# union

• Sep 6th 2008, 07:08 PM
particlejohn
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.
• Sep 6th 2008, 07:25 PM
Plato
Quote:

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?
• Sep 6th 2008, 07:28 PM
particlejohn
Yes. For example, $L_{1} = \mathbb{R} \ \backslash \{1 \}$, $L_{1/2} = \mathbb{R} - \backslash \ \{1/2\}$ etc...
• Sep 6th 2008, 07:45 PM
particlejohn
I think its similar to Cantor's diagonal method. At the last step you will still have one hole to fill for the union?
• Sep 6th 2008, 09:42 PM
particlejohn
or because it is infinite and so you will not have a hole?
• Sep 7th 2008, 05:22 AM
Plato
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 $\bigcup\limits_{r \in \mathbb{Q}} {L_r } = \mathbb{R}$.