MATH 242 – Principles of Analysis Solutions for Problem Set 1 – due: Feb. 1

‘A‘ Section

1. Let A = {x ∈ R | x^{2} − 5x + 6 = 0}, B = (0, 4) = {x ∈ R | 0 < x < 4} and C = \{\frac{x}{x^{2}+1} | x ∈ R\} (Note: C is the range of the function f defined by f(x) = \frac{x}{x^{2}+1}.) a. Express the set C as a union of one or more closed intervals [a, b] in R. (Note:

You should use facts from calculus to solve this. Don’t worry that we have not justified them yet.) Solution: The function f(x) = \frac{x}{x^{2}+1} has f′(x) = \frac{1−x^{2}}{x^{2}+1}. This is = 0 at x = ±1. Moreover f′(x) < 0 for x < −1, f′(x) > 0 for −1 < x < 1 and f′(x) < 0 for x > 1. Therefore, at x = −1, f has a local minimum with f(−1) = −1/2. Similarly, at x = 1, f has a local maximum with f(1) = 1/2. We also see lim_{x→±∞} f(x) = 0. Hence f(−1) = −1/2 is also an absolute minimum, and f(1) = 1/2 is also an absolute maximum. We will show later in the course that every y with −1/2 < y < 1/2 is also in the range. Hence C = [−1/2, 1/2].

b. Find the sets A ∩ C and B ∩ C.

Solution: Since A = {2, 3}, we see that A ∩ C = ∅ and B ∩ C = (0, 1/2]. c. Find the sets B ∪ A and B ∪ C and express as unions of intervals in R.

Solution: We have B ∪ A = (0, 4) = B, since A ⊂ B. Then by part a, B ∪ C = (0, 4) ∪ [−1/2, 1/2] = [−1/2, 4).

2. Let B_{n} = {1, 1/4, 1/9, …, 1/n^{2}} for each natural number n ≥ 1. What are ∩_{n=1}^{∞}B_{n} and ∪_{n=1}^{∞}B_{n}? Solution: The union, ∪_{n=1}^{∞}B_{n}, is the set

{1/n^{2} | n ≥ 1}.

The intersection, ∩_{n=1}^{∞}B_{n}, is the set {1}, since that is the only element in B_{n} for all n ≥ 1.

3. Let I_{n} = [−1/n, 1/n] for any n ≥ 1. What are ∩_{n=1}^{∞}I_{n} and ∪_{n=1}^{∞}I_{n}. (Explain your reasoning intuitively.) Solution: Note first that I_{m} ⊂ I_{n} whenever m > n. This shows that the union is the same as I_{1} = [−1, 1]. The intersection contains only 0. We will see in about a week how to justify the claim that for any real a > 0, there is some n ≥ 1 such that 1/n < a. Hence a is not in the intersection. The same is true on the negative side: for any b < 0, there exists some n ≥ 1 such that b < −1/n. Hence b is not in the intersection either. This leaves only 0 which does satisfy −1/n < 0 < 1/n for all n ≥ 1.

4. Let f : R → R be the function defined by f(x) = x^{2} − 4x + 1.

a. Is f one-to-one? Why or why not? Solution: By completing the square, we see x^{2} − 4x + 1 = (x − 2)^{2} − 3. From this we can see for instance that f(3) = −2 = f(1). Therefore f is not one-to-one. We also see that the graph y = f(x) is a shifted parabola with vertex at (2, −3). This fact can be used to see parts of what we are saying in the later parts of the problem.

b. Is f onto? Why or why not?

Solution: By the same computations as for part a, we see that f(x) ≥ −3 for all x. Therefore f is not onto R.

c. If I = (1, 3), what is the set f(I)? Explain. Solution: f has a local and global minimum at f(2) = −3. Hence f((1, 3)) = [−3, −2).

d. If J = (5, 6), what is the set f^{−1}(J). Explain. Solution: We have f(x) = x^{2} − 4x + 1 = (x − 2)^{2} − 3 = 5 when x = 2 ± \sqrt{8}. Similarly, f(x) = 6 when x = 2 ± 3 = −1, 5. Hence f^{−1}(J) is the union of the two intervals f^{−1}(J) = (−1, 2 − \sqrt{8}) ∪ (2 + \sqrt{8}, 5).

