Dr. Zachary Scherr
1
Math 370 HW 1
Solutions
Book Problems
1. 1.1.4
(a) (A ∩ B)0 = A0 ∪ B 0
Solution: Suppose x ∈ (A ∩ B)0 . Then x 6∈ A ∩ B meaning x cannot be an element of both
A and B. If x ∈ A then x 6∈ B so x ∈ B 0 . Otherwise, x 6∈ A so x ∈ A0 . In either case we see
x ∈ A0 ∪ B 0 and so (A ∩ B)0 ⊆ A0 ∪ B 0 . Conversely, if y ∈ A0 ∪ B 0 then y ∈ A0 or y ∈ B 0 . If
y ∈ A0 then y 6∈ A so y 6∈ A ∩ B as A ∩ B ⊆ A and thus y ∈ (A ∩ B)0 . A similar argument shows
that if y ∈ B 0 then y ∈ (A ∩ B)0 and hence A0 ∪ B 0 ⊆ (A ∩ B)0 . Therefore (A ∩ B)0 = A0 ∪ B 0 .
(b) (A ∪ B)0 = A0 ∩ B 0
Solution: The easiest way to prove this is to use part a). Part a) says that
(A0 ∩ B 0 )0 = (A0 )0 ∪ (B 0 )0 = A ∪ B.
Take the complement of both sides to see that A0 ∩ B 0 = (A ∪ B)0 . (For this to be a rigorous
argument you need to know that given a set X we have X 00 = X. Check this!)
2. 1.1.6
Solution: We prove this by induction on n. When n = 0, we know that the empty set ∅ has exactly
1 = 20 subsets, namely ∅ itself. Assume that any set of n elements has 2n distinct subsets and let S
be a set of n + 1 elements. Pick s ∈ S. Then any subset of S either contains s or it doesn’t. There
is a one-to-one correspondence between subsets of S containing s and subsets of S not containing s
(check this). Since S −{s} is a set containing n elements, the induction hypothesis tells us that S has
exactly 2n subsets not containing s. Thus the total number of subsets of S is 2n + 2n = 2 · 2n = 2n+1
and our induction is complete.
3. 1.1.10
(a) S is the set of people in the world today, a ∼ b if a and b have a common ancestor.
Solution: One could go either way on this question...genealogy is a tough subject. I would
say this relation is probably not transitive. I am related to my cousins on my dad’s side of the
family and I am also related to my cousins on my mom’s side of the family. Hopefully though,
my cousins on each side of the family are not related to one another.
(b) S is the set of people in the world today, a ∼ b if a lives within 100 miles of b.
Solution: No, ∼ is not transitive. For example, person a and live 100 miles from person b,
and person b could live 100 miles from person c, yet person a and person c could live 200 miles
apart.
(c) S is the set of people in the world today, a ∼ b if a and b have the same father.
Solution: Yes. Let’s check the three properties:
• (reflexive) every a has the same father as themselves so a ∼ a.
Dr. Zachary Scherr
Math 370 HW 1
Solutions
• (symmetric) If a ∼ b then a and b share the same father. Thus b ∼ a as well.
• (transitive) If a ∼ b and b ∼ c then a and b share the same father and b and c share the
same father. Since a person has at most one biological father, this means that a and c
must share the same father and hence a ∼ c.
(d) S is the set of real numbers, a ∼ b if a = ±b.
Solution: Yes. Let’s check the three properties:
• (reflexive) For every a ∈ R we have a = a so by definition a ∼ a.
• (symmetric) If a ∼ b then a = ±b. Multiplying both sides of the equality by −1 shows
b = ±a and hence b ∼ a.
• (transitive) If a ∼ b and b ∼ c then by defintion we have a = ±b and b = ±c. Plugging in
shows a = ±(±c) = ±c and hence a ∼ c.
(e) S is the set of integers, a ∼ b if both a > b and b > a.
Solution: No, ∼ is not reflexive. In fact, there is no a for which a ∼ a.
