By Randall R. Holmes
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B) {{1, 2, 3}} ⊆ A, but {1, 2, 3} A. (c) {X ∈ A | 1 ∈ X} in list form is {{1, 2, 3}, {1, 2, 5, 6}}. (d) {X ∪ Y | X, Y ∈ A} in list form is {{1, 2, 3}, {1, 2, 3, 4}, {1, 2, 3, 5, 6}, {2, 4}, {1, 2, 4, 5, 6}, {1, 2, 5, 6}}. 38 Discussion: In part (d), we let X and Y range through all elements of A and in each case write X ∪ Y . With A = {1, 2, 3}, B = {2, 4}, and C = {1, 2, 5, 6} as before, the indicated set is {A ∪ A, A ∪ B, A ∪ C, B ∪ B, B ∪ C, C ∪ C}. This yields all of the possibilities since, for instance, B ∪ A = A ∪ B.
4 – Exercises 4–1 Let n ∈ Z. Prove: If n ≥ 3, then 4n − 5 ≥ 7. 4–2 Let y ∈ R. Prove: If y ∈ {2x + 3 | x ∈ (−∞, 1)}, then y − 1 < 4. 4–3 Prove: For every n ∈ Z, if n ≤ 2, then n ∈ {x ∈ Z | 4 − 3x > −3}. 60 4–4 Let n ∈ N. Prove, using a string of implications: If n > 2, then 2n+1 > 8. 4–5 Put A = {x ∈ R | 2(x − 1)2 < 1}. Let r ∈ R. Prove, using a string of implications: r + 1 ∈ A =⇒ r4 < 1/4. 4–6 Define h : R → R by h(x) = 5x − 2. Prove: For every r ∈ R, if h(r) ∈ Q, then r ∈ Q. 4–7 Let r ∈ R. Prove: If r2 − 2r ≥ 3, then r ≥ 3 or r ≤ −1.
We have x < n0 + 1 < n0 + 2, so ∞ x ∈ [n0 , n0 + 2) = An0 . Therefore, x ∈ n=0 An since An0 is one of the An in this union. The containment (⊇) follows. 6 Example For r ∈ R, put Ar = [r − 1, r + 1]. We have Ar = (−1, ∞), r∈R+ Ar = [0, 3], r∈[1,2] Ar = [1, 2]. r∈[1,2] 43 Discussion: As in the preceding example, these equalities are arrived at by picturing the intervals Ar on the number line as r ranges through the indicated sets. Here is a formal proof of the equation r∈[1,2] Ar = [1, 2]: (⊆) Let x ∈ r∈[1,2] Ar .
Introduction to Advanced Mathematics by Randall R. Holmes
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