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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 → B is a function.
(a) Define what is meant by the pre-image f ∗(V ).
Answer. For V ⊆ B, f ∗(V ) := {a ∈ A : f (a) ∈ V }.
(b) Define what is meant by the image f∗(U ).
Answer. For U ⊆ A, f∗(U ) := {b ∈ B : b = f (u) for some u ∈ U } = {f (u) : u ∈ U }.
(c) For a collection of subsets {Vj }j∈J of B, prove that f ∗^
j∈J
Vj =
j∈J
f ∗(Vj ).
Proof. Observe that
x ∈ f ∗
j∈J
Vj
⇐⇒ f (x) ∈
i∈J
Vj
⇐⇒ f (x) ∈ Vj for all j ∈ J ⇐⇒ x ∈ f ∗(Vj ) for all j ∈ J ⇐⇒ x ∈
j∈J
f ∗(Vj )
(d) Is it true that f∗(U 1 ∩ U 2 ) = f∗(U 1 ) ∩ f∗(U 2 )? Prove or provide a counterexample.
Answer. This is not true. For example, let f : R → R be defined by f (x) = |x|. Let U 1 = [− 1 , 0] and U 2 = [0, 1]. Then U 1 ∩ U 2 = { 0 } and so f∗(U 1 ∩ U 2 ) = { 0 } while f∗(U 1 ) = [0, 1] = f∗(U 2 ) and so f∗(U 1 ∩ U 2 ) = [0, 1].
- (a) Define the terms injection, surjection and bijection.
Answer. Let f : A → B be a function. We say that f is an injection if f (a 1 ) 6 = f (a 2 ) whenever a 1 , a 2 ∈ A and a 1 6 = a 2. We say that f is a surjection if for each b ∈ B there exists an a ∈ A such that f (a) = b; in otherwords, f∗(A) = B. We say that f is a bijection if it is both an injection and a surgection.
(b) Suppose that both f : A → C and g : B → D are bijections. Show that the function h : A × B → C × 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) ∈ C × D, because f and g are surjections, we choose a ∈ A and b ∈ 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 : (−∞, 0] → [1, ∞) 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 −^1 : [1, ∞) → (−∞, 0] defined by f −^1 (x) = −
x − 1. To verify this, we check
(f −^1 ◦ f )(x) = −
f (x) − 1 = −
x^2 + 1 − 1 = −
x^2 = x for all x ∈ (−∞, 0]
and
(f ◦ f −^1 )(x) = (f −^1 (x))^2 + 1 = (−
x − 1)^2 + 1 = (x − 1) + 1 = x for all x ∈ [1, ∞)
This verifies that f −^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 ∼ on R^2 by (a, b) ∼ (c, d) iff a^2 + b^2 = c^2 + d^2. Is ∼ an equivalence relation? Verify your assertion.
Answer. Yes, ∼ is an equivalence relation. First, (a, b) ∼ (a, b) because a^2 + b^2 = a^2 + b^2. Hence ∼ is reflexive. Also, if (a, b) ∼ (c, d), then c^2 + d^2 = a^2 + b^2 and so (c, d) ∼ (a, b), thus ∼ is symmetric. Finally, if (a, b) ∼ (c, d), and (c, d) ∼ (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) ∼ (e, f ). Therefore, ∼ is transitive. This completes the verification that ∼ 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.
