Question 409468
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If *[tex \Large x\ \in\ B] then *[tex \Large B\ \cup\ {x}\ =\ B] and then *[tex \Large (B\ \cup\ {x})\ -\ {x}\ =\ B\ -\ {x}] and then *[tex \Large B\ -\ {x}] contains all the elements of *[tex \Large B] except *[tex \Large x].  Then *[tex \Large B\ -\ {x}] is a proper subset of B, and is therefore a subset of B by definition of subset.


If *[tex \Large x\ \notin\ B] then *[tex \Large (B\ \cup\ {x})\ -\ {x}\ =\ B].  But any set is a subset of itself by definition of subset, therefore *[tex \Large (B\ \cup\ {x})\ -\ {x}\ =\ B\ \subseteq\ B].  




John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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