‘B’ Section

1. Prove part (f) of Theorem 1.1.3 in the text. These are the De Morgan Laws for complements. Solution: We show (A ∩ B)^{c} = A^{c} ∪ B^{c}. Let x ∈ (A ∩ B)^{c}, then x ∉ A ∩ B, which says x ∉ A or x ∉ B. But then x ∈ A^{c} ∪ B^{c}, and it follows that (A ∩ B)^{c} ⊂ A^{c} ∪ B^{c}. Conversely, if x ∈ A^{c} ∪ B^{c}, then x ∉ A or x ∉ B. This shows x ∉ A ∩ B, so x ∈ (A ∩ B)^{c}, and it follows that A^{c} ∪ B^{c} ⊂ (A ∩ B)^{c}. Since we have both inclusions, (A ∩ B)^{c} = A^{c} ∪ B^{c}. The second statement (A ∪ B)^{c} = A^{c} ∩ B^{c} is proved similarly.

2. Let A and B be arbitrary sets. Does B = A−(A−B), as we might expect if we looked at the formula through the lens of ordinary algebra? If this is always true, prove it; if it is not, give both a counterexample (an example where the formula is not true), and a correct statement with proof. Solution: This is not true in general as the following counterexample shows. Let A = {a} and let B = {b} (with a ≠ b). Then A−B = {a} = A, so A−(A−B) = ∅ ≠ B. The statement that is true here is that A − (A − B) = A ∩ B. To prove this quickly, the best way is probably to use the De Morgan Laws from question 1 and other parts of Theorem 1.1.3 in the text. We have

A − (A − B) = A ∩ (A − B)^{c}
= A ∩ (A ∩ B^{c})^{c}
= A ∩ (A^{c} ∪ (B^{c})^{c}) (by 1.1.3 (f))
= A ∩ (A^{c} ∪ B) (by 1.1.3 (a))
= (A ∩ A^{c}) ∪ (A ∩ B) (by 1.1.3 (e))
= ∅ ∪ (A ∩ B)
= A ∩ B.

3. Let f : A → B be a function.

a. Let C, D be subsets of A. Is it always true that f(C ∪ D) = f(C) ∪ f(D)? If this is always true prove it; if it is not, give a counterexample. Solution: This statement is always true. We can prove it as follows. If x ∈ C ∪D, then x ∈ C or x ∈ D. Hence f(x) ∈ f(C) or f(x) ∈ f(D). It follows that f(x) ∈ f(C) ∪ f(D), so f(C ∪ D) ⊂ f(C) ∪ f(D). Conversely, if y ∈ f(C) ∪ f(D), then y ∈ f(C) or y ∈ f(D). So y = f(x) for some x ∈ C or y = f(x) for some x ∈ D. It follows that y ∈ f(C ∪ D), so f(C) ∪ f(D) ⊂ f(C ∪ D). This shows the equality.

b. Show that f is onto if and only if f(f^{−1}(E)) = E for all subsets E of B. Solution: Suppose that f(f^{−1}(E)) = E for all subsets E ⊂ B. Let b ∈ B and E = {b}, then f^{−1}(E) ≠ ∅ since f(f^{−1}(E)) = E. Thus there is some a ∈ f^{−1}(E), so f(a) = b. Since this is true for all b ∈ B, f is onto. Conversely, if f is onto, we must show f(f^{−1}(E)) = E for all subsets E ⊂ B. So let E be an arbitrary subset of B. The definition of the inverse image says f(f^{−1}(E)) ⊂ E for all mappings f (that is, even without the assumption that f is onto). If in addition we know that f is onto, we have that for all b ∈ E, there exist a ∈ A such that f(a) = b, and hence that those a ∈ f^{−1}(E). It follows that E ⊂ f(f^{−1}(E)) when f is onto. Hence if f is onto, then f(f^{−1}(E)) = E.

4. Let f : A → B and g : B → C.

a. Show that if f and g are both one-to-one, then g ∘ f : A → C is also one-to-one. Solution: Let (g ∘ f)(x) = (g ∘ f)(y) for some x, y ∈ A. Then g(f(x)) = g(f(y)). Since g is assumed to be one-to-one, we have f(x) = f(y). But then, because f is assumed to be one-to-one, x = y. Therefore, g ∘ f is one-to-one. b. Is the converse of the statement in part a true? That is, if you know that g ∘

f is one-to-one, does it follow that f and g are one-to-one? Prove or find a counterexample. Solution: This statement is not true. For instance, consider f : {b, c} → {b, c} defined by f(b) = b and f(c) = c. Also let g : {a, b, c} → {b, c} by g(a) = g(b) = b and g(c) = c. Then g ∘ f : {b, c} → {b, c} satisfies (g ∘ f)(b) = b and (g ∘ f)(c) = c so g ∘ f is one-to-one. However, g is not one-to-one. (The statement that is true here is that f must be one-to-one and g must be one-to-one when restricted to the range of f.)