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  1. #1
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    sets

    Show that AnA'= nullset
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  2. #2
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    Quote Originally Posted by dcapdogg
    Show that AnA'= nullset
    If $\displaystyle A=\{ \}$ then proof is complete.
    If $\displaystyle A\not = \{\}$ then $\displaystyle x\in A$
    Then, $\displaystyle x\not \in A'$ thus, $\displaystyle A\mbox{ and }A'$ are disjoint thus, $\displaystyle A\cap A'=\{\}$
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  3. #3
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    Hello, dcapdogg!

    The proof depends upon what axioms and definitions are established.


    Show that $\displaystyle A \cap A' \:=\:\emptyset$

    $\displaystyle A \cap A'$ is the set of elements that are in $\displaystyle A$and in $\displaystyle A'$.

    By definition of complement: .If $\displaystyle x \in A$, then $\displaystyle x \not\!\!{\in} A'$
    . . Hence, there are no elements common to $\displaystyle A$and $\displaystyle A'.$ .They are disjoint.

    Therefore: .$\displaystyle A \cap A' \:= \:\emptyset$

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