By Randall R. Holmes

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Then x ∈ A∈A A, so that x ∈ A for every A ∈ A. In particular, x ∈ A0 , so x ∈ X ∪ A0 . This shows that x ∈ X ∪ A for every A ∈ A, that is, x ∈ A∈A (X ∪ A). The inclusion (⊆) follows. (⊇) Let x ∈ A∈A (X ∪ A). Then x ∈ X ∪ A for every A ∈ A. Assume that x ∈ / X. Let A0 ∈ A. We have x ∈ X ∪ A0 and x ∈ / X, so x ∈ A0 . This shows that x ∈ A for every A ∈ A, that is, x ∈ A∈A A. Therefore, x∈X∪ A∈A A . The containment (⊇) follows. Discussion: In the proof of the inclusion (⊆), after letting x be an arbitrary element of the left-hand side and writing what this means we turn to the task of showing that x must be an element of the right-hand side, that is, x ∈ X ∪ A for every A ∈ A.

Prove: n = 3m − 9 for some m ∈ Z iff 2n ∈ 6Z, where 6Z = {6x | x ∈ Z}. Proof (⇒) Assume that n = 3m − 9 for some m ∈ Z. We have 2n = 6m − 18 = 6(m − 3) ∈ 6Z. (⇐) Assume that 2n ∈ 6Z. Then 2n = 6x for some x ∈ Z. Put m = x + 3. Then m ∈ Z and n = 3x = 3(x + 3) − 9 = 3m − 9. 3 Example P(X) ⊆ P(Y ). Let X and Y be sets. Prove: X ⊆ Y if and only if Proof (⇒) Assume that X ⊆ Y . Let S ∈ P(X). Then S ⊆ X ⊆ Y . Therefore, S ∈ P(Y ). It follows that P(X) ⊆ P(Y ). 51 (⇐) Exercise 4–13. Occasionally the most efficient way to prove an if-and-only-if statement is to use a string of double implications (⇔).

Multiplying both sides of this inequality by 2 gives 2n ≤ 10, and then subtracting 1 from both sides gives 2n − 1 ≤ 9. The statement says [ If n ≤ 5, . . ] , so it is making a claim only in the situation where n ≤ 5. Therefore, the proof begins by explicitly making this assumption: Assume that n ≤ 5. If we let P stand for the statement n ≤ 5 and Q stand for the statement 2n − 1 ≤ 9, then the statement can be written [ If P , then Q ] . This is the general form of an if-then statement. We record here the method for proving such a statement.