[I\TEXquestion: submit a printout of your answer, typeset in INTEX] Let A = {|ne N, m e N, m > n}. (a) Prove that inf A = 0. (b) Prove that sup A = 1. Solution: 0 is a lower bound of A since > 0. We need to show that there is no larger lower bound. So for any e> 0 there exist n, m eN with m >n and 2 with m >e. Such an m exists becatuse of the Archimedean property of R. Since in this case 0 is not a lower bound. Hence, inf A = 0, as 0 is the greatest lower bound. Now since n0 the number 1-e is not an upper bound. We do this by finding n, m e N with n < m and >1-e. One chooses (after trying a bit) an m e N with m>e and m > 1. Then choose n= m - 1. In this case one obtains m-1 1 >1-e, which is what we had to show. We have used here that
[I\TEXquestion: submit a printout of your answer, typeset in INTEX] Let A = {|ne N, m e N, m > n}. (a) Prove that inf A = 0. (b) Prove that sup A = 1. Solution: 0 is a lower bound of A since > 0. We need to show that there is no larger lower bound. So for any e> 0 there exist n, m eN with m >n and 2 with m >e. Such an m exists becatuse of the Archimedean property of R. Since in this case 0 is not a lower bound. Hence, inf A = 0, as 0 is the greatest lower bound. Now since n0 the number 1-e is not an upper bound. We do this by finding n, m e N with n < m and >1-e. One chooses (after trying a bit) an m e N with m>e and m > 1. Then choose n= m - 1. In this case one obtains m-1 1 >1-e, which is what we had to show. We have used here that
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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![[LATEXquestion: submit a printout of your answer, typeset in LTEX]
Let A = {" |n € N, m e N, m > n}.
m
(a) Prove that inf A = 0.
(b) Prove that sup A = 1.
Solution: 0 is a lower bound of A since > 0. We need to show that there is no
larger lower bound. So for anye>0 there exist n, m eN with m >n and <e.
This is not hard to do. Simply choose n = 1 and m > 2 with m > el. Such an m
exists because of the Archimedean property of R. Since in this case <e, any e>0
is not a lower bound. Hence, inf A = 0, as 0 is the greatest lower bound. Now since
n<m we get < 1, so 1 is an upper bound. To show that there is no smaller upper
bound we proceed in exactly the same way as before. We show that for any e> 0 the
number 1 - e is not an upper bound. We do this by finding n, m eN with n < m
and >1-e. One chooses (after trying a bit) an m eN with m > e and m > 1.
Then choose n = m - 1. In this case one obtains
1
>1-e,
m-1
which is what we had to show. We have used here that < e and therefore ->](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F986f6c25-9d5d-4e3f-93bc-35b073c4428b%2F4f5f9ff6-d212-4ea0-8959-a79d00adc3e4%2Fwdvezj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[LATEXquestion: submit a printout of your answer, typeset in LTEX]
Let A = {" |n € N, m e N, m > n}.
m
(a) Prove that inf A = 0.
(b) Prove that sup A = 1.
Solution: 0 is a lower bound of A since > 0. We need to show that there is no
larger lower bound. So for anye>0 there exist n, m eN with m >n and <e.
This is not hard to do. Simply choose n = 1 and m > 2 with m > el. Such an m
exists because of the Archimedean property of R. Since in this case <e, any e>0
is not a lower bound. Hence, inf A = 0, as 0 is the greatest lower bound. Now since
n<m we get < 1, so 1 is an upper bound. To show that there is no smaller upper
bound we proceed in exactly the same way as before. We show that for any e> 0 the
number 1 - e is not an upper bound. We do this by finding n, m eN with n < m
and >1-e. One chooses (after trying a bit) an m eN with m > e and m > 1.
Then choose n = m - 1. In this case one obtains
1
>1-e,
m-1
which is what we had to show. We have used here that < e and therefore ->
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