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Thread: Set Theory.

  1. #1
    Junior Member
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    Set Theory.

    Hi, could anyone help me start off this proof:

    Notation: For a set $\displaystyle X$ we define $\displaystyle \bigcup\ X \;=\;\{x \mid x\in y\ for\ some\ y\in X \}$

    Then show if $\displaystyle X_{ij}$ for $\displaystyle i,j\in\mathbb{N}$ are sets then:

    $\displaystyle \bigcap\limits^\infty_{i=0}(\bigcup\limits^\infty_ {j=0}X_{ij})\;=\;\bigcup\{( \bigcap\limits^\infty_{i=0}X_{ih(i)})\mid h\in\mathbb{N^{N}}\}$
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  2. #2
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    If $\displaystyle x\in LHS$ then $\displaystyle \left( {\forall i} \right)\left( {\exists j} \right)\left[ {x \in X_{i~j} } \right]$.
    We are going to use that to define $\displaystyle h:\mathBB{N}\to\mathBB{N}$ by $\displaystyle h:i\mapsto j$.
    So $\displaystyle x\in X_{i~h(i)}$.
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  3. #3
    Junior Member
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    Thanks Plato, I think understand what you're saying but doesn't this only give $\displaystyle \bigcap\limits^\infty_{i=0}(\bigcup\limits^\infty_ {j=0}X_{ij})\;\subseteq\;\bigcup\{( \bigcap\limits^\infty_{i=0}X_{ih(i)})\mid h\in\mathbb{N^{N}}\}$?
    If so, how would I prove $\displaystyle \supseteq$
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