Mathematics 242 – Principles of Analysis Selected Solutions – Problem Set 1, “B portion” February 6, 2011

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3. (a) A question like this is asking for a specific counterexample to the statement (A − B) ∪ B = A – an explicit example that shows the statement is not true. You can find a simple example where this fails whenever A − B and B − A are nonempty. For instance, let

A = {1, 2}, B = {2, 3}

Then A − B = {1} so (A − B) ∪ B = {1, 2, 3} ≠ A.

(b) From examples like this (and/or Venn diagrams), we can see that the correct statement should be (A − B) ∪ B = A ∪ B. Here are two different “styles” of proofs of this, equally correct.

Proof 1: We will show the two inclusions (A − B) ∪ B ⊂ A ∪ B and A ∪ B ⊂ (A − B) ∪ B. First, let x ∈ (A − B) ∪ B. then x ∈ A − B or x ∈ B by the definition of the union. Hence by part (d) of problem 2 (part (g) of Theorem 1.1.3), x ∈ A and x ∉ B, or x ∈ B. This shows x ∈ A or x ∈ B, so x ∈ A ∪ B. Hence (A − B) ∪ B ⊂ A ∪ B. For the other inclusion, let x ∈ A ∪ B. Then x ∈ A or x ∈ B. If x ∈ B, then x ∈ (A − B) ∪ B by definition of the union. On the other hand if x ∉ B, then it must be true that x ∈ A, and hence x ∈ A ∩ B^{c} = A − B (using part (d) of problem 2 again). This shows A ∪ B ⊂ (A − B) ∪ B and hence the equality we want. //

Proof 2: For this proof, we will use several parts of Theorem 1.1.3 that incorporate the sort of element-by-element arguments in Proof 1. We have

(A − B) ∪ B = (A ∩ B^{c}) ∪ B (by 1.1.3(g))

= (A ∪ B) ∩ (B^{c} ∪ B) (by 1.1.3(e)) = (A ∪ B) ∩ X (by 1.1.3(b)) = A ∪ B (by def. of universal set)

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5. Let f : A → B be a function and C, D ⊂ A.

(a) We want to show f(C ∪ D) = f(C) ∪ f(D) (subsets of B), and we will do this by showing each set is contained in the other. First, let y ∈ f(C ∪ D). Then y = f(x) for some x ∈ C ∪ D. Hence y = f(x) for some x ∈ C or some x ∈ D. If x ∈ C, then y = f(x) ∈ f(C); if x ∈ D, then y = f(x) ∈ f(D). Hence y ∈ f(C) ∪ f(D). This shows f(C ∪ D) ⊂ f(C) ∪ f(D). For the reverse inclusion, suppose y ∈ f(C) ∪ f(D). Then y ∈ f(C) or y ∈ f(D). In the first case, y = f(x) for some x ∈ C; in the second y = f(x) for some x ∈ D. Either way, y = f(x) for some x ∈ C ∪ D since then x ∈ C or x ∈ D. It follows that y ∈ f(C ∪ D) and f(C) ∪ f(D) ⊂ f(C ∪ D).

(b) We want to show f(C ∩ D) ⊂ f(C) ∩ f(D). Let y ∈ f(C ∩ D). Then y = f(x) for some x ∈ C ∩ D. Hence y = f(x) for some x ∈ C which also satisfies x ∈ D. Hence y ∈ f(C) and y ∈ f(D), so y ∈ f(C) ∪ f(D). This shows f(C ∩ D) ⊂ f(C) ∩ f(D).

(c) If we think carefully about how the reverse inclusion might fail, then it should be clear that knowing y ∈ f(C) ∩ f(D) just says that we have some x_{1} ∈ C and some x_{2} ∈ D such that f(x_{1}) = y = f(x_{2}). There’s no reason why x_{1} or x_{2} must come from C ∩ D. Here is an explicit example. Let f : R → R be the function defined by f(x) = x^{2}. Let C = {1} and D = {−1}. Then f(C) = f(D) = {1}, so f(C) ∩ f(D) = {1}. But C ∩ D = ∅, so f(C ∩ D) = ∅ ≠ f(C) ∩ f(D).

6. We did one part of this in class(!) You should be reading your class notes in addition to working on the problem sets.

12. Comments: This problem is really a continuation of 5 (b) and (c) above. There will be interconnected problems on many assignments for this course; be sure you can spot them and can see what steps are necessary for the later problems, without repeating work you already did! Also, if a problem states an extra hypothesis (like the hypothesis that f is one-to-one in part (a)), you should ask yourself: Did I use that anywhere? How did I use it?

