For each $\displaystyle r \in \mathbb{Q} $, let $\displaystyle L_r = \{r \}_{\mathbb{R}}^{\mathcal{C}} $ (e.g. the complement of the set of a rational).

(a) Find $\displaystyle \bigcup_{r \in \mathbb{Q}} L_{r} $.

(b) Find $\displaystyle \bigcap_{r \in \mathbb{Q}} L_{r} $.

If we arrange the set of rationals in an ordered sequence, then $\displaystyle \bigcup_{r \in \mathbb{Q}} L_{r} = \mathbb{R} \ \backslash \ \{r_{n}\} $ and $\displaystyle \bigcap_{r \in \mathbb{Q}} L_{r} = I $ where $\displaystyle I $ are the irrational numbers?

Is this correct? Because for the union, you can fill in all the previous "holes" but will still be left with a hole.