Suppose A and B are each bounded above. Define the set A+Bas follows A+B = {a +b: a E A, bE B} We claim that sup(A + B) < sup A+ sup B To prove this, we first let m = supA and n = supB. The existence of the supremums is guaranteed by axiom 5.2.4. We need to show that m + nis an upper bound for the set A + B. If this is true, then the least upper bound of A + B (which we know is sup(A+ B)) must be no larger than m + n. Now for any a E A, we know a < m, and for any b e B, we know b < n,hence a + b
Suppose A and B are each bounded above. Define the set A+Bas follows A+B = {a +b: a E A, bE B} We claim that sup(A + B) < sup A+ sup B To prove this, we first let m = supA and n = supB. The existence of the supremums is guaranteed by axiom 5.2.4. We need to show that m + nis an upper bound for the set A + B. If this is true, then the least upper bound of A + B (which we know is sup(A+ B)) must be no larger than m + n. Now for any a E A, we know a < m, and for any b e B, we know b < n,hence a + b
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Suppose A and B are each bounded above. Define the set A + B as follows:**
\[ A + B = \{a + b: a \in A, \, b \in B\} \]
We claim that
\[ \sup(A + B) \leq \sup A + \sup B \]
To prove this, we first let \( m = \sup A \) and \( n = \sup B \). The existence of the supremums is guaranteed by axiom 5.2.4. We need to show that \( m + n \) is an upper bound for the set \( A + B \). If this is true, then the least upper bound of \( A + B \) (which we know is \( \sup(A + B) \)) must be no larger than \( m + n \). Now for any \( a \in A \), we know \( a \leq m \), and for any \( b \in B \), we know \( b \leq n \), hence \( a + b \leq m + n \) for any \( a + b \in A + B \). This shows that \( m + n \) is an upper bound for \( A + B \) and hence \( \sup(A + B) \leq m + n = \sup A + \sup B \).
*Using the notation above, show \(\inf(A + B) \geq \inf A + \inf B\).*
*Provide examples of sets A and B where strict inequality \(\sup(A + B) < \sup A + \sup B\) holds.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45ccc3c2-853e-4ee6-91f9-7e4723419012%2F0bbb98a7-ab9b-4f02-b99a-c30fdee7e8da%2Fyaxotrg_processed.png&w=3840&q=75)
Transcribed Image Text:**Suppose A and B are each bounded above. Define the set A + B as follows:**
\[ A + B = \{a + b: a \in A, \, b \in B\} \]
We claim that
\[ \sup(A + B) \leq \sup A + \sup B \]
To prove this, we first let \( m = \sup A \) and \( n = \sup B \). The existence of the supremums is guaranteed by axiom 5.2.4. We need to show that \( m + n \) is an upper bound for the set \( A + B \). If this is true, then the least upper bound of \( A + B \) (which we know is \( \sup(A + B) \)) must be no larger than \( m + n \). Now for any \( a \in A \), we know \( a \leq m \), and for any \( b \in B \), we know \( b \leq n \), hence \( a + b \leq m + n \) for any \( a + b \in A + B \). This shows that \( m + n \) is an upper bound for \( A + B \) and hence \( \sup(A + B) \leq m + n = \sup A + \sup B \).
*Using the notation above, show \(\inf(A + B) \geq \inf A + \inf B\).*
*Provide examples of sets A and B where strict inequality \(\sup(A + B) < \sup A + \sup B\) holds.*
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