Mathematics 242 – Principles of Analysis Selected Solutions – Problem Set 1 September 13, 2000

§1.3/12 c. We want to find the smallest c such that

C = {x||x − 1| + |x − 13| + |x − 31| + |x − 131| = c}

is not empty, and determine the set of x satisfying |x−1|+|x−13|+|x−31|+|x−131| = c for this smallest c value. Thinking of |x − a| as the distance from x to a along the real number line, we can see that if x ∈ [13, 31], then |x − 1| + |x − 13| + |x − 31| + |x − 131| equals the total distance between 1 and 131, plus the total distance between 13 and 31, or 130 + 18 = 148. We claim that c = 148 is the smallest value for which C ≠ ∅. To see this, suppose first that x < 13. Then the distance from x to 13, plus the distance from x to 31 is |x − 13| + |x − 31| > 31 − 13 = 18. Similarly, |x − 1| + |x − 131| ≥ 131 − 1 = 130. Hence |x − 1| + |x − 13| + |x − 31| + |x − 131| > 130 + 18 = 148. Similarly, if x > 31, we get |x − 1| + |x − 13| + |x − 31| + |x − 131| > 148. Hence c = 148, and the set C for this c value is the closed interval [13, 31].

§1.5/5. Suppose that A is bounded above in R, and let α = lub(A). Let B = {x+k|x ∈ A}. We want to show that lub(B) = α + k. To do this we need to show first that α + k is an upper bound for B, and second that any other upper bound for B is greater than or equal to α + k.

Suppose x ∈ A. Then by definition α is an upper bound for A, so α ≥ x for all x ∈ A. If we add k to both sides of this inequality, we obtain another true inequality: α + k ≥ x + k, valid for all x ∈ A. Hence α + k is an upper bound for B.

To show that every other upper bound of B is greater than or equal to α + k, we can argue by contradiction. Namely suppose that β ≥ x + k for all x ∈ A, but β < α + k. This implies β − k < α. Since α = lub(A), this implies that β − k is not an upper bound for A, so there exists x ∈ A with x > β − k. But this shows x + k > β. This is a contradiction since we assumed that β was an upper bound for the set B. Hence every upper bound of B is greater than or equal to α + k, and lub(B) = α + k = lub(A) + k.

The proof for the greatest lower bound is similar, reversing all inequalities.

§1.5/6. a) Suppose that A is bounded above in R, and let α = lub(A). Let k > 0, and let B = {kx|x ∈ A}. We want to show that lub(B) = kα. To do this we need to show first that kα is an upper bound for B, and second that any other upper bound for B is greater than or equal to kα.

Suppose x ∈ A. Then by definition α is an upper bound for A, so α ≥ x for all x ∈ A. Since we assume k > 0, when we multiply k on both sides of this inequality, we obtain another true inequality: kα ≥ kx, valid for all x ∈ A. Hence kα is an upper bound for B.

To show that every other upper bound of B is greater than or equal to kα, we can argue by contradiction. Namely suppose that β ≥ kx for all x ∈ A, but β < kα. This implies β/k < α (again, k > 0, so 1/k > 0, so multiplying both sides by 1/k preserves the inequality). Since α = lub(A), this implies that β/k is not an upper bound for A, so there exists x ∈ A with x > β/k. But this shows kx > β. This is a contradiction since we assumed that β was an upper bound for the set B. Hence every upper bound of B is greater than or equal to kα, and lub(B) = kα = k · lub(A).

The proof for the greatest lower bound is similar, reversing all inequalities.

b) If k < 0, then lub(B) = k · glb(A) and glb(B) = k · lub(A).

§1.5/8 b. We are given that A and B are bounded subsets of R such that for all x ∈ A and all y ∈ B, x ≤ y. Let α = lub(A) and β = glb(B). We want to show that α ≤ β. Suppose on the other hand that β < α. Since α = lub(A), this implies that β is not an upper bound for A. Hence there exists x ∈ A such that β < x ≤ α. But then since β = glb(B), x is not a lower bound for B, so there exists some y ∈ B such that β ≤ y < x ≤ α. But this is a contradiction of the hypothesis that x ≤ y for all x ∈ A and all y ∈ B. Hence α ≤ β.