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

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Description: **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 existen

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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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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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.*

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