Solutions for Real Analysis and Foundations, 4th Edition Krantz (All Chapters included)
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Analysis
Institution
Analysis
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 ...
Instructor
Solutions Manual
for
Real Analysis and Foundations
Fourth Edition
by Steven G. Krantz
Complete Chapter Solutions Manual
are included (Ch 1 to 12)
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** 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
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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