Math 201
Introduction to Proofs
Instructor: Alex Roitershtein
Iowa State University
Fall 2015
Solutions to selected problems from Homework #1
Section 1.2.
A.1.(a) In the lexicographic order:
A × B = {(1, a), (1, c), (2, a), (2, c), (3, a), (3, c), (4, a), (4, c)}.
A.1.(g) The elements of the set B × B are ordered pairs of the elements of B. That is, an
element of B × B takes the form (x, y), where x, y ∈ B. An element of A × (B × B) is a
pair whose first element is a member of the set A and the second element is in ordered
pair
of elements of B. Therefore, an element of A × (B × B) takes the form x, (y, z) , where
x ∈ A, y, z ∈ B. Notice that this is different from a triple (x, y, z). The latter represents
three singletons while the former is composed of a singleton and a pair. In the lexicographic
order:
A × (B × B) =
1, (a, a) , 1, (a, c) , 1, (c, a) , 1, (c, c) , 1, (a, a) , 1, (a, c) ,
1, (c, a) , 1, (c, c) , 2, (a, a) , 2, (a, c) , 2, (c, a) , 2, (c, c) ,
3, (a, a) , 3, (a, c) , 3, (c, a) , 3, (c, c) , 4, (a, a) , 4, (a, c) ,
4, (c, a) , 4, (c, c) , .
A.1.(h) In the lexicographic order:
B 3 = {(a, a, a), (a, a, c), (a, c, a), (a, c, c), (c, a, a), (c, a, c), (c, c, a), (c, c, c)}.
Section 1.3.
A.1 ∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 3}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4},
{2, 3, 4}, {1, 2, 3, 4}.
A.2 ∅, {1}, {2}, {∅}, {1, 2}, {1, ∅}, {2, ∅}, {1, 2, ∅}.
A.3 ∅, {R}.
A.4 ∅.
A.5 ∅, {∅}.
A.6 ∅, {R}, {Q}, {N}, {R, Q}, {R, N}, {Q, N}, {R, Q, N}.
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A.7 ∅, {R}, {Q, N}, R, {Q, N} .
C.13 True. A ⊆ A for any A.
C.14 False. This exercise emphasizes the difference between the Euclidean space Rd we
studied in calculus and the space of ordered d-tuples of reals Rd = R × · · · × R as it discussed
in the text book. While the former has a structure beyond just the lexicographic order of its
elements (say, distance is defined between any two of its elements), the former is a space of
ordered “lists” of reals lacking any additional structure. In particular, Rs is a space of pairs
while R3 is a space of triples of reals. Hence, R2 6⊆ R3 using the conventions introduced in
the text.
C.15 True. Indeed, if x − 1 = 0 then x2 − x = x(x − 1) = 0.
C.16 False. Indeed, if x2 − x = x(x − 1) = 0 then either x = 0 or x − 1 = 0. Thus the pairs
(x, y) = 0 with x = 0 (the 0y axis in the xy-plane) belong to the set {(x, y) : x2 − x = 0}
but don’t belong to the set {(x, y) : x − 1 = 0}.
Section 1.5.
3(a) (A × B) ∩ (B × B) = (A ∩ B) × B = {(1, 1), (1, 2)}.
3(c) (A × B) − (B × B) = (A − B) × B = {(0, 1), (0, 2)}.
3(e) (A × B) ∩ B = ∅ since A × B consists of pairs of numbers while B is a set of single
numbers.
3(f ) P(A) ∩ P(B) = ∅, {1} .
3(h) P(A ∩ B) = P {1} = ∅, {1} .
Section 1.6.
1(a) A = {0, 2, 5, 8, 10}.
1(c) A ∩ A = ∅.
1(e) A − A = {x ∈ A : x 6∈ A} = A.
1(f ) A − B = A ∩ B = {4, 6}.
1(h) A ∩ B = B − A = {5, 8}.
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Section 1.8.
1(a) ∪4i=1 Ai = {a, b, c, d, e, f, g, h}.
1(b) ∩4i=1 Ai = {a, b}.
7(a) ∪i∈N R × [i, i + 1] = R × ∪i∈N R × [i, i + 1] = R × [1, ∞).
7(b) ∩i∈N R × [i, i + 1] = R × ∩i∈N R × [i, i + 1] = R × ∅ = ∅.
12(h) If ∩α∈I Aa = ∪α∈I Aα then, in particular, ∪α∈I Aa ⊆ ∩α∈I Aα . The latter inclusion
implies that whenever x ∈ Aα for some α ∈ I, in fact x ∈ Aα for all α ∈ I. That is, Aα = Aβ
for any α, β ∈ I. In other words, all sets Aα , α ∈ I, are identical. Thus ∩α∈I Aa = ∪α∈I Aα if
and only if all the sets Aα , α ∈ I, are identical.
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