(f) S is the set of all straight lines in the plane, `1 ∼ `2 if `1 is parallel to `2 .
Solution: This depends on your definition of parallel. I would say that two lines in the plane
y = mx + b and y = nx + c are parallel if m = n. Under this definition parallel is an equivalence
relation. Let’s check:
• (reflexive) Let ` be given by the equation y = mx + b. Since m = m we have that ` ∼ `.
• (symmetric) Suppose `1 ∼ `2 . Let the equation for `1 be y = mx + b and the equation
for `2 be y = nx + c. Then m = n so in fact `2 ∼ `1 .
• (transitive) Suppose `1 ∼ `2 and `2 ∼ `3 . If the equations for these three lines are
y = mx + b, y = nx + c and y = rx + d respectively, then our assumnption says that
m = n and n = r. Thus m = r and so `1 ∼ `3 .
4. 1.1.12
Solution: This question is hard to do before learning about division. My purpose in assigning it
was to get you thinking about how you might prove that there are n equivalence classes. Here is a
complete solution.
Define a ∼ b if a − b is a multiple of n. First we prove that ∼ is an equivalence relation.
• (reflexive): a ∼ a since a − a = 0 is a multiple of n for any integer a.
• (symmetric): If a ∼ b then a − b = kn for some integer k. Then b − a = (−k)n so b ∼ a.
• (transitive): If a ∼ b and b ∼ c then a − b = kn and b − c = `n for some integers k and `. Then
a − c = (a − b) + (b − c) = kn + `n = (k + `)n
and so a ∼ c.
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Dr. Zachary Scherr
Math 370 HW 1
Solutions
Now we need to find the equivalence classes of ∼. For any integer a we know there exist integers
q, r for which
a = nq + r
and 0 ≤ r < n. Thus a − r is a multiple of n and hence a ∼ r. Thus we see that every integer
is equivalent to an integer in the range 0, 1, . . . , n − 1. This says that every integer is contained in
at least one of cl(0), cl(1), . . . , cl(n − 1). It only remains to show that these equivalence classes are
disjoint. To do so, assume we have two integers i, j for which 0 ≤ i ≤ j < n. Then 0 ≤ j − i < n.
Since the smallest positive multiple of n is n itself, we see that i ∼ j precisely when j − i = 0 or
i = j. This proves that when i 6= j are in the range 0, 1, . . . , n − 1 we have that cl(i) ∩ cl(j) = ∅.
This completes the proof that there are exactly n equivalence classes cl(0), cl(1), . . . , cl(n − 1).
5. 1.2.1
(a) σ : R → R≥0 and σ(s) = s2 .
Solution:
The function σ is onto but not one-to-one. The inverse image of any t ∈ R≥0 consists
√
of ± t.
(b) σ : R≥0 → R≥0 and σ(s) = s2 .
Solution: Now σ is both one-to-one and onto. Here the inverse image of any t ∈ R≥0 is
√
t.
(c) σ : Z → Z and σ(s) = s2 .
Solution: This function is neither one-to-one
√ nor onto. The inverse image of t ∈ Z is empty if
t is not a perfect square and consists of ± t if t is a perfect square.
(d) σ : Z → Z and σ(s) = 2s.
Solution: The function σ is one-to-one but not onto. The inverse image of t ∈ Z is empty if t
is odd and consists of 2t if t is even.
6. 1.2.4
(a) Prove that the one-to-one correspondence relation is symmetric.
Solution: Let σ : S → T be a one-to-one correspondence. By Lemma 1.2.3 of Herstein there
exists a map µ : T → S such that σ ◦ µ is the identity map on T and µ ◦ σ is the identity map
on S. We prove that µ is a one-to-one correspondence.
• (one-to-one) Suppose µ(t1 ) = µ(t2 ). Applying σ to both sides we have σ(µ(t1 )) =
σ(µ(t2 )). Since σ ◦ µ is the identity map on T we get t1 = t2 and so µ is one-to-one.
