1. ## union

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

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

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

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

2. Originally Posted by particlejohn
For each $r \in \mathbb{Q}$, let $L_r = \{r \}_{\mathbb{R}}^{\mathcal{C}}$ (e.g. the complement of the set of a rational).
Are you saying that $\left( {\forall r \in Q} \right)\left[ {L_r = \left( { - \infty ,r} \right) \cup \left( {r,\infty } \right)} \right]$?
If not, what in the world are you trying to ask?

3. Yes. For example, $L_{1} = \mathbb{R} \ \backslash \{1 \}$, $L_{1/2} = \mathbb{R} - \backslash \ \{1/2\}$ etc...

4. I think its similar to Cantor's diagonal method. At the last step you will still have one hole to fill for the union?

5. or because it is infinite and so you will not have a hole?

6. If you consider this $2 \in L_1 \;\& \;1 \in L_2 \; \Rightarrow \;L_1 \cup L_2 = \mathbb{R}$, it will follow at once that $\bigcup\limits_{r \in \mathbb{Q}} {L_r } = \mathbb{R}$.