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Let $(r_n)_{n \ge 1}$ be the set of rational numbers. For any given $k \ge 1$, define:

$$G_k = \bigcup_{n = 1}^{\infty} (r_n - d_{k, n} / 2, r_n + d_{k, n} / 2)$$

where

$$d_{k, n} = \frac{1}{k}\frac{1}{2^n}$$

then clearly $G_k$ has finite measure, indeed $m(G_k) \lt \frac{1}{k}$. Also, note that $(G_k)_{k \ge 1}$ is a decreasing sequence of open sets. Since $\mathbb{Q}$ is not a $\mathcal{G}_{\delta}$ set, we cannot have $G = \bigcap_{k = 1}^{\infty} G_k = \mathbb{Q}$, however, $m(G) = 0$ by the limit property of measure.

My question is, what lies in the set $G \setminus \mathbb{Q}$ (which is a non-empty zero measure set)?

I suppose the answer should rely on the ordering of rational numbers. So examples of $G \setminus \mathbb{Q}$ with any ordering of rational numbers are welcomed.

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A lot is in there! Since each $G_k$ is open dense, their intersection $G$ is a dense $G_\delta$ set, i.e. comeager. So in particular it is an uncountable superset of the rationals. In fact $G\setminus \mathbb{Q}$ is also comeager, so it is still dense.

If your answer is what number specifically are in there, that is harder to say.

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  • $\begingroup$ Hi, thanks for your reply! And yes, I would like a solid example. BTW, I don't quite follow here, I know that $G \setminus \mathbb{Q}$ is comeager and dense, but why is it uncountable? $\endgroup$
    – Hans
    Commented Jun 16, 2023 at 2:54
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    $\begingroup$ Countable sets are meager. $\endgroup$ Commented Jun 16, 2023 at 3:46

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