• (onto) Let s ∈ S. We must show there exists t ∈ T for which µ(t) = s. We know that
µ ◦ σ is the identity map on S so µ(σ(s)) = s. Letting t = σ(s) proves that µ(t) = s and
so µ is onto.
(b) Prove that the one-to-one correspondence relation is transitive.
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Dr. Zachary Scherr
Math 370 HW 1
Solutions
Solution: Let σ : S → T be a one-to-one correspondence and let τ : T → U be a one-to-one
correspondence. We will prove that the composition
τ ◦ σ: S → U
is a one-to-one correspondence which will prove the assertion. This follows from Lemma 1.2.2
of Herstein, which we reprove here for completeness.
• (one-to-one) Suppose that τ ◦ σ(s1 ) = τ ◦ σ(s2 ). Then τ (σ(s1 )) = τ (σ(s2 )). Since τ is
one-to-one we get that σ(s1 ) = σ(s2 ). The map σ is also one-to-one so indeed s1 = s2
and thus τ ◦ σ is one-to-one.
• (onto) Let u ∈ U . Then since τ is onto there exists t ∈ T so that τ (t) = u. The map σ is
similarly onto so there exists s ∈ S so that σ(s) = t. Thus
τ ◦ σ(s) = τ (σ(s)) = τ (t) = u
and hence τ ◦ σ is onto.
7. 1.2.7
Solution: We could be rigorous and prove this claim by induction on the size of S. Instead, we will
use a combinatorial (counting) proof. Let #S = n and denote the elements of S as
S = {s1 , s2 , . . . , sn }.
Any map from S to itself must send s1 to one of the n elements s1 , . . . , sn . Thus there are n choices
for the image of s1 . In order for our map to be one-to-one, once we’ve chosen an image for s1 , the
image of s2 must be different. Thus there are n − 1 choices for the image of s2 . Continuing in this
fashion we see that once we have chosen the images of s1 , s2 , . . . , si then there are n − i choices for
si+1 . Thus in total there are
n × (n − 1) × (n − 2) × . . . × 1 = n!
one-to-one maps from S to itself.
8. 1.2.8
(a) An onto map from a finite set to itself is also one-to-one.
(b) A one-to-one map from a finite set to itself is also onto.
Solution: We prove both of these assertions. Let f : S → S be any map from the finite set
S to itself and let #S = n. By definition of a map we have f −1 (S) = S. We can write this
statement in terms of unions by:
[
S=
f −1 (s).
s∈S
Each of these subsets is disjoint so
#S =
X
s∈S
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#f −1 (s)
Dr. Zachary Scherr
Math 370 HW 1
Solutions
If f is onto then for every s ∈ S we have #f −1 (s) ≥ 1. In this case we get
X
X
n = #S =
#f −1 (s) ≥
1 = n.
s∈S
s∈S
Since n is not greater than n we must have #f −1 (s) = 1 for every s ∈ S which implies that f
is also one-to-one.
Conversely, if f is one-to-one then #f −1 (s) ≤ 1 for every s ∈ S. Thus
X
X
n = #S =
#f −1 (s) ≤
1 = n.
s∈S
s∈S
Again since we must have equality we see that #f −1 (s) = 1 for every s ∈ S and so indeed f is
onto.
(c) a) and b) may be false in the case S is infinite.
Solution: We give examples. Let σ : Z → Z be defined by σ(s) = 2s. Then σ is one-to-one
but not onto. Define τ : Z → Z by τ (s) = b 2s c. For those of you unfamiliar, bxc denotes the
largest integer which is at most x. For example, b1.5c = b1c = 1 and b−2.3c = −3. Then τ is
onto since τ (2n) = n for every integer n, but τ is certainly not one-to-one.
9. 1.2.11
(a) σA is a mapping of A into T .
