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Thread: Algebra of sets problem

  1. #1
    Nov 2007

    Algebra of sets problem

    I'm having some difficulty with these two problems,any help is appreciated.
    1)Prove that A \cup \bigcap B=\bigcap \{A\cup X \mid X\in B\} given B is not empty
    Since B is not empty, let X\in B, then it follows X \in \bigcap B, so LHS= A \cup X such that X \in B. This doesn't look right, so anyone can help me figure out what I did wrong here.
    2) Is it true that A \cup \bigcup B= \bigcup \{A\cup X \mid X \in B\}? When does equality hold?
    Here, can I take the empty set to be B, and the equality will be false because the empty set doesn't contain any element, but we need X \in B. Is my argument correct? I can't find condition such that equality holds.
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  2. #2
    Senior Member
    Nov 2008
    \bigcap B=\bigcap_{x\in B}x isn't it?

    If that's the definition, A\cup \bigcap B=A\cup \bigcap_{x\in B}x=\{y;\ y\in A \vee \forall x\in B, y\in x\}= \{y;\ \forall x\in B\ (y\in A \vee  y\in x)\}=\bigcap_{x\in B}A\cup X

    The difficulty is to show that \forall y\ ((y\in A \vee \forall x\in B,\ y\in x)\Leftrightarrow \forall x\in B\ (y\in A \vee  y\in x)) which is true because x has no occurence in the formula y\in A. (in the two cases, y is either in A or has to be in every x\in B)
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