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Math Help - union

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    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?
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  3. #3
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    Yes. For example,  L_{1} = \mathbb{R} \ \backslash \{1 \} ,  L_{1/2} = \mathbb{R} -  \backslash \ \{1/2\} etc...
    Last edited by particlejohn; September 6th 2008 at 08:46 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 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}.
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