# A set slightly larger than rational number?

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.

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.
• 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?