Math 220, Section 201/202
Solutions to Study Questions for First Midterm
I. (D’Angelo and West, p. 21, #1.7) The statement below is not always true for x, y ∈ R. Give an
example where it is false, and and a hypothesis on y that makes it a true statement.
“If x and y are nonzero real numbers and x > y, then (−1/ x) > (−1/ y).
The example x = 2, y = −3 shows that the statement is false: both 2 and −3 are
nonzero real numbers and 2 > −3, but it is false that − 12 > 13 . (Notice that a single counterexample is enough to prove the statement false, as it is implicitly a universal statement
about all real numbers x, y.) If we add the hypothesis y > 0, then the statement becomes
true. To prove this, suppose that x and y are nonzero real numbers with x > y and y > 0.
Then x > 0 as well. When we take the reciprocals of the positive real numbers in the inequality x > y, the inequality reverses: 1x < 1y . Now multiplying through by −1 reverses
the inequality again: − 1x > − 1y as claimed.
II. (D’Angelo and West, p. 21, #1.13) Let A be the set of integers expressible as 2k − 1 for some
k ∈ Z. Let B be the set of integers expressible as 2k + 1 for some k ∈ Z. Prove that A = B.
First of all, the following proof is not correct: “An integer is in A if and only if it is odd.
An integer is in B if and only if it is odd. Therefore A = B.” This is incorrect because
the writer has used two different definitions for the term “odd”: an integer that can be
written as 2k − 1, and alternatively an integer that can be written as 2k + 1. Are these
two definitions equivalent? Yes, but that equivalence is exactly what is supposed to be
proved!
Now for the correct proof. Suppose first that x ∈ A. Then we can choose an integer
k such that x = 2k − 1. This is the same as x = 2(k − 1) + 1. In particular, there exists
an integer (namely m = k − 1) such that x = 2m + 1. Hence x ∈ B. This proves that
A ⊆ B. Similarly, suppose now that x ∈ B. Then we can choose an integer k such that
x = 2k + 1. This is the same as x = 2(k + 1) − 1. In particular, there exists an integer
(namely m = k + 1) such that x = 2m − 1. Hence x ∈ A. This proves that B ⊆ A, and we
conclude that A = B as desired.
If you are uncomfortable with the ks changing to ms and so on, consider that the definition of A (for example) is A = { x ∈ Z : (∃k ∈ Z)( x = 2k − 1)}. But this means exactly
the same thing as A = { x ∈ Z : (∃m ∈ Z)( x = 2m − 1)}. The variables appearing in
quantifiers are “dummy variables”; their names are not important.
III. (D’Angelo and West, p. 21, #1.14) Let a, b, c, d be real numbers with a < b < c < d. Express
the set [ a, b] ∪ [c, d] as the difference of two sets.
The most straightforward answer is [ a, b] ∪ [c, d] = [ a, d] − (b, c). (Other answers are
possible, like [ a, b] ∪ [c, d] = R − S where S = (−∞, a) ∪ (b, c) ∪ (d, ∞).)
IV. (D’Angelo and West, p. 21, #1.15) For what conditions on A and B does A − B = B − A
hold?
The answer is that A − B = B − A if and only if A and B are equal. Suppose that
A − B = B − A. Then in particular, A − B ⊆ B − A. Since B − A ⊆ B, we conclude
that A − B ⊆ B. This means that for every x, we have x ∈ ( A − B) ⇒ x ∈ B; from the
definition of set complement, this is equivalent to ( x ∈ A) ∧ ( x ∈
/ B) ⇒ x ∈ B. Now
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the statements ( P ∧ ¬ Q) ⇒ Q and P ⇒ Q are logically equivalent (one way to see this is
to change both implications to their or-forms). Therefore ( x ∈ A) ∧ ( x ∈
/ B) ⇒ x ∈ B is
equivalent to ( x ∈ A) ⇒ ( x ∈ B). Since this is true for every x, we conclude that A ⊆ B.
