MATH 16100-33: HONORS CALCULUS-1
HOMEWORK 4: SOLUTIONS MANUAL
Exercise 6-1(iv). Let f (x) = 1/q, x = p/q rational in lowest terms. Is there
a continuous function F with domain R such that F (x) = f (x) for all x ∈ Q?
Solution. If such an F existed, then
lim F (x) = F (1/2) = f (1/2) = 1/2.
x→1/2
Now let ε = 1/3, and let δ > 0. Choose an odd prime number n large enough
so that 1/n < δ, and set x = 1/2 + 1/n. Then |x − 1/2| = 1/n < δ, but
f (x) − 1 = F n + 2 − 1 = 1 − 1 ≥ 1 − 1 = 1 = ε,
2
2n
2
2n 2
6 2 3
where the inequality comes from n ≥ 3. Thus the limit is not 1/2, and F is not
continuous.
Exercise 6-14.
(a) Suppose that g and h are continuous at a, and that g(a) = h(a). Define
f (x) to be g(x) if x ≥ a and h(x) if x ≤ a. Prove that f is continuous at a.
Proof. Let ε > 0 be given. By continuity, there exist δg > 0 such that |x −
a| < δg implies |g(x) − g(a)| < ε, and δh > 0 such that |x − a| < δh implies
|h(x) − h(a)| < ε. Put δ = min(δg , δh ), and note that
(
|g(x) − g(a)| if x ≥ a
|f (x) − f (a)| =
.
|h(x) − h(a)| if x ≤ a
In particular, when −δ ≤ x − a ≤ 0, |f (x) − f (a)| = |g(x) − g(a)| < ε, and when
0 ≤ |x − a| < δ, |f (x) − g(a)| = |h(x) − h(a)| < ε.
(b) Suppose g is continuous on [a, b] and h is continuous on [b, c] and g(b) =
h(b). Let f (x) be g(x) for x in [a, b] and h(x) for x in [b, c]. Show that f is
continuous on [a, c].
Proof. Let x ∈ [a, c]. If x = a, then f (x) = g(x) for some half-open interval
[a, a + δ) so f is continuous at a. If x ∈ (a, b), then f (x) = g(x), so f is
continuous at x. If x = b, then f is continuous at b by Exercise 6-14(b). If
x ∈ (b, c), then f (x) = h(x), so f is continuous at x. If x = c, then f (x) = h(x)
on some half-open interval (c − δ, c] so f is continuous at c.
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Exercise 6-17. If limx→a f (x) exists but is 6= f (a), then f is said to have a
removable discontinuity at a.
(a) If f (x) = sin 1/x for x 6= 0 and f (0) = 1, does f have a removable discontinuity at 0? What if f (x) = x sin 1/x for x 6= 0 and f (0) = 1?
Solution. Let f (x) = sin 1/x for x 6= 0 and f (0) = 1. Let L ∈ R be given, and
set ε = 1 if L = 1 and ε = |1 − L| if L 6= 1. For any δ > 0, we can choose n ∈ N
large enough so that 0 < 2/(π(4n + 3)) < 2/(π(4n + 1)) < δ. Define
(
2/(π(4n + 3)) if L = 1
x=
2/(π(4n + 1)) if L 6= 1.
Then 0 < |x| < δ, but |f (x) − L| ≥ ε. Thus limx→0 sin 1/x does not exist, and
f does not have a removable discontinuity at 0.
Now put f (x) = x sin 1/x for x 6= 0 and f (0) = 1. Let ε > 0, and choose δ = ε.
Then whenever 0 < |x| < δ, we have
x sin 1 = |x| sin 1 < δ = ε,
x
x where the first inequality comes from | sin(α)| ≤ 1 for all α. Thus f has a removable discontinuity at x = 0.
(b) Suppose f has a removable discontinuity at a. Let g(x) = f (x) for x 6= a,
and let g(a) = limx→a f (x). Prove that g is continuous at a.
Proof. Since g(x) = f (x) on any deleted neighborhood of a,
lim g(x) = lim f (x) = g(a).
x→a
x→a
(c) Let f (x) = 0 if x is irrational, and let f (p/q) = 1/q if p/q is in lowest
terms. What is the function g defined by g(x) = limy→x f (y)?
