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Solutions for Real Analysis and Foundations, 4th Edition Krantz (All Chapters included)
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Complete Solutions Manual for Real Analysis and Foundations, 4th Edition by Steven G. Krantz ; ISBN13: 9781315181592. (Full Chapters included Chapter 1 to 12)....Chapter 1.Number Systems Chapter 2.Sequences Chapter 3.Series of Numbers Chapter 4.Basic Topology Chapter 5.Limits and Continuity of ...

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  • March 13, 2024
  • 170
  • 2016/2017
  • Exam (elaborations)
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  • solutions
  • manual
  • real analysis
  • foundation
  • 4th edition
  • krantz
  • 9781315181592
  • solutions manual
  • 4e

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1


Instructor
Solutions Manual
for
Real Analysis and Foundations
Fourth Edition


by Steven G. Krantz




Complete Chapter Solutions Manual
are included (Ch 1 to 12)




** Immediate Download
** Swift Response
** All Chapters included

,Chapter 1

Number Systems

1.1 The Real Numbers
1. The set (0, 1] contains its least upper bound 1 but not its greatest lower
bound 0. The set [0, 1) contains its greatest lower bound 0 but not its
least upper bound 1.

2. The set Z ⊆ R has neither a least upper bound nor a greatest lower
bound.

3. We know that α ≥ a for every element a ∈ A. Thus −α ≤ −a for
every element a ∈ A hence −α ≤ b for every b ∈ B. If b0 > −α is a
lower bound for B then −b0 < α is an upper bound for A, and that is
impossible. Hence −α is the greatest lower bound for B.
Likewise, suppose that β is a greatest lower bound for A. Define
B = {−a : a ∈ A}. We know that β ≤ a for every element a ∈ A.
Thus −β ≥ −a for every element a ∈ A hence −β ≥ b for every b ∈ B.
If b0 < −β is an upper bound for B then −b0 > β is a lower bound for
A, and that is impossible. Hence −β is the least upper bound for B.
√
4. The least upper bound for S is 2.

5. We shall treat the least upper bound. Let α be the least upper bound
for the set S. Suppose that α0 is another least upper bound. It α0 > α
then α0 cannot be the least upper bound. If α0 < α then α cannot be
the least upper bound. So α0 must equal α.

1

, 2 CHAPTER 1. NUMBER SYSTEMS

6. Certainly S is bounded above by the circumference of C. The least
upper bound of S is π. This exercise cannot work in the rational
number system because π is irrational.
7. Let x and y be real numbers. We know that

(x + y)2 = x2 + 2xy + y 2 ≤ |x|2 + 2|x||y| + |y|2 .

Taking square roots of both sides yields

|x + y| ≤ |x| + |y| .

8. We treat the supremum. Notice that, since the empty set has no ele-
ments, then −∞ ≥ x for all x ∈ ∅ vacuously. There are no real numbers
less than −∞, so −∞ is the supremum of ∅.

9. We treat commutativity. According to the definition in the text, we
add two cuts C and D by

C + D = {c + d : c ∈ C, d ∈ D} .

But this equals
{d + c : c ∈ C, d ∈ D}
and that equals D + C.
11. Consider the set of all numbers of the form
j
√
k 2
for j, k relatively prime natural numbers and j < k. Then certainly
each of these numbers lies between 0 and 1 and each is irrational.
Furthermore, there are countably many of them.
* 12. Let x be in the domain of f. Then x is a local minimum, so there are
rational numbers αx < x < βx so that

f(x) ≤ f(t)

for every t ∈ (αx , βx). Thus we associate to each value f(x) of the
function f a pair of rational numbers (αx , βx). But the set of such
pairs is countable. So the set of values of f is countable.

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