The same argument with the roles of A and B reversed shows that B ⊆ A as well. We
conclude that A = B. Note that so far we have only proved that ( A − B = B − A) ⇒
( A = B), but the problem asks for an exact characterization, so we still need to prove that
( A = B) ⇒ ( A − B = B − A). But if A = B then it is easy to see that both A − B and
B − A equal the empty set, and hence they are equal to each other as desired.
V. (D’Angelo and West, p. 22, #1.21) Let a, b, c be real numbers with a 6= 0. Find the flaw in the
book’s “proof” that −b/2a is a solution to ax2 + bx + c = 0.
Consider this excerpt from the proof: “. . . which we rewrite as a( x + y)( x − y) + b( x −
y) = 0. Hence a( x + y) + b = 0, . . . ”. Here the author has divided by x − y, which is only
valid when x − y 6= 0. However, later y is set equal to x, thus ensuring that x − y = 0 and
making the given excerpt invalid. The moral is to always note the circumstances under
which your manipulations are not valid.
VI. (D’Angelo and West, p. 22, #1.27) Determine the set of real solutions to | x/( x + 1)| ≤ 1.
The answer is [− 21 , ∞). We might guess this by interpreting | x| as the distance from x to
0 and | x + 1| = | x − (−1)| as the distance from x to −1; then the inequality | x/( x + 1)| ≤ 1
is the same as | x| ≤ | x + 1|, which means that x is closer to 0 than to −1. In any case, we
prove formally that
| x/( x + 1)| ≤ 1 ⇔ x ∈ [− 12 , ∞),
splitting the proof into four cases according to the location of x.
• First suppose that x ≥ 0. Then x is nonnegative and x + 1 is positive, so | x/( x +
1)| = x/( x + 1), and x/( x + 1) ≤ 1 if and only if x ≤ x + 1. But this is always true
for this case, and so every x ≥ 0 satisfies the original inequality.
• Now suppose that −1 < x < 0. Then x is negative and x + 1 is positive, so
| x/( x + 1)| = − x/( x + 1), and − x/( x + 1) ≤ 1 if and only if − x ≤ x + 1. But this
is equivalent to −1 ≤ 2x or − 12 ≤ x, so the numbers in this case that satisfy the
original inequality are those x for which − 12 ≤ x < 0.
• If x = −1 then the expression x/( x + 1) is not defined, and hence the inequality
cannot be true.
• Finally, if x < −1 then both x and x + 1 are negative, so | x/( x + 1)| = (− x)/(−( x +
1)) = x/( x + 1), and x/( x + 1) ≤ 1 if and only if x ≥ x + 1. But this is never true
for this case, and so no x < −1 satisfies the original inequality.
We conclude that | x/( x + 1)| ≤ 1 if and only if either x ≥ 0 or − 12 ≤ x < 0, that is, if and
only if x ≥ − 21 , which is the same as saying x ∈ [− 12 , ∞).
VII. (D’Angelo and West, p. 23, #1.34) Let S = {( x, y) ∈ R2 : (1 − x)(1 − y) ≥ 1 − x − y}.
Give a simple description of S involving the signs of x and y.
We haven’t talked about ordered pairs and Cartesian products of sets yet, so this problem is a little ahead of where we are in class. But for those who are interested, the answer
is that S equals the set of ordered pairs ( x, y) ∈ R2 such
that either x ≥ 0 and y ≥ 0, or
x ≤ 0 and y ≤ 0. In other words, S = [0, ∞) × [0, ∞) ∪ (−∞, 0] × (−∞, 0] .
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VIII. (D’Angelo and West, p. 23, #1.40) Let A and B be sets. Explain why the two sets ( A − B) ∪
( B − A) and ( A ∪ B) − ( A ∩ B) must be equal.
We prove the two inclusions separately.
• ( A − B) ∪ ( B − A) ⊆ ( A ∪ B) − ( A ∩ B). Suppose that x ∈ ( A − B) ∪ ( B − A); we
want to prove that x ∈ ( A ∪ B) − ( A ∩ B). We know that either x ∈ ( A − B) or
x ∈ ( B − A). In the former case, we have x ∈ A and x ∈
/ B. Since x ∈ A, we have
x ∈ A ∪ B; since x ∈
/ B, we have x ∈
/ A ∩ B. Therefore x ∈ ( A ∪ B) − ( A ∩ B) in
the case x ∈ ( A − B). The case x ∈ ( B − A) is the same with the roles of A and B
reversed.