(a) We know from 5 (b) above that f(C ∩D) ⊂ f(C)∩f(D) for all functions f : A → B and all subsets C, D ⊂ A. So, what must be proved for this part is that if f is one-to-one, then f(C)∩f(D) ⊂ f(C ∩D) for all C, D. So let C, D be subsets of A, and let y ∈ f(C)∩f(D). Then by definition we know y ∈ f(C) and y ∈ f(D). Hence, y = f(x_{1}) for some x_{1} ∈ C and also (the same) y = f(x_{2}) for some x_{2} ∈ D. Since f is assumed to be one-to-one, and f(x_{1}) = y = f(x_{2}), it must be the case that x_{1} = x_{2}. But then this single element is in both C and D, hence in C ∩ D. It follows that y ∈ f(C ∩ D), so f(C) ∩ f(D) ⊂ f(C ∩ D).

(b) Note: A statement of the form “p if and only if q” means “if p then q, and if q then p.” To prove a statement of this form, you must supply arguments for both implications. (This is like the standard plan for showing A = B, since A ⊂ B is equivalent to “if x ∈ A, then x ∈ B” and B ⊂ A is equivalent to “if x ∈ B, then x ∈ A”).

Part (a) shows the statement “If f is one-to-one, then f(C ∩ D) = f(C) ∩ f(D) for all C, D ⊂ A.” So that is already known, and what we need to show for this part is the reverse implication “If f(C ∩ D) = f(C) ∩ f(D) for all subsets C, D of A, then f is one-to-one.”

Let x_{1}, x_{2} ∈ A and assume f(x_{1}) = f(x_{2}) = y ∈ B. Consider the subsets C = {x_{1}} and D = {x_{2}} in A. By our hypothesis, we have

f(C ∩ D) = f(C) ∩ f(D) = {y}.

But in particular, this says C ∩ D ≠ ∅ (since otherwise f(C ∩ D) = ∅ too). Therefore, it must be true that x_{1} = x_{2}, and hence f is one-to-one.

14. (a) If ad − bc = 0 and a = 0, then bc = 0, so b = 0 or c = 0. In the first case we have f(x) = \frac{0x+0}{cx+d} = 0 for all x so f is constant. In the second case we have f(x) = \frac{0x+b}{0x+d} is constant. On the other hand, if a ≠ 0, then we have d = \frac{bc}{a} and

f(x) = \frac{ax + b}{cx + \frac{bc}{a}}

= \frac{a(ax + b)}{c(ax + b)}

= \frac{a}{c}

is constant again.

(b) Now assume ad − bc ≠ 0. We can do most of this part with one algebraic calculation. Let y = f(x), and try to solve for x as a function of y. If we can do that uniquely, then f is one-to-one, and we can determine the range and the formula for the inverse function all at once! The computations:

\begin{aligned}y & = \frac{ax + b}{cx + d} \\ (cx + d)y & = ax + b \\ x(cy − a) & = −dy + b \\ x & = \frac{−dy + b}{cy − a} \end{aligned}

This shows that as long as y ≠ a/c, there is a unique x such that f(x) = y. If y = a/c, then we get ad − bc = 0 which contradicts our hypothesis. Hence f is one-to-one. The inverse function is also found from the result of this same calculation:

f^{−1}(x) = \frac{−dx + b}{cx − a}.

The domain of f is

domain(f) = \{ \begin{matrix}R − \{−d/c\} & if c ≠ 0 \\ R & if c = 0 \end{matrix}

and the domain of f^{−1} is

domain (f^{−1}) = \{ \begin{matrix}R − \{a/c\} & if c ≠ 0 \\ R & if c = 0 \end{matrix}

(c) Finally, we ask, when is f(x) = f^{−1}(x) for all x. If we set f(x) = f^{−1}(x), we get

\frac{ax + b}{cx + d} = \frac{−dx + b}{cx − a}

If this is true for all x in the domains, then

(ax + b)(cx − a) = (cx + d)(−dx + b)

which implies

c(a + d)x^{2} − (a^{2} − d^{2})x − b(a + d) = 0.

The only way this can be satisfied for infinitely many real x is if c(a + d) = 0, a^{2} − d^{2} = 0, and b(a + d) = 0. If b ≠ 0 or c ≠ 0, this implies a = −d and the middle equation is also satisfied. If b = c = 0, then a = ±d. Hence the f for which f = f^{−1} are all linear fractional mappings of the form

f(x) = \frac{ax + b}{cx − a}

with b ≠ 0 or c ≠ 0, and

f(x) = ±x.