Algebra of sets problem

**namelessguy** · January 26th 2009, 11:58 PM

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.

**clic-clac** · January 27th 2009, 01:39 AM

$\bigcap B=\bigcap_{x\in B}x$ 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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