbb{Q}$ would have been Baire, contradiction.


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