• ( A ∪ B) − ( A ∩ B) ⊆ ( A − B) ∪ ( B − A). Suppose that x ∈ ( A ∪ B) − ( A ∩ B);
we want to prove that x ∈ ( A − B) ∪ ( B − A). We know that x ∈ ( A ∪ B) and
x∈
/ ( A ∩ B). Because x ∈ ( A ∪ B), we have that either x ∈ A or x ∈ B. In the first
case, since x ∈ A and x ∈
/ ( A ∩ B), we conclude that x ∈
/ B; therefore x ∈ ( A − B).
In the second case, since x ∈ B and x ∈
/ ( A ∩ B), we conclude that x ∈
/ A; therefore
x ∈ ( B − A). In either case, we can conclude that x ∈ ( A − B) ∪ ( B − A) as desired.
(Note the similarities between these two set expressions and two of the logical expressions
in the solution to problem XI).
IX. (D’Angelo and West, p. 46, #2.24) In simpler language, describe the meaning of the following
two statements and their negations. Which one implies the other, and why?
(a) There is a number M such that, for every x in the set S, | x| ≤ M.
(b) For every x in the set S, there is a number M such that | x| ≤ M.
The first sentence says that the set S is bounded above by the number M; the negation
would say that the set S is unbounded above. The second sentence says that each element
of S is bounded above by some number M; the negation would say that some element of S
is not bounded above. The first sentence implies the second. In general, (∃ x)(∀ y) P( x, y)
always implies (∀ y)(∃ x) Q( x, y) (don’t memorize this, just think about what they mean).
Note that the second statement is trivially true for any set S ⊆ R—we can just take M
to be | x| for every x ∈ S. Remember, we read quantifiers left to right, so in the second
sentence, we are allowed to wait until x has been given before we choose M. This isn’t
possible in the first sentence: we have to choose a single M first, and then hope it works
for all x ∈ S.
X. (D’Angelo and West, p. 49, #2.49) Let S = { x ∈ R : x2 > x + 6}. Let T = { x ∈ R : x > 3}.
Determine whether the following statements are true, and interpret these results in words:
(a) T ⊆ S.
(b) S ⊆ T.
(a) This statement is true; it is equivalent to “For every real number x, if x > 3 then
x2 > x + 6.” To prove this, note that the hypothesis implies x − 3 > 0. Since 5 > 0,
we can add to obtain ( x − 3) + 5 = x + 2 > 0. Multiplying these two inequalities
yields ( x − 3)( x + 2) = x2 − x − 6 > 0, which implies x2 > x + 6.
(b) This statement is false, as the counterexample x = −3 shows: −3 ∈ S but −3 ∈
/ T.
(Note that T = (3, ∞) in our notation.) The statement is equivalent to “For every
real number x, if x2 > x + 6 then x > 3.”
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XI. Using the logical symbols we have been working with, find a formula that expresses the “exclusive or”. That is, find a formula F ( P, Q) involving the sentences P and Q that has the following
truth table:
P Q F ( P, Q)
T T
F
T F
T
F T
T
F F
F
One way is to start with “or” and then rule out the possibility that P and Q are both
true: ( P ∨ Q) ∧ (¬( P ∧ Q)), which is equivalent to ( P ∨ Q) ∧ (¬ P ∨ ¬ Q). Another way is to
simply list the two possible situations that make the statement true: ( P ∧ ¬ Q) ∨ (¬ P ∧ Q),
corresponding to the second and third rows of the truth table, respectively. Finally, there
is a very slick formula expressing “exclusive or”: P ⇔ ¬ Q.
XII. Prove that Z ∩ ( 41 , 34 ) = ∅.
We have to prove (∀ x)( x ∈
/ Z ∩ ( 14 , 34 )). By the definition of intersection,
( x ∈ Z ∩ ( 14 , 34 )) ⇔ ( x ∈ Z) ∧ ( x ∈ ( 14 , 34 )).
