Page 317 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Closures Again 303
b a−1 kb ab
Induction step: Suppose a ≥ 1 and (2 − 1) · 2 = 2 − 1. Then
k=0

a a−1
b kb b kb ab
(2 − 1) · 2 = (2 − 1) · 2 + 2
k=0 k=0
a−1
b kb b ab ab
= (2 − 1) · 2 + 2 · 2 − 2
k=0
= 2 ab − 1 + 2 b+ab − 2 ab
= 2 (a+1)b − 1.

Exercises

1. Suppose f : A → A. A set C ⊆ A is said to be closed under f if
∗
∀x ∈ C( f (x) ∈ C). Now suppose B ⊆ A. The closure of B under f is
the smallest set C such that B ⊆ C ⊆ A and C is closed under f, if there
is such a smallest set. In this problem you will give two different proofs
that the closure of B under f exists.
(a) Let F ={C | B ⊆ C ⊆ A and C is closed under f}. Prove that
F = ∅, and then prove that ∩F is the closure of B under f.
(b) Define sets B n , for n ≥ 1, as follows:
B 1 = B;
for every n ≥ 1, B n+1 = f (B n ) ={ f (x) | x ∈ B n }.

Prove that ∪ n∈Z B n is the closure of B under f.
+
2. Let f : R → R be defined by the formula f (x) = x + 1. What is the
closure of the set {0} under f ? (See exercise 1.)
3. Suppose F is a family of functions from A to A, and B ⊆ A. The closure of
B under F is the smallest set C such that B ⊆ C ⊆ A and for all f ∈ F, C
is closed under f, if there is such a smallest set. Prove that the closure of
B under F exists.
4. Suppose f : A × A → A.If(x, y) ∈ A × A, then the result of applying
f to (x, y) should be written f ((x, y)), but it is customary to leave out
one set of parentheses and just write f (x, y). A set C ⊂ A is said to be
closed under f if ∀x ∈ C∀y ∈ C( f (x, y) ∈ C). Now suppose B ⊆ A.
The closure of B under f is the smallest set C such that B ⊆ C ⊆ A and
C is closed under f , if there is such a smallest set. Prove that the closure
of B under f exists.
5. Let f : Z × Z → Z be defined by the formula f (x, y) = xy. Let P be
∗
the set of all prime numbers. What is the closure of P under f ? (See
exercise 4.)