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Math 415, Test 2, November 19, 2003
Instructions: Do all five of the following problems. Please do your best, and show all appropriate details in your solutions.
- Suppose f : A โ B is a function.
(a) Define what is meant by the pre-image f โ(V ).
Answer. For V โ B, f โ(V ) := {a โ A : f (a) โ V }.
(b) Define what is meant by the image fโ(U ).
Answer. For U โ A, fโ(U ) := {b โ B : b = f (u) for some u โ U } = {f (u) : u โ U }.
(c) For a collection of subsets {Vj }jโJ of B, prove that f โ^
jโJ
Vj =
jโJ
f โ(Vj ).
Proof. Observe that
x โ f โ
jโJ
Vj
โโ f (x) โ
iโJ
Vj
โโ f (x) โ Vj for all j โ J โโ x โ f โ(Vj ) for all j โ J โโ x โ
jโJ
f โ(Vj )
(d) Is it true that fโ(U 1 โฉ U 2 ) = fโ(U 1 ) โฉ fโ(U 2 )? Prove or provide a counterexample.
Answer. This is not true. For example, let f : R โ R be defined by f (x) = |x|. Let U 1 = [โ 1 , 0] and U 2 = [0, 1]. Then U 1 โฉ U 2 = { 0 } and so fโ(U 1 โฉ U 2 ) = { 0 } while fโ(U 1 ) = [0, 1] = fโ(U 2 ) and so fโ(U 1 โฉ U 2 ) = [0, 1].
- (a) Define the terms injection, surjection and bijection.
Answer. Let f : A โ B be a function. We say that f is an injection if f (a 1 ) 6 = f (a 2 ) whenever a 1 , a 2 โ A and a 1 6 = a 2. We say that f is a surjection if for each b โ B there exists an a โ A such that f (a) = b; in otherwords, fโ(A) = B. We say that f is a bijection if it is both an injection and a surgection.
(b) Suppose that both f : A โ C and g : B โ D are bijections. Show that the function h : A ร B โ C ร D defined by h(a, b) = (f (a), g(b)) is a bijection.
Proof. If (a 1 , b 1 ) 6 = (a 2 , b 2 ), then either a 1 6 = a 2 or b 1 6 = b 2 and so either f (a 1 ) 6 = f (a 2 ) or g(b 1 ) 6 = g(b 2 ) because f and g are injections. In either case, h(a 1 , b 1 ) 6 = h(a 2 , b 2 ), and so h is an injection. To show that h is a surjection, let (c, d) โ C ร D, because f and g are surjections, we choose a โ A and b โ B so that f (a) = c and g(b) = d. Then h(a, b) = (f (a), g(b)) = (c, d), and so h is a surjection. Because h is both an injection and surjection, h is a bijection.
(c) Find the inverse function of f : (โโ, 0] โ [1, โ) of f (x) = x^2 + 1. Be sure to state the domain and range of the inverse function, and to verify it is the inverse function of f.
Answer. The inverse of f is f โ^1 : [1, โ) โ (โโ, 0] defined by f โ^1 (x) = โ
x โ 1. To verify this, we check
(f โ^1 โฆ f )(x) = โ
f (x) โ 1 = โ
x^2 + 1 โ 1 = โ
x^2 = x for all x โ (โโ, 0]
and
(f โฆ f โ^1 )(x) = (f โ^1 (x))^2 + 1 = (โ
x โ 1)^2 + 1 = (x โ 1) + 1 = x for all x โ [1, โ)
This verifies that f โ^1 is the inverse of f.
- (a) Define what is meant by an equivalence relation on a set A.
Answer. An equivalence relation on a set A is a relation that is reflexive, symmetric and transitive.
(b) Define the relation โผ on R^2 by (a, b) โผ (c, d) iff a^2 + b^2 = c^2 + d^2. Is โผ an equivalence relation? Verify your assertion.
Answer. Yes, โผ is an equivalence relation. First, (a, b) โผ (a, b) because a^2 + b^2 = a^2 + b^2. Hence โผ is reflexive. Also, if (a, b) โผ (c, d), then c^2 + d^2 = a^2 + b^2 and so (c, d) โผ (a, b), thus โผ is symmetric. Finally, if (a, b) โผ (c, d), and (c, d) โผ (e, f ), then
a^2 + b^2 = c^2 + d^2 and c^2 + d^2 = e^2 + f 2
and so a^2 + b^2 = e^2 + f 2 which means (a, b) โผ (e, f ). Therefore, โผ is transitive. This completes the verification that โผ is an equivalence relation on R^2.
(c) Do the relation classes from the relation in (b) form a partition of R^2? If not, explain which properties of a partition are violated. If so, describe the partition, and explain why it is a partition.