Negating both sides yields
(x ∈
/ Z ∩ ( 41 , 34 )) ⇔ ( x ∈
/ Z) ∨ ( x ∈
/ ( 14 , 34 )),
which is the or-form of
(x ∈
/ Z ∩ ( 41 , 34 )) ⇔ ( x ∈ Z) ⇒ ( x ∈
/ ( 14 , 34 )) .
Therefore we have to prove (∀ x) ( x ∈ Z) ⇒ ( x ∈
/ ( 14 , 34 )) . We need to use some known
fact about the integers, such as “there is no integer between 0 and 1”. In other words, we
use
(∀ x) ( x ∈ Z) ⇒ ( x ∈
/ (0, 1)) .
(∗)
Since ( 41 , 34 ) ⊆ (0, 1) (prove it from the definitions of the intervals!), we have
(∀ x) ( x ∈ ( 41 , 34 )) ⇒ ( x ∈ (0, 1)) ,
which is logically equivalent to its contrapositive
(∀ x) ( x ∈
/ (0, 1)) ⇒ ( x ∈
/ ( 14 , 34 )) .
(∗∗)
Now the two implications (∗) and (∗∗) link together to give (∀ x) ( x ∈ Z) ⇒ ( x ∈
/ ( 41 , 43 ))
as desired.
XIII. Find set identities that are analogous to each of the following tautologies:
(a) ( P ∧ Q) ⇒ P
(b) ( P ∧ ¬ P) ⇒ Q
(c) P ∧ ( Q ∨ R) ⇔ ( P ∧ Q) ∨ ( P ∧ R)
(d) P ∨ (¬ P)
(a) A ∩ B ⊆ A.
(b) A ∩ Ac ⊆ B. Or, since A ∩ Ac always equals the empty set (prove it!), ∅ ⊆ B for
every set B.
4
(c) A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C ). The other “distributive law for ands/ors” gives
a similar “distributive law for unions/intersections”.
(d) This is tricky because P ∨ (¬ P) is more of a “noun” then a “sentence”. I think
the closest analogue is A ∪ Ac = U, where U is the universe implicit in the notation for complement. This set identity follows immediately from the definition of
complement.
XIV. In class, we learned that to prove A = ∅, we should show that
(∀ x)( x ∈
/ A).
(∗)
On the other hand, we learned that to prove that any two sets A and B are equal, we need to prove
(∀ x) ( x ∈ A ⇒ x ∈ B) ∧ ( x ∈ B ⇒ x ∈ A) .
It seems we should be able to take B = ∅, so that proving that A = ∅ requires proving
(∀ x) ( x ∈ A ⇒ x ∈ ∅) ∧ ( x ∈ ∅ ⇒ x ∈ A) .
(∗∗)
Is this logically equivalent to what we learned in class?
Yes, it is. Let P( x) be the statement x ∈ A, and let Q( x) be the statement x ∈ ∅. The
first statement (∗) is simply (∀ x)(¬ P( x)), while the second statement (∗∗) is
(∀ x) ( P( x) ⇒ Q( x)) ∧ ( Q( x) ⇒ P( x)) .
We have another fact at our disposal: no object is a member of the empty set, that is,
¬ Q( x) is a true statement for every x. Our hope is that this should imply the equivalence
of (∗) and (∗∗); in other words, we want
¬ Q( x) ⇒ ¬ P( x) ⇔ ( P( x) ⇒ Q( x)) ∧ ( Q( x) ⇒ P( x)) .
Fortunately, this statement is a tautology, as can be checked using a truth table.
XV. (D’Angelo and West, p. 24, #1.46) Determine the images of the functions f : R → R defined
as follows:
(a) f ( x) = x2 /(1 + x2 )
(b) f ( x) = x/(1 + | x|)
(a) The image of f is [0, 1). We first prove that the image of f is contained in [0, 1),
and then prove that [0, 1) contains the image of f .
