Math 414
Professor Lieberman
February 5, 2003
HOMEWORK #2 SOLUTIONS
Section 1.2
4. (a) Strictly increasing and bounded. sup f = 1, inf f = min f = 0.
(b) Strictly decreasing and bounded. sup f = max f = 1, inf f = 0.
(c) Not increasing or decreasing, but bounded. sup f = max f = 3/2, inf f = min f = 1.
(d) Strictly decreasing and bounded below. inf f = 0.
13. (a) If g ◦ f (a1 ) = g ◦ f (a2 ), then f (a1 ) = f (a2 ) because g is one-to-one. But then a1 = a2
because f is one-to-one.
(b) For any c ∈ C, there is b ∈ B such that g(b) = c because g is onto. Then there is a ∈ A
such that f (a) = b because f is onto. Therefore g ◦ f (a) = c, so g ◦ f is onto.
Section 1.3
2. (b) The statement P (1) is
1
X
1(1 + 1)(2 · 1 + 1)
,
6
k2 =
k=1
which is true because both sides of the equation are equal to 1.
Now suppose that P (j) is true for some j ∈ N. Then
j+1
X
k=1
2
j
X
j(j + 1)(2j + 1)
+ (j + 1)2
6
k=1
2
(2j + 1) + (6j + 6)
(j + 1)(j + 2)(2j + 3)
= (j + 1)
=
,
6
6
k =
k 2 + (j + 1)2 =
which is P (j + 1).
(q) The statement P (1) is
1
X
k(k!) = 2! − 1,
k=1
which is true because both sides of this equation are equal to 1.
Now suppose that P (j) is true for some j ∈ N. Then
j+1
X
k=1
k(k!) =
j
X
k(k!) + (j + 1)(j + 1)! = (j + 1)! − 1 + (j + 1)(j + 1)!
k=1
= [1 + (j + 1)](j + 1)! − 1 = (j + 2)(j + 1)! − 1 = (j + 2)! − 1,
which is P (j + 1).
1
2
6
= 15.
4
(f) The coefficient of x3 is
4. (a)
7
(−1) 2
= −560.
3
3 4
(j) For each n, we prove the inequality by induction on k. (There are other ways to do this
part.) For k = 1, we have
n1
n
.
=n=
1
1!
Now, we suppose that the inequality is true when k = j for some j ∈ N. Then
nj (n − j)
nj (n − j)
nj+1
n
n n−j
=
≤
=
≤
.
j+1
j j+1
j!(j + 1)
(j + 1)!
(j + 1)!
6. (a) By direct calculation, the statement is true for n = 1 and n = 2. (This means that
7
a1 = 1 < ,
4
and
2
49
7
a2 = 2 <
=
.)
16
4
If n ≥ 2, then
n n−1 n−1
n+1
7
7
7
7
49 7 n−1
7
an+1 = an + an−1 <
+
= 1+
<
=
.
4
4
4
4
16 4
4
Section 1.4
4. (a) If A is a finite subset of N, then it is countable by definition, so let A be an infinite
subset of N, and define the function f : N → A inductively by f (1) is the smallest element
of A and f (j) is the smallest element of A \ {f (1), . . . , f (j − 1)} for j ≥ 2. It is easy to see
that f is one-to-one. To show that f is onto, we use induction in the form of Theorem 1.3.9
on the statement P (n) is “if n ∈ A, then n = f (m) for some m ∈ N”. For n = 1, we note
that 1 = f (1) if 1 ∈ A. If P (n) is true for some n, we consider three cases: (i) n + 1 ∈
/ A,
(ii) n + 1 ∈ A and i ∈
/ A for all i ≤ n, (iii) n + 1 ∈ A and i ∈ A for some i ≤ n. In the
first case, P (n + 1) is certainly true. In the second case, we have n = f (1), so P (n + 1) is
true. In the third case, because An = {i ∈ A : i ≤ n} is finite, there is a positive integer m
such that An = {f (1), . . . , f (m − 1)}. By definition, n + 1 = f (m), so P (n + 1) is true in
this case as well. Note that it follows from this proof that, if A is a countable set and B is
a subset of A, then B is also countable.
(b) First we show that the product of two countably infinite sets A and B is countable.
Note that there is a bijection f : N × N → A × B, so we only have to show that N × N
is countable. Define the function g : N × N → N by g(i, j) = i + (i + j − 1)(i + j)/2. To
see that g is a bijection, we compute its inverse. Given a number k, let m be the largest
triangular number, that is, a number of the form n(n + 1)/2 for some n ∈ N, which is less
than k. Then we take i = k − m and j = n + 1 − i. It’s easy to check that g(i, j) = k, so
g has an inverse, which means it’s a bijection. If A or B is finite (this hypothesis includes
the possibility that both are finite), then A × B is equivalent to a subset of N × N so it’s
countable by the remark at the end of part (a) and the first half of the proof of this part.
3
(c) We label the sets A1 , A2 , . . . (where the dots may indicate only finitely many sets) and
we label the elements of the set Ai by aij with j = 1, . . . . Then we define
B = {(i, j) ∈ N × N : there is an element aij }.
(So, if A6 has 3 elements, then (6, 1), (6, 2), and (6, 3) are in B but (6, 5) is not in B.) Then
g : B → ∪Ai defined by g(i, j) = aij is a bijection. By the remark at the end of the solution
to part (a), the set B is countable and then Theorem 1.4.4(c) implies that the union of the
Ai s is countable.
(d) First Z is the union of the three countable sets N, {0}, and Z− = {−1, −2, −3, . . . }.
Then Q is equivalent to a subset of Z × N, specifically, the set of all pairs (p, q) with p 6= 0
which have no common factors or with p = 0 and q = 1.
(e) Done in class
(f) We argue by contradiction. If R were countable, then every subset would be countable.
But the set (0, 1) ⊂ R is uncountable by part (e), so R is uncountable.
(g) We know that R is the union of Q and I, the set of all irrational numbers. If I were
countable, then parts (c) and (d) would imply that R is also countable. This contradiction
(with part (f)) shows that I is uncountable.