Solution. As shown in lecture, f approaches 0 at any a ∈ [0, 1]. It is clear by
construction that f is 1-periodic, so g(x) = 0 for all x ∈ R.
(d) Let f be a function with the property that every point of discontinuity is
a removable discontinuity. This means that limy→x f (y) exists for all x, but
f may be discontinuous at some (even infinitely many) numbers x. Define
g(x) = limy→x f (y). Prove that g is continuous.
Proof. Let a ∈ R and ε > 0. Then there exists δ > 0 such that whenever
0 < |y − a| < δ, we have |f (y) − g(a)| < ε, or
−ε + g(a) < f (y) < ε + g(a).
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Letting y → x, we have
−ε + g(a) ≤ g(x) ≤ ε + g(a),
or |g(x) − g(a)| ≤ ε. Thus g is continuous at a, and since a was arbitrary, g is
continuous on the real line.
Exercise 8-1. Find the least upper bound and the greatest lower bound (if they
exist) of the following sets. Also decide which sets have greatest and least elements.
√
(iv) A = {x : 0 ≤ x ≤ 2 and x is rational}.
Solution. √
Since 0 ∈ A, the set is nonempty, and by construction A is bounded
above by 2 and below by 0. By the completeness axiom, the least upper bound
and greatest lower√bound exist.
√
To show that 2 is the least upper bound, suppose y < 2 was an upper
bound. By the
√ density of the rationals in the reals, there is some q ∈ Q such
that y < q < 2. But then q ∈ A, so y is not an upper bound. This argument
also shows that there is no greatest element in the set.
The greatest upper bound is clearly 0, as any positive number z admits a
positive rational r so that 0 < r < z. Since 0 is in the set, it is the greatest
lower bound.
1
+ (−1)n : n ∈ N .
n
Solution. The set is clearly nonempty, bounded above by 2, and bounded below
by −1, so the completeness axiom furnishes a least upper bound and greatest
lower bound.
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Let an denote the nth element of the set, and note that {a2n }n∈N = 2n
+ 1 n∈N
n
o
1
is positive and decreasing and {a2n+1 }n∈N = −1 + 2n+1
is negative and
(viii)
n∈N
decreasing. Thus a2 = 3/2 is the least upper bound and greatest element.
To show that −1 is the greatest lower bound, let y > −1 so that y + 1 > 0.
By the Archimedean property, there exists n ∈ N odd such that n1 < y + 1.
Rearranging and rewriting, y > n1 − 1 = n1 + (−1)n , so y is not a lower bound.
Exercise 8-2(b). If A 6= ∅ is bounded below, let B be the set of all lower bounds
of A. Show that B 6= ∅, that B is bounded above, and that sup B is the greatest
lower bound of A.
Proof. Since A is bounded below, B is not empty. Since B is the set of all y
which satisfy the inequality y ≤ x for every x ∈ A, every x ∈ A is an upper
bound of L. By the least upper bound axiom, sup B exists; call it β.
To show that β is the greatest lower bound of A, we will show that it is a
lower bound for A and that if y > β, y is not lower bound for A.
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If y < β, then by definition of least upper bound, y is not an upper bound
of B, and hence y ∈
/ A. It follows that if x ∈ A, then x ≥ β, so β is a lower
bound for A, and in particular, β ∈ B.
If y > β, then y ∈
/ B, since β is an upper bound and in the set.
Thus β is a lower bound for A, and y is not if y > β. It follows that
β = inf A.
Exercise 8-6. A set A of real numbers is said to be dense if every open interval
contains a point of A.
(a) Prove that if f is continuous and f (x) = 0 for all numbers x in a dense set
A, then f (x) = 0 for all x.
Proof. Let ε > 0, and choose x0 ∈ A{ . By continuity, there exists δ > 0 such
|x − x0 | < δ implies |f (x) − f (x0 )| < ε. Since A is dense, there exists y ∈ A
such that |y − x0 | < δ. In particular, |f (y) − f (x0 )| = |f (x0 )| < ε. Thus for
every ε > 0, we have |f (x0 )| < ε, implying f (x0 ) = 0.
(b) Prove that if f and g are continuous and f (x) = g(x) for all x in a dense
set A, then f (x) = g(x) for all x.
Proof. Put h = f − g so that h is continuous and h(x) = 0 for all x ∈ A. By
part (a), h(x) = 0 for all x, so f (x) = g(x) for all x.
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