Let y be an element of the image of f . By the definition of image, this means that
y = f ( x) = x2 /(1 + x2 ) for some x ∈ R. Now x2 ≥ 0, and so 1 + x2 ≥ 1 > 0, so
x2 /(1 + x2 ) must be nonnegative; hence y ≥ 0. Also, x2 < 1 + x2 , and so (since
both numbers are nonnegative) x2 /(1 + x2 ) < 1. Therefore y < 1. We conclude
that y ∈ [0, 1) as desired.
p
Now let y ∈ [0, 1). Define x =
y/(1 − y). (Note that y is nonnegative and
1 − y is positive, since y ∈ [0, 1); therefore y/(1 − y) is nonnegative and hence x is
well-defined.) Then
p
( y/(1 − y))2
x2
y/(1 − y)
p
f ( x) =
=
=
2
1 + y/(1 − y)
1+x
1 + ( y/(1 − y))2
5
(again using the fact that y/(1 − y) is nonnegative). Multiplying the numerator
and denominator by the nonzero quantity 1 − y, this becomes
f ( x) =
y/(1 − y) 1 − y
y
y
=
= = y.
1 + y/(1 − y) 1 − y
(1 − y) + y
1
In particular, there exists a real number x such that f ( x) = y. Therefore y is in the
image of f as desired.
(b) The image of f is (−1, 1). We first prove that the image of f is contained in [0, 1),
and then prove that (−1, 1) contains the image of f .
Let y be an element of the image of f . By the definition of image, this means
that y = f ( x) = x/(1 + | x|) for some x ∈ R. Then | y| = | x/(1 + | x|)| = | x|/|1 +
| x|| = | x|/(1 + | x|) since 1 + | x| is always positive. Because | x| < 1 + | x|, we
have | x|/(1 + | x|) < 1, and so | y| < 1. But this implies that −1 < y < 1, and so
y ∈ (−1, 1) as desired.
Now let y ∈ (−1, 1). Define x = y/(1 − | y|). Then
f ( x) =
x
y/(1 − | y|)
y/(1 − | y|)
y/(1 − | y|)
=
=
=
.
1 + | x|
1 + | y/(1 − | y|)|
1 + | y|/|1 − | y||
1 + | y|/(1 − | y|)
(Here we have used the fact that | y| < 1, so that 1 − | y| is positive). Multiplying
the numerator and denominator by the nonzero quantity 1 − | y|, this becomes
f ( x) =
y/(1 − | y|) 1 − | y|
y
y
=
= = y.
1 + | y|/(1 − | y|) 1 − | y|
(1 − | y|) + | y|
1
In particular, there exists a real number x such that f ( x) = y. Therefore y is in the
image of f as desired.
XVI. (D’Angelo and West, p. 24, #1.49) Let f and g be functions from R to R. For the sum and
product of f and g, determine which statements below are true. If true, provide a proof; if false,
provide a counterexample.
(a) If f and g are bounded, then f + g is bounded.
(b) If f and g are bounded, then f g is bounded.
(c) If f + g is bounded, then f and g are bounded.
(d) If f g is bounded, then f and g are bounded.
(e) If both f + g and f g are bounded, then f and g are bounded.
(a) This statement is true. Since f is bounded, there is a real number M such that
| f ( x)| ≤ M for all x ∈ R. Similarly, since g is bounded, there is a real number N
such that | f ( x)| ≤ N for all x ∈ R. If we can prove that |( f + g)( x)| ≤ M + N for
all x ∈ R, then we will have shown that the function f + g satisfies the definition
of a bounded function (using the real number M + N). So let x ∈ R. By the
definition of f + g, we have |( f + g)( x)| = | f ( x) + g( x)|. The triangle inequality
(Proposition 1.3 of D’Angelo and West) tells us that | f ( x) + g( x)| ≤ | f ( x)| + | g( x)|.
Since | f ( x)| ≤ M and | g( x)| ≤ N, we conclude that | f ( x)| + | g( x)| ≤ M + N.
Stringing these inequalities together gives |( f + g)( x)| ≤ M + N as desired.
