to restate part (ii). Lemma 1.3.8. Assume s ER is an upper bound for a set A CR. Then, sup A if and only if, for every choice of e> 0, there exists an element a Є A satisfying s - € < a. 18 Chapter 1. The Real Numbers Proof. Here is a short rephrasing of the lemma: Given that s is an upper bound, s is the least upper bound if and only if any number smaller than s is not an upper bound. Putting it this way almost qualifies as a proof, but we will expand on what exactly is being said in each direction. () For the forward direction, we assume s = sup A and consider s-e, where € > 0 has been arbitrarily chosen. Because s-e 0 is chosen, s-e is no longer an upper bound for A. Notice that what this implies is that if b is any number less than s, then b is not an upper bound. (Just let e = s-b.) To prove that s = sup A, we must verify part (ii) of Definition 1.3.2. (Read it again.) Because we have just argued that any number smaller than s cannot be an upper bound, it follows that if b is some other upper bound for A, then s sup B. (b) A finite set that contains its infimum but not its supremum. (c) A bounded subset of Q that contains its supremum but not its infimum. Exercise 1.3.3. (a) Let A be nonempty and bounded below, and define B = {bЄR: b is a lower bound for A}. Show that sup B = inf A. (b) Use (a) to explain why there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness. Exercise 1.3.4. Let A1, A2, A3,... be a collection of nonempty sets, each of which is bounded above.
to restate part (ii). Lemma 1.3.8. Assume s ER is an upper bound for a set A CR. Then, sup A if and only if, for every choice of e> 0, there exists an element a Є A satisfying s - € < a. 18 Chapter 1. The Real Numbers Proof. Here is a short rephrasing of the lemma: Given that s is an upper bound, s is the least upper bound if and only if any number smaller than s is not an upper bound. Putting it this way almost qualifies as a proof, but we will expand on what exactly is being said in each direction. () For the forward direction, we assume s = sup A and consider s-e, where € > 0 has been arbitrarily chosen. Because s-e 0 is chosen, s-e is no longer an upper bound for A. Notice that what this implies is that if b is any number less than s, then b is not an upper bound. (Just let e = s-b.) To prove that s = sup A, we must verify part (ii) of Definition 1.3.2. (Read it again.) Because we have just argued that any number smaller than s cannot be an upper bound, it follows that if b is some other upper bound for A, then s sup B. (b) A finite set that contains its infimum but not its supremum. (c) A bounded subset of Q that contains its supremum but not its infimum. Exercise 1.3.3. (a) Let A be nonempty and bounded below, and define B = {bЄR: b is a lower bound for A}. Show that sup B = inf A. (b) Use (a) to explain why there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness. Exercise 1.3.4. Let A1, A2, A3,... be a collection of nonempty sets, each of which is bounded above.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please answer Exercise 1.3.1
Transcribed Image Text:to restate part (ii).
Lemma 1.3.8. Assume s ER is an upper bound for a set A CR. Then,
sup A if and only if, for every choice of e> 0, there exists an element a Є A
satisfying s - € < a.
Transcribed Image Text:18
Chapter 1. The Real Numbers
Proof. Here is a short rephrasing of the lemma: Given that s is an upper bound,
s is the least upper bound if and only if any number smaller than s is not an
upper bound. Putting it this way almost qualifies as a proof, but we will expand
on what exactly is being said in each direction.
() For the forward direction, we assume s = sup A and consider s-e, where
€ > 0 has been arbitrarily chosen. Because s-e<s, part (ii) of Definition 1.3.2
implies that s-e is not an upper bound for A. If this is the case, then there
must be some element a Є A for which s-e <a (because otherwise 8 - € would
be an upper bound). This proves the lemma in one direction.
(<) Conversely, assume s is an upper bound with the property that no
matter how € > 0 is chosen, s-e is no longer an upper bound for A. Notice
that what this implies is that if b is any number less than s, then b is not an
upper bound. (Just let e = s-b.) To prove that s = sup A, we must verify part
(ii) of Definition 1.3.2. (Read it again.) Because we have just argued that any
number smaller than s cannot be an upper bound, it follows that if b is some
other upper bound for A, then s <b.
It is certainly the case that all of our conclusions to this point about least
upper bounds have analogous versions for greatest lower bounds. The Axiom of
Completeness does not explicitly assert that a nonempty set bounded below has
an infimum, but this is because we do not need to assume this fact as part of
the axiom. Using the Axiom of Completeness, there are several ways to prove
that greatest lower bounds exist for nonempty bounded sets. One such proof is
explored in Exercise 1.3.3.
Exercises
D9D
Exercise 1.3.1. (a) Write a formal definition in the style of Definition 1.3.2
for the infimum or greatest lower bound of a set.
(b) Now, state and prove a version of Lemma 1.3.8 for greatest lower bounds.
Exercise 1.3.2. Give an example of each of the following, or state that the
request is impossible.
(a) A set B with inf B > sup B.
(b) A finite set that contains its infimum but not its supremum.
(c) A bounded subset of Q that contains its supremum but not its infimum.
Exercise 1.3.3. (a) Let A be nonempty and bounded below, and define B =
{bЄR: b is a lower bound for A}. Show that sup B = inf A.
(b) Use (a) to explain why there is no need to assert that greatest lower bounds
exist as part of the Axiom of Completeness.
Exercise 1.3.4. Let A1, A2, A3,... be a collection of nonempty sets, each of
which is bounded above.
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