Page 385 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 371
B
Now define F : A C → D by the formula F( f ) = h ◦ f ◦ j.To
A A
see that F is one-to-one, suppose that f 1 ∈ C, f 2 ∈ C, and F( f 1 ) =
F( f 2 ), which means h ◦ f 1 ◦ j = h ◦ f 2 ◦ j. Let a ∈ A be arbitrary.
Since j is onto, there is some b ∈ B such that j(b) = a. Therefore
h( f 1 (a)) = (h ◦ f 1 ◦ j)(b) = (h ◦ f 2 ◦ j)(b) = h( f 2 (a)), and since h
is one-to-one, it follows that f 1 (a) = f 2 (a). Since a was arbitrary, this
shows that f 1 = f 2 .
(b) Yes. (You should be able to justify this answer with a counterexample.)
8. (a) Let n be arbitrary, and then proceed by induction on m. The base case
is m = n + 1, and it is taken care of by exercise 7. For the induction
step, apply exercise 2(b).
(b) ∪ n∈Z A n is an infinite set that is not equinumerous with A n for any
+
+
n ∈ Z . In fact, for every positive integer n, A n ≺∪ n∈Z A n . Can you
+
find even larger infinite sets?
+
10. (a) Note that E ⊆ P (Z × Z ). It follows, using exercise 5 of Sec-
+
+
+
+
tion 7.1, that E P (Z × Z ) ∼ P (Z ).
(b) Suppose f (X) = f (Y). Then X ∪{1}∈ f (X) = f (Y) ={Y ∪{1},
(A \ Y) ∪{2}}, so either X ∪{1}= Y ∪{1} or X ∪{1}= (A \ Y) ∪
{2}. But clearly 2 /∈ X ∪{1}, so the second possibility can be ruled
out. Therefore X ∪{1}= Y ∪{1}. Since neither X nor Y contains 1, it
follows that X = Y.
(c) Clearly A is denumerable, and we showed at the end of Section 5.3
+
that P ∼ E. It follows that P (Z ) ∼ P (A) P ∼ E. Combining this
with part (a) and applying the Cantor–Schr¨oder–Bernstein Theorem
gives the desired conclusion.
14. (a) According to the definition of function, R R ⊆ P (R × R), and
therefore by exercise 12(b) and exercise 5 of Section 7.1, R R
P (R × R) ∼ P (R).
Clearly {yes, no} R, so by exercise 5(c) of Section 7.2 and
R
R
exercise 5, P (R) ∼ {yes, no} R. Since we have both R R
R
P (R) and P (R) R, by the Cantor–Schr¨oder–Bernstein theorem,
R R ∼ P (R).
(b) By Theorems 7.1.6 and 7.3.3, exercise 21(a) of Section 7.1, and exer-
Q Z + + +
cise 5(d) of Section 7.2, R ∼ P (Z ) ∼ P (Z ) ∼ R.
Q
(c) Define F : C → R by the formula F( f ) = f Q. (See exercise 7
of Section 5.1 for the meaning of the notation used here.) Suppose
f ∈ C, g ∈ C, and F( f ) = F(g). Then f Q = g Q, which means
that for all x ∈ Q, f (x) = g(x). Now let x be an arbitrary real num-
ber. Use Lemma 7.3.4 to construct a sequence x 1 , x 2 ,... of rational
numbers such that lim n→∞ x n = x. Then since f and g are continuous,