(b) This statement is true. Since f is bounded, there is a real number M such that
| f ( x)| ≤ M for all x ∈ R. Similarly, since g is bounded, there is a real number N
such that | f ( x)| ≤ N for all x ∈ R. If we can prove that |( f g)( x)| ≤ MN for all
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x ∈ R, then we will have shown that the function f g satisfies the definition of a
bounded function (using the real number MN). So let x ∈ R. By the definition of
f g, we have |( f g)( x)| = | f ( x) g( x)| = | f ( x)| · | g( x)|. Now M and N must both
be nonnegative: for instance, M ≥ | f (1)| ≥ 0 since absolute values are always
nonnegative, and the same argument works for N. Since | f ( x)| ≤ M and | g( x)| ≤
N and all quantities in these inequalities are nonnegative, we conclude that | f ( x)| ·
| g( x)| ≤ MN. Stringing these inequalities together gives |( f g)( x)| ≤ MN as
desired.
(c) This statement is false. Let f ( x) = x − 2 and g( x) = − x + 3; then ( f + g)( x) =
( x − 2) + (− x + 3) = 1 for all x ∈ R. In particular, |( f + g)( x)| ≤ 1 for all x ∈ R,
and so f + g is bounded. However, neither f nor g is bounded in this case. It is
only necessary to prove that f is unbounded
(why?). We need to prove false the
statement (∃ M ∈ R)(∀ x ∈ R) | f ( x)| ≤ M , and so we need to prove its negation
(∀ M ∈ R)(∃ x ∈ R) | f ( x)| > M true. So let M be a real number. If M ≤ 0 then
choose x = 3; certainly | f (3)| = |3 − 2| = 1 > 0 ≥ M. On the other hand, if M > 0
then choose x = M + 3; in this case
| f ( x)| = | x − 2| = |( M + 3) − 2| = | M + 1| = M + 1 > M.
(The equality | M + 1| = M + 1 is valid since M, and hence M + 1, is positive.) In
either case, there exists a real number x such that | f ( x)| > M as desired.
(d) This statement is false. Its negation is equivalent to “ f g is bounded and either f
or g is unbounded”, which we prove true by the following rexample. Let f ( x) =
x2 + 1 and let g( x) = 1/( x2 + 1). Then ( f g)( x) = f ( x) g( x) = 1 for all x ∈ R,
and so f g is certainly bounded. However, we can show that f is unbounded.
Let M be a real number. If M ≤ 0 then choose x = 0; then f ( x) =
√ f (0) =
2
0 + 1 = 1 > 0 ≥√M. On the other hand, if M > 0 then
√ choose x = M; then
f ( x) = x2 + 1 = ( M)2 + 1 = M + 1 > M. (Replacing ( M)2 by M is valid since
M is positive.) In either case, there exists a real number x such that | f ( x)| > M as
desired.
(e) Following the book’s hint, we first show that the function f 2 + g2 is bounded.
Notice that f 2 + g2 = ( f 2 + 2 f g + g2 ) − 2 f g = ( f + g)2 − h( f g), where h is the
constant function h( x) = −2. Now we are assuming that f + g is bounded. By part
(b) we see that ( f + g)( f + g) is bounded as well. We are also assuming that f g
is bounded. Since the constant function h is clearly bounded as well, we see from
part (b) again that h( f g) is bounded. Since the sum of two bounded functions is
still bounded by part (a), we conclude that ( f + g)2 − h( f g) is indeed bounded,
and hence f 2 + g2 is bounded.
Now we prove that f is bounded; the proof that g is bounded is exactly similar.
Since f 2 + g2 is bounded, choose a real number M such that |( f 2 + g2 )( x)| ≤ M
for all x ∈ R. Of course it must
that M ≥ 0 as in part (b). Now for
p be truep
2
any x ∈ R, we have | f ( x)| =
f ( x) ≤
f ( x)2 + g( x)2 since g( x)2 is nonnegative. But f ( x)2 + g( x)2 = | f ( x)2 + g( x)2 | since it is a nonnegative quantity, and
| f ( x)2 + g( x)2 | = |( f 2 + g2 )( x)| by the definition of the function f 2 + g2 . We
have chosen M so that |( f 2 + g2 )( x)| ≤ M. Stringing these facts together yields
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p
p
√
2 =
| f ( x)| ≤ f ( x)2 + g( x)√
|( f 2 + g2 )( x)| ≤ M. In particular, there exists a
real number N (namely M) such that | f ( x)| ≤ N for all x ∈ R, which proves that
f is bounded.
