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Thread: [SOLVED] set proof help

  1. #1
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    [SOLVED] set proof help

    Can anyone prove or disprove
    $\displaystyle X \cap Z \subseteq Y \Leftrightarrow (X - Y) \cup (Y-Z) \subseteq Z^c$

    I think you have to split it up into two parts, but whenever I do it one way I can't get the same exact argument to go back the other way.
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  2. #2
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    Quote Originally Posted by horan View Post
    Can anyone prove or disprove
    $\displaystyle X \cap Z \subseteq Y \Leftrightarrow (X - Y) \cup (Y-Z) \subseteq Z^c$

    I think you have to split it up into two parts, but whenever I do it one way I can't get the same exact argument to go back the other way.
    Hi

    Suppose that $\displaystyle X \cap Z \subseteq Y$
    Let's prove that $\displaystyle (X - Y) \cup (Y-Z) \subseteq Z^c$

    Let $\displaystyle x \in (X - Y) \cup (Y-Z)$
    We have to prove that $\displaystyle x \in Z^c$

    $\displaystyle x \in (X - Y) \cup (Y-Z)$ means that or $\displaystyle x \in (Y-Z)$ or $\displaystyle x \in (X - Y)$
    If $\displaystyle x \in (Y-Z)$ then $\displaystyle x \notin Z$ and therefore $\displaystyle x \in Z^c$ => OK
    If $\displaystyle x \in (X - Y)$ then $\displaystyle x \in X$ and $\displaystyle x \notin Y$. If $\displaystyle x \in Z$ then $\displaystyle x \in X \cap Z$. But $\displaystyle X \cap Z \subseteq Y$ therefore $\displaystyle x \in Y$ which is not possible since $\displaystyle x \notin Y$. Threfeore $\displaystyle x \notin Z$.
    In both cases $\displaystyle x \in Z^c$.
    Therefore $\displaystyle (X - Y) \cup (Y-Z) \subseteq Z^c$.
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  3. #3
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    Set Theory Proof

    Hello horan
    Quote Originally Posted by horan View Post
    Can anyone prove or disprove
    $\displaystyle X \cap Z \subseteq Y \Leftrightarrow (X - Y) \cup (Y-Z) \subseteq Z^c$

    I think you have to split it up into two parts, but whenever I do it one way I can't get the same exact argument to go back the other way.
    running-gag has proved that $\displaystyle X \cap Z \subseteq Y \Rightarrow (X - Y) \cup (Y-Z) \subseteq Z^c$. Here's the proof of $\displaystyle X \cap Z \subseteq Y \Leftarrow (X - Y) \cup (Y-Z) \subseteq Z^c$.

    Suppose that $\displaystyle (X - Y) \cup (Y-Z) \subseteq Z^c$and that $\displaystyle x \in X\cap Z$. Therefore we must show that $\displaystyle x \in Y$.

    $\displaystyle x \in X \cap Z \Rightarrow x \in Z $

    $\displaystyle \Rightarrow x \notin Z^c$

    $\displaystyle \Rightarrow x \notin (X - Y) \cup (Y-Z)$

    $\displaystyle \Rightarrow x \notin (X - Y)$

    But $\displaystyle x \in X$

    Therefore $\displaystyle x \in Y$

    Grandad
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  4. #4
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    hello running-gag
    ..If then ..
    how can you say like that..
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  5. #5
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    Hello doresa
    Quote Originally Posted by doresa View Post
    hello running-gag

    ..If then ..

    how can you say like that..
    You're just quoting part of the whole proof here. Obviously you can't say that, in general, $\displaystyle x \in Z \Rightarrow x \in X \cap Z$.

    But the bit you're quoting is from a line that starts: If $\displaystyle x \in (X-Y)$... In other words, it pre-supposes that $\displaystyle x \in X$. Therefore if in addition $\displaystyle x \in Z$, then $\displaystyle x \in X \cap Z$.

    Is that OK now?

    Grandad
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  6. #6
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    ok.thank you
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