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Thread: Algebra of sets problem

  1. #1
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    Algebra of sets problem

    I'm having some difficulty with these two problems,any help is appreciated.
    1)Prove that $\displaystyle A \cup \bigcap B=\bigcap \{A\cup X \mid X\in B\}$ given $\displaystyle B$ is not empty
    Since B is not empty, let $\displaystyle X\in B$, then it follows $\displaystyle X \in \bigcap B$, so LHS=$\displaystyle A \cup X$ such that $\displaystyle X \in B$. This doesn't look right, so anyone can help me figure out what I did wrong here.
    2) Is it true that $\displaystyle A \cup \bigcup B= \bigcup \{A\cup X \mid X \in B\}$? 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 $\displaystyle X \in B$. Is my argument correct? I can't find condition such that equality holds.
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  2. #2
    Senior Member
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    $\displaystyle \bigcap B=\bigcap_{x\in B}x$ isn't it?

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

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