Solution: This is obvious, but let’s prove it in full rigor. The map σ correspondes to a subset
M ⊆ S × T with the property that for all s ∈ S there is a unique t ∈ T such that (s, t) ∈ M .
Let
N = {(s, t) ∈ M | s ∈ A}.
Then N ⊆ M ∩ (A × T ) and by definition, for every s ∈ A there is a unique t ∈ T so that
(s, t) ∈ N . Hence N definites a function A → T which agrees with the definition of σA .
(b) The restriction of a one-to-one map is one-to-one
Solution: Let a1 , a2 ∈ A and suppose that σA (a1 ) = σA (a2 ). Then by definition of σA we
have σ(a1 ) = σ(a2 ) and so a1 = a2 since σ is one-to-one.
(c) The restriction map may be one-to-one even if the original map is not.
Solution: We give an example. Let S = R and T = R≥0 . Then the map σ : S → T defined by
σ(x) = x2 is not one-to-one. If A = R≥0 ⊆ S then the restriction map σA is one-to-one.
10. 1.2.13
(a) Z is infinite.
Solution: Let 2Z denote the set of even integers of Z. Then the function σ : Z → 2Z defined
by σ(s) = 2s is a one-to-one correspondence. Thus Z is infinite since 2Z ⊆ Z is a proper subset.
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Dr. Zachary Scherr
Math 370 HW 1
Solutions
(b) R is infinite.
Solution: Consider the map σ : R → R>0 defined by σ(x) = ex . Then σ is a one-to-one
correspondence and so R is an infinite set.
(c) If A ⊆ S and A is infinite then S is infinite as well.
Solution: The fact that A is infinite means there exits a proper subset T ⊆ A and a map
σ : A → T which is a one-to-one correspondence. Define a map τ : S → S by
(
s,
s∈S−A
τ (s) =
σ(s), s ∈ A.
Check that τ is one-to-one and that its image is S − (A − T ), which is a proper subset of S.
Thus τ is a one-to-one correspondence of S onto a proper subset of itself and hence S is infinite.
11. In class I proved that if S is a set with an equivalence relation ∼, then S equals a disjoint union of
equivalence classes. This problem explores the converse (see Theorem 1.1.1 in Herstein).
Let S be a set and suppose
S=
[
Aα
α∈I
where Aα ∩ Aβ = ∅ whenever α 6= β. Define a relation on S via x ∼ y if x and y are in the same subset
Aα . Prove that ∼ is an equivalence relation.
Solution: Let’s prove that ∼ is reflexive, symmetric and transitive.
• (reflexive) The fact that S = ∪α∈I Aα means that every a ∈ S lies in some Aα . Thus by
definition a ∼ a for every a ∈ S.
• (symmetric) If a ∼ b then a and b lie in the same Aγ for some γ ∈ I so indeed b ∼ a.
• (transitive) Suppose a ∼ b and b ∼ c. Then a, b ∈ Aα for some α ∈ I and b, c ∈ Aβ for some
β ∈ I. But then b ∈ Aα ∩ Aβ which implies α = β. Thus a, b, c ∈ Aα and so by definition
a ∼ c.
2
For Fun (i.e. very hard)
1. 1.2.6
Solution: Let f be any map f : S → S ∗ . We will prove that f is not onto by constructing an
element of S ∗ not in the image of f . To do so, let
T = {s ∈ S | s 6∈ f (s)}.
Then T , being a subset of S is an element of S ∗ . I claim that there is no element of S mapping to
T . Assume for the sake of contradiction that there is some t ∈ S for which f (t) = T . Then either
t ∈ T or t 6∈ T . If t ∈ T , then by definition of T we have that t 6∈ f (t) = T which is a contradiction.
If t 6∈ T then by definition of T we have t ∈ f (t) = T which is also a contradiction. Thus in all cases
we get a contradiction and hence there can be no element of S which maps to T .
