1. Show that sup {1−1n:n∈N}=1{1−1n:n∈N}=12. If S={1n−1m:n,m∈N}S={1n−1m:n,m∈N},f∈d∈fSand⊃S.end{taba―r}

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Aniqa O&#039;Neill · Accepted Answer

Let S=1−1n:n∈NSince1−1n&amp;lt;1, therefore 1 is an upper bound of S. We have to prove that 1 is the least upper bound. Let aa be another upper bound of SS such that a&amp;lt;1. Since 1−a&amp;gt;01−a&amp;gt;0 by Archimedean Property there exist a natural number N0 such that N0 (1−a)&amp;gt;1⇒1−a&amp;gt;1/N0⇒1−1/N0&amp;gt;a. This contradicts that aa is not an uper bound of SS. Hence 11 is the least upper bound of SS consequently sup1−1/n:n∈N=1. (2) Let S=1n−1/m:m,n∈N. Since 1n−1m&amp;gt;1n−1&amp;gt;1−1, therefore −1 is a lower bound of S. We have to prove that 1 is the gratest upper bound. Let ϵ&amp;gt;0, by Archimedean Property there exist a natural number N0 such that ​ N0ϵ&amp;gt;1⇒1N0&amp;lt;ϵ⇒1N0−1&amp;lt;ϵ−1=−1+ϵ Now since 1N0−1&amp;lt;−1+ϵ for every ϵ&amp;gt;0 and 1N0−1∈S, thus inf1n−1/m:m,n∈N=−1 Similarly we can prove that sup1n−1m:m,n∈N=1

1. Show that sup {1−1/n:nin N}=1{1−1/n:nin N}=frac{1}{2}. If S:={frac{1}{n}−frac{1}{m}:n,min N}S:={frac{1}{n}−frac{1}{m}:n,min N}, find inf S and sup S. end{tabular}

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