1
Student
Solutions Manual
for
Real Analysis and Foundations
Fourth Edition
by Steven G. Krantz
,Preface
This Manual contains the solutions to selected exercises in the book Real
Analysis and Foundations by Steven G. Krantz, hereinafter referred to as
“the text.”
The problems solved here have been chosen with the intent of covering the
most significant ones, the ones that might require techniques not explicitly
presented in the text, or the ones that are not easily found elsewhere.
The solutions are usually presented in detail, following the pattern in
the text. Where appropriate, only a sketch of a solution may be presented.
Our goal is to illustrate the underlying ideas in order to help the student to
develop his or her own mathematical intuition.
Notation and references as well as the results used to solve the problems
are taken directly from the text.
Steven G. Krantz
St. Louis, Missouri
,
, 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.
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.
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.
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 .
1
Student
Solutions Manual
for
Real Analysis and Foundations
Fourth Edition
by Steven G. Krantz
,Preface
This Manual contains the solutions to selected exercises in the book Real
Analysis and Foundations by Steven G. Krantz, hereinafter referred to as
“the text.”
The problems solved here have been chosen with the intent of covering the
most significant ones, the ones that might require techniques not explicitly
presented in the text, or the ones that are not easily found elsewhere.
The solutions are usually presented in detail, following the pattern in
the text. Where appropriate, only a sketch of a solution may be presented.
Our goal is to illustrate the underlying ideas in order to help the student to
develop his or her own mathematical intuition.
Notation and references as well as the results used to solve the problems
are taken directly from the text.
Steven G. Krantz
St. Louis, Missouri
,
, 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.
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.
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.
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 .
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