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Dr. Zachary Scherr
Math 370 HW 1
Solutions
NOTE: One can use this result to prove that there is no one-to-one correspondence between Z and
R. One can prove that Z∗ is in one-to-one correspondence with R and so the result follows from this
question. There are other, more clever, ways to prove that there is no one-to-one correspondence between
the integers and the real numbers but this approach will work.
2. 1.2.14
Solution: Let N denote the set of positive integers. In class I showed that N can be put into
one-to-one correspondence with Z. Thus it suffices to prove that there is a one-to-one correspondence between N and N × N. The following construction gives an example of such a one-to-one
correspondence.
Think about putting the elements of N × N into an infinitely large table as follows
1
2
3
4
..
.
1
(1,1)
(2,1)
(3,1)
(4,1)
..
.
2
(1,2)
(2,2)
(3,2)
(4,2)
..
.
3
(1,3)
(2,3)
(3,3)
(4,3)
..
.
4
(1,4)
(2,4)
(3,4)
(4,4)
..
.
...
...
...
...
...
..
.
We can enumerate the elements of this table by moving diagonally along the table from left to right
and bottom to top. To get the idea, the first few elements of this list would be
(1, 1), (2, 1), (1, 2), (3, 1), (2, 2), (1, 3), (4, 1), (3, 2), (2, 3), (1, 4), . . .
Notice that if we could continue this list infinitely we would hit every single element of N × N. This
list defines a one-to-one correspondence of N onto N × N since for example we map
1
7→
(1, 1)
2
7→
(2, 1)
3
7→
(1, 2)
4
7→
..
.
(3, 1)
NOTE: Solving the next problem gives an easier way to prove a one-to-one correspondence between N
and N × N. Check that the function f : N → N × N defined by f (a) = (a, a) is one-to-one and that
the function g : N × N → N defined by g(a, b) = 2a 3b is also one-to-one. Thus there exists a one-to-one
correspondence between N and N × N.
One can use this same proof technique to prove that there is a bijection between Z, the integers, and Q,
the rational numbers.
3. Let X and Y be sets. Suppose there exists a one-to-one map f : X → Y and a one-to-one map g : Y → X.
Prove there exists a one-to-one correspondence of X to Y .
Solution: This is known as the Cantor-Bernstein-Schröder Theorem. This is a hard result, but I
hope the following proof makes the ideas somewhat intuitive.
For x ∈ X we call f (x) ∈ Y the child of x and similarly for y ∈ Y we call g(y) ∈ X the child of y.
The fact that f is one-to-one means that every y ∈ Y has at most one preimage in X. If y has a
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Dr. Zachary Scherr
Math 370 HW 1
Solutions
preimage we call f −1 (y) the parent of y. Otherwise we call y an orphan. Similarly, if x ∈ X has a
preimage under g we call g −1 (x) the parent of x and otherwise we call x an orphan.
There are three types of elements of X. Those elements whose ancestry (parent, grandparent, great
grandparent, etc.) begin with an orphan in X, those whose ancestry begins with an orphan in Y ,
and those ancestry goes on forever. Denote these sets Xx , Xy and X∞ . We can similarly define
Yx , Yy and Y∞ .
If x ∈ Xx then x’s ancestry begins with an orphan in X. Thus the ancestry of the child of x also
begins with an orphan in X. Mathematically, this means that f maps Xx into Yx . No element
of Yx is an orphan (the ancestry of an element in Yx cannot begin with an orphan in Y ) so every
element of Yx has a parent in Xx . This means that f : Xx → Yx is onto and hence is a one-to-one
correspondence. A similar argument shows that g : Yy → Xy is also a one-to-one correspondence.
Lastly, since no element of Y∞ is an orphan we see that f : X∞ → Y∞ is onto and thus gives a
one-to-one correspondence between X∞ and Y∞ . Since X is a disjoint union of Xx , Xy , X∞ and
Y is a disjoint union of Yx , Yy , Y∞ we can paste these one-to-one correspondences together to get a
one-to-one correspondence from X to Y .
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