Solution to 5.2.4

Solution to 5.2.4 - the purported identity isn't always...

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INFS 501 - Solution to 5.2.4 Problem 5.2.4 asks us to fill in the missing parts of an "is an element of" proof of an obvious fact about sets - It's obvious when you draw a picture. She is asking for a rigorous proof, using the definitions of union and subset. The "is an element of" approach works well here. However, I personally find counterproductive her practice of injecting into a proof parenthetical remarks about what the proof is trying to do. (The Venn Diagram approach that we did in class is best for solving a different type of problem: either calculating whether a not-so- obvious identity holds "for all sets" or finding a counterexample if
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Unformatted text preview: the purported identity isn't always true.) 5.2.4 The following is a proof that for all sets A and B, if A B, then A B B. Proof: Suppose A and B are any sets and A B. [ We must show that A B B. ] ] Let x A B. [ We must show that x B. ] By the definition of the union A B, either: (i) x A, or ii) x B. In the first case, x A, so by the definition of A B, x B. In the second case, x B. Thus, in both cases, x B. We have shown in general that if x A B, then x B. Therefore, by the definition of subset, A B B....
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This note was uploaded on 11/01/2010 for the course INFS 501 taught by Professor Ellis,w during the Spring '08 term at George Mason.

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