XVII. (D’Angelo and West, p. 24, #1.51) When f : A → B and S ⊆ B, we define I f ( S) = { x ∈
A : f ( x) ∈ S}. Let X and Y be subsets of B.
(a) Determine whether I f ( X ∪ Y ) must equal I f ( X ) ∪ I f (Y ).
(b) Determine whether I f ( X ∩ Y ) must equal I f ( X ) ∩ I f (Y ).
(a) These two sets are equal, which we prove by showing that x ∈ I f ( X ∪ Y ) if and
only if x ∈ I f ( X ) ∪ I f (Y ). By the definition of I f , we have x ∈ I f ( X ∪ Y ) if and
only if x ∈ A and f ( x) ∈ X ∪ Y. This is equivalent to
x ∈ A ∧ ( f ( x) ∈ X ∨ f ( x) ∈ Y ).
On the other hand, we have x ∈ I f ( X ) ∪ I f (Y ) if and only if x ∈ I f ( X ) or x ∈ I f (Y ).
By the definition of I f , this is equivalent to
( x ∈ A ∧ f ( x) ∈ X ) ∨ ( x ∈ A ∧ f ( x) ∈ Y ).
These two conditions are equivalent by the distributive law of logic, and hence
x ∈ I f ( X ∪ Y ) if and only if x ∈ I f ( X ) ∪ I f (Y ) as desired.
(b) These two sets are equal as well, which we prove by showing that x ∈ I f ( X ∩ Y )
if and only if x ∈ I f ( X ) ∩ I f (Y ). By the definition of I f , we have x ∈ I f ( X ∩ Y ) if
and only if x ∈ A and f ( x) ∈ X ∩ Y. This is equivalent to
x ∈ A ∧ ( f ( x) ∈ X ∧ f ( x) ∈ Y ).
On the other hand, we have x ∈ I f ( X ) ∩ I f (Y ) if and only if x ∈ I f ( X ) and x ∈
I f (Y ). By the definition of I f , this is equivalent to ( x ∈ A ∧ f ( x) ∈ X ) ∧ ( x ∈
A ∧ f ( x) ∈ Y ), which is the same as
( x ∈ A ∧ x ∈ A) ∧ ( f ( x) ∈ X ∧ f ( x) ∈ Y )
since we can change order in a multiple “and” statement. Since x ∈ A ∧ x ∈ A is
equivalent to x ∈ A, these two conditions are equivalent, and hence x ∈ I f ( X ∩ Y )
if and only if x ∈ I f ( X ) ∩ I f (Y ) as desired.
XVIII. Determine which statements below are true. If true, provide a proof; if false, provide a
counterexample.
(a) If f : R → R and g : R → R are increasing functions, then the function f + g is
increasing.
(b) If f : R → R and g : R → R are increasing functions, then the function f g is increasing.
(c) If f : [0, ∞) → R and g : [0, ∞) → R are increasing functions, then the function f g is
increasing.
(a) This statement is true. Let x1 , x2 be real numbers with x1 < x2 ; we need to prove
that ( f + g)( x1 ) < ( f + g)( x2 ). Since f is increasing and x1 < x2 , we know
that f ( x1 ) < f ( x2 ). Similarly, since g is increasing we know that g( x1 ) < g( x2 ).
Adding these two inequalities together yields f ( x1 ) + g( x1 ) < f ( x2 ) + g( x2 ). By
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the definition of the function f + g, this inequality is equivalent to ( f + g)( x1 ) <
( f + g)( x2 ) as desired.
(b) This statement is false. Let f ( x) = x − 2 and g( x) = x − 3; then both f and g are
increasing functions. However, f g is the function ( f g)( x) = ( x − 2)( x − 3), which
is not increasing; for example, 1 < 3 but ( f g)(1) = 2 > 0 = ( f g)(3).
