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Thread: union

  1. #1
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    union

    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.
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    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).
    Are you saying that $\displaystyle \left( {\forall r \in Q} \right)\left[ {L_r = \left( { - \infty ,r} \right) \cup \left( {r,\infty } \right)} \right]$?
    If not, what in the world are you trying to ask?
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  3. #3
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    Yes. For example, $\displaystyle L_{1} = \mathbb{R} \ \backslash \{1 \} $, $\displaystyle L_{1/2} = \mathbb{R} - \backslash \ \{1/2\} $ etc...
    Last edited by particlejohn; Sep 6th 2008 at 07:46 PM.
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  4. #4
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    I think its similar to Cantor's diagonal method. At the last step you will still have one hole to fill for the union?
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  5. #5
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    or because it is infinite and so you will not have a hole?
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  6. #6
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    If you consider this $\displaystyle 2 \in L_1 \;\& \;1 \in L_2 \; \Rightarrow \;L_1 \cup L_2 = \mathbb{R}$, it will follow at once that $\displaystyle \bigcup\limits_{r \in \mathbb{Q}} {L_r } = \mathbb{R}$.
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