MATH 242 – Principles of Analysis Solutions for Problem Set 1 – due: Jan. 31

‘A‘ Section

1. Assume that A, B are sets of integers.

a. What is the contrapositive of the statement: “If x is even then x ∈ A ∪ B”? Express without using not. Solution: The contrapositive of “if p then q” is “if not q then not p.” Here, by the DeMorgan Law, x ∉ A ∪ B is equivalent to x ∈ A^{c} and x ∈ B^{c}. Also, “not even” is equivalent to “odd.” So, without using not, we can state the contrapositive as “If x ∈ A^{c} and x ∈ B^{c}, then x is odd.”

b. What is the converse of the statement in part a?

Solution: The converse is: “If x ∈ A ∪ B, then x is even.”

2. Let A = {x ∈ R | x^{2} − 5x + 4 = 0}, B = (0, 1) = {x ∈ R | 0 < x < 1} and C = \{\frac{x}{x^{2}+9} | x ∈ R\} (Note: C is the range of the function f defined by f(x) = \frac{x}{x^{2}+9}.) 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}+9} has f′(x) = \frac{9−x^{2}}{(x^{2}+9)^{2}} . This is = 0 at x = ±3. Moreover f′(x) < 0 for x < −3, f′(x) > 0 for −3 < x < 3 and f′(x) < 0 for x > 3. Therefore, at x = −3, f has a local minimum with f(−3) = −1/6. Similarly, at x = 3, f has a local maximum with f(3) = 1/6. We also see lim_{x→±∞} f(x) = 0. Hence f(−3) = −1/6 is also an absolute minimum, and f(3) = 1/6 is also an absolute maximum. We will prove a general theorem later in the course that shows that every y with −1/6 < y < 1/6 must also be in the range, but this can also be checked directly here since the equation

y = \frac{x}{x^{2} + 9}

can be rearranged to yx^{2} − x + 9y = 0. If y = 0, then x = 0. Otherwise, by the quadratic formula this has roots

x = \frac{1 ± \sqrt{1 − 36y^{2}}}{2y} .

The expression in the square root is nonnegative exactly when −1/6 ≤ y ≤ 1/6 and we get x with f(x) = y (two of them in fact for y ≠ 0, −1/6, 1/6). Hence C = [−1/6, 1/6]. b. Find the sets B ∩ A and B ∩ C.

Solution: Since A = {1, 4}, we see that B ∩ A = ∅ and B ∩ C = (0, 1/6]. c. Find the sets B ∪ A and B ∪ C and express using set notation.

Solution: We have B ∪ A = (0, 1] ∪ {4} = B. Then by part a, B ∪ C = (0, 1) ∪ [−1/6, 1/6] = [−1/6, 1).

3. For n a general natural number, let B_{n} = {0, 2n}. What are ∩_{n=1}^{∞}B_{n} and ∪_{n=1}^{∞}B_{n}? Solution: The union, ∪_{n=1}^{∞}B_{n}, is the set

{0, 2, 4, ⋯} = {2n | n ≥ 0},

or the set of nonnegative even integers. The intersection, ∩_{n=1}^{∞}B_{n}, is the set {0}, since that is the only element in B_{n} for all n ≥ 1.

4. 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.

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

a. Is f one-to-one? Why or why not? Solution: Yes, the inverse tangent of x is defined as the unique angle θ in the interval (−π/2, π/2) such that tan(θ) = x. So tan^{−1}(x) = θ = tan^{−1}(x′) implies that x = x′.

b. Is f onto? Why or why not?

Solution: No, since the range is just the interval (−π/2, π/2). c. If I = (0, \sqrt{3}), what is the set f(I)? Explain.

Solution: f(I) = (0, π/3) since tan(0) = 0 and tan(π/3) = \sqrt{3}. d. If J = (−π/4, π/4), what is the set f^{−1}(J). Explain.

Solution: f^{−1}(J) = {x | −π/4 < tan^{−1}(x) < π/4}, which is the same as tan(−π/4) < x < tan(π/4), so −1 < x < 1. Hence f^{−1}(J) is the open interval (−1, 1).

‘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 A = A−(B −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 true since for any set B, we have B − B = ∅. This follows from part g of Theorem 1.1.3, for instance: B − B = B ∩ B^{c} = ∅. But then A − ∅ = A, since A − ∅ = A ∩ ∅^{c} = A ∩ U = A (where U denotes the universal set).

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 is not true. For instance, let f : R → R be defined by f(x) = x^{2}. Let C = (−1, 0) and D = (0, 1). Then f(C) = f(D) = (0, 1), so f(C) ∩ f(D) = (0, 1). But C ∩ D = ∅, so f(C ∩ D) = ∅ as well. Note that other similar examples can be constructed any time that f is not one-to-one.

b. Show that f is one-to-one if and only if f^{−1}(f(C)) = C for all subsets C of A. Solution: First note that C ⊆ f^{−1}(f(C)) for all f and all C since if x ∈ C, then f(x) ∈ f(C), so x ∈ f^{−1}(f(C)) and hence C ⊆ f^{−1}(f(C)). So what we need to show here can be restated as follows: (1) if f is one-to-one, then we need to show f^{−1}(f(C)) ⊆ C for all C. And conversely (2) if f^{−1}(f(C)) ⊆ C for all C, then we need to show that f is one-to-one. To prove (1), let f be one-to-one. For each y ∈ f(C), there is some x ∈ C such that f(x) = y. But if f(x′) = y, then f being one-to-one implies that x = x′. Hence the only elements of A that map to f(C) are the elements of C, so f^{−1}(f(C)) ⊆ C. To prove (2), let f^{−1}(f(C)) ⊆ C for all subsets C of A. In particular, let C = {x} for some particular element x ∈ A. Suppose that f(x′) = f(x). Then by definition, x and x′ are both elements of f^{−1}(f(C)). But by assumption f^{−1}(f(C)) = {x}, so x = x′. This shows that f must be one-to-one.

4. Let f : A → B and g : B → C.

a. Show that if f and g are both onto, then g ∘ f : A → C is also onto. Solution: Let z ∈ C. Since g is onto, there exists y ∈ B such that g(y) = z. But then since f is onto, there exists x ∈ A such that f(x) = y. Combining these statements, we see that g(f(x)) = (g ∘ f)(x) = z. Since z was arbitrary, this shows that g ∘ f is onto.

b. Is the converse of the statement in part a true? That is, if you know that g ∘ f is onto, does it follow that f and g are onto? Prove or find a counterexample. Solution: This statement is not true. Let A = B = R and C = [0, ∞), and let f : A → B be defined by f(x) = x^{2} and g(y) = \sqrt{|y|}. Then for all z ∈ C we have z = g(f(z)), so g ∘ f is onto. However, f is not onto since its range contains no negative numbers. (It does follow in general that g must be onto, but as in the counterexample, if g is not one-to-one, the range of f only needs to contain one inverse image of each element z ∈ C.)