(c) This statement is false; the counterexample in part (b) works here too if we restrict
the domains of f and g to be [0, ∞).
XIX.
(a) Let S1 = { x ∈ R : xn ≤ x for every n ∈ N}. Prove that S1 = [0, 1]. (Don’t neglect
negative numbers x!)
(b) Let S2 = { x ∈ R : xn ≥ x for every n ∈ N}. Find (with proof) a simple expression for S2
in terms of intervals.
(c) Prove that for all real numbers x, if ( x123 − x)( x124 − x) < 0 then x < −1.
(a) We first prove that x ∈ [0, 1] ⇒ x ∈ S1 and then prove that x ∈ S1 ⇒ x ∈ [0, 1].
Let x ∈ [0, 1]. Let n ∈ N. If n = 1 then xn = x1 ≤ x trivially, so we can suppose
n ≥ 2. Since the product of positive numbers that are less than or equal to 1 is
still a positive number that is less than or equal to 1, we see that xn−1 is positive
and less than or equal to 1. Multiplying the inequality xn−1 ≤ 1 through by the
positive number x gives xn ≤ x. This shows that x ∈ S1 as desired.
For the second half, we prove the contrapositive, namely x ∈
/ [0, 1] ⇒ x ∈
/ S1 .
Let x ∈
/ [0, 1]; then either x > 1 or x < 0. If x > 1 then, multiplying both sides by
the positive number x, we see that x2 > x. If x < 0 then x2 > 0 > x. In either case,
x2 > x, and so it is false that xn ≤ x for every natural number n. Therefore x ∈
/ S1
as desired.
(b) We claim that S2 = [−1, 0] ∪ [1, ∞). We first prove that x ∈ [−1, 0] ∪ [1, ∞) ⇒ x ∈
S2 and then prove that x ∈ S2 ⇒ x ∈ [−1, 0] ∪ [1, ∞).
Let x ∈ [−1, 0] ∪ [1, ∞); then either x ≥ 1 or −1 ≤ x ≤ 0. In the first case, we
have xn−1 ≥ 1 for all n ∈ N, since the product of numbers that are greater than or
equal to 1 is still a positive number that is greater than or equal to 1. Multiplying
through by the positive number x, we see that xn ≥ x as desired. Now suppose
that −1 ≤ x ≤ 0. Clearly x1 ≤ x, and for every even n ∈ N, clearly xn ≥ 0 ≥ x. If
n ≥ 3 is odd, write n = 2k + 1 with k ∈ N. Then x2k = ( x2 )k is a positive number
in [0,1], and so ( x2 )k ≤ 1 for every k ∈ N. Multiplying through by the negative
number x gives xn = x2k+1 = ( x2 )k x ≥ x as desired.
For the second half, we prove the contrapositive, namely x ∈
/ [−1, 0] ∪ [1, ∞) ⇒
x∈
/ S2 . Let x ∈
/ [−1, 0] ∪ [1, ∞); then either x < −1 or 0 < x < 1. If x < −1, then
x2 > 1, and multiplying by the negative number x yields x3 < x. If 0 < x < 1 then
x2 < 1, and multiplying by the positive number x yields x3 < x. In either case,
x3 < x, and so it is false that xn ≥ x for every natural number n. Therefore x ∈
/ S2
as desired.
(c) We prove the contrapositive, namely x ≥ −1 ⇒ ( x123 − x)( x124 − x) ≥ 0. Suppose
x ≥ −1. Then x ∈ [−1, ∞) = [0, 1] ∪ [−1, 0] ∪ [1, ∞) = S1 ∪ S2 . If x ∈ S1 , then
xn ≤ x for every n ∈ N. In particular, x123 and x124 are both less than or equal
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to x; hence ( x123 − x)( x124 − x), being the product of two nonpositive numbers,
is nonnegative as desired. On the other hand, if x ∈ S2 , then xn ≥ x for every
n ∈ N. In particular, x123 and x124 are both greater than or equal to x; hence ( x123 −
x)( x124 − x), being the product of two nonnegative numbers, is nonnegative as
desired.
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