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satellite73Best ResponseYou've already chosen the best response.0
hard to write because saying it says the same thing. but i think the way to do it is to show that each is contained in the other pick some element on the left, say \(x\) and show that it is in the right an vice versa if it is in the left, then either \(x\in B\) or \(x\in \cap A_{\alpha\in I}\) meaning \(\exists\beta\in I\) such that \(x\in A_\beta\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
then if \(x\in B\) it must be in \(A_{\alpha}\cup B\) for any \(\alpha\) and if \(a\notin B\) then by the proceeding it must be in some \(A_{\beta}\) oh crap i got my modifier wrong!!
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
if \(x\in \cap A_{\alpha \in I}\) it is in ALL \(A_{\alpha}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
so \(\forall \alpha \in I\) we know \(x\in A_{\alpha}\) and therefore \(x\in \cap (A_{\alpha \in I}\cup B)\)
 one year ago
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