Page 383 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 369
one-to-one, onto function from I n to A i }. By inductive hypothesis, F i is

finite and |F i |= n!. Now let F ={ f ∈ F | f (n + 1) = g(i)}. Define
i
a function h : F i → F by the formula h( f ) = f ∪{(n + 1, g(i))}.

i
It is not hard to check that h is one-to-one and onto, so F is

i
F and

i i
finite and |F |=|F i |= n!. Finally, notice that F =∪ i∈I n+1

∀i ∈ I n+1 ∀ j ∈ I n+1 (i = j → F ∩ F = ∅). It follows, by exercise
i j
n+1

19, that F is finite and |F|= |F |= (n + 1) · n! = (n + 1)!.
i=1 i
(b) Hint: Define h : F → L by the formula h( f ) ={(a, b) ∈ A ×
A | f −1 (a) ≤ f −1 (b)}. (You should check that this set is a total order on
A.) To see that h is one-to-one, suppose that f ∈ F, g ∈ F, and f = g.
Let i be the smallest element of I n for which f (i) = g(i). Now show
that ( f (i), g(i)) ∈ h( f ) but ( f (i), g(i)) /∈ h(g), so h( f ) = h(g). To see
that h is onto, suppose R is a total order on A. Define g : A → I n by the
formula g(a) =|{x ∈ A | xRa}|. Show that ∀a ∈ A∀b ∈ A(aRb ↔
−1
g(a) ≤ g(b)), and use this fact to show that g −1 ∈ F and h(g ) = R.
(c) 5! = 120.
25. Base case: n = 1. Then I n ={1}, P ={{1}}, and A {1} = A 1 . Therefore
|S|+1 2
A i |=|A 1 | and (−1) |A S |= (−1) |A {1} |=|A 1 |.
|∪ i∈I n S∈P
Induction step: Suppose the Inclusion–Exclusion Principle holds for n
sets, and suppose A 1 , A 2 ,..., A n+1 are finite sets. Let P n = P (I n ) \{∅}
and P n+1 = P (I n+1 ) \{∅}. By exercise 24(a), exercise 22(a) of Section
3.4, and the inductive hypothesis,
A i ) ∪ A n+1 |
|∪ i∈I n+1 A i |=|(∪ i∈I n
A i ) ∩ A n+1 |
=|∪ i∈I n A i |+|A n+1 |−|(∪ i∈I n
|S|+1
= (−1) |A S |+|A n+1 |−|∪ i∈I n (A i ∩ A n+1 )|.
S∈P n
Now notice that for every S ∈ P n , ∩ i∈S (A i ∩ A n+1 ) = (∩ i∈S A i ) ∩ A n+1 =
A S∪{n+1} . Therefore, by another application of the inductive hypothesis,
|S|+1
(A i ∩ A n+1 )|= (−1) |A S∪{n+1} |. Thus
|∪ i∈I n
S∈P n
|S|+1 |S|+1
A i |= (−1) |A S |+|A n+1 |− (−1) |A S∪{n+1} |
|∪ i∈I n+1
S∈P n S∈P n
|S|+1 2
= (−1) |A S |+ (−1) |A {n+1} |
S∈P n
|S∪{n+1}|+1
+ (−1) |A S∪{n+1} |.
S∈P n