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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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