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Thread: Small Set Theory Proof

  1. #1
    Aug 2008

    Small Set Theory Proof

    Let I be a nonempty set and for each $\displaystyle i\epsilon I $ let $\displaystyle X_{i} $ be a set.

    Prove that if each $\displaystyle X_{i} $ is a subset of a given set S, then $\displaystyle e( \bigcup_{i\epsilon I} X_{i}) = \bigcap_{i\epsilon I} e X_{i} $

    Please note that e represents the complement of a set.

    Thank you as always!

    Edit: Also note, of course, that we must show inclusion both ways.
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  2. #2
    Moo is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Hello !

    If $\displaystyle x \in e \left( \bigcup_{i\in I} X_i \right)$ we can say that $\displaystyle x \not \in \bigcup_{i \in I} X_i$.
    This means that $\displaystyle \forall i \in I ~,~ x \not \in X_i$ (by the definition of the union)
    This is equivalent to saying that $\displaystyle \forall i \in I~,~ x \in e\left(X_i\right)$
    Therefore $\displaystyle x \in \bigcap_{i \in I} e \left(X_i\right)$ (by the definition of the intersection)

    Conclusion : $\displaystyle x \in e \left( \bigcup_{i\in I} X_i \right) \Longleftrightarrow x \in \bigcap_{i \in I} e \left(X_i\right)$
    Thus $\displaystyle e \left( \bigcup_{i\in I} X_i \right) = \bigcap_{i \in I} e \left(X_i\right)$

    You have to check it, but I'm quite sure the reasoning works with equivalence and not implication, and hence the conclusion that doesn't go through the steps $\displaystyle A \subseteq B$ and $\displaystyle B \subseteq A$
    If you want to do the reverse order, it's the same reasoning.
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  3. #3
    MHF Contributor Matt Westwood's Avatar
    Jul 2008
    Reading, UK
    More rigorous to prove it by induction, of course.

    This is a standard result and it pays to study it. This is known as (one of) "De Morgan's Laws" - the other being with the union and intersection the other way round. I recommend a google session on it.
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