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}

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}

Question
Alternate coordinate systems
asked 2021-02-13
1. Show that sup \(\displaystyle{\left\lbrace{1}−\frac{{1}}{{n}}:{n}\in{N}\right\rbrace}={1}{\left\lbrace{1}−\frac{{1}}{{n}}:{n}\in{N}\right\rbrace}={\frac{{{1}}}{{{2}}}}\). If \(\displaystyle{S}\:={\left\lbrace{\frac{{{1}}}{{{n}}}}−{\frac{{{1}}}{{{m}}}}:{n},{m}\in{N}\right\rbrace}{S}\:={\left\lbrace{\frac{{{1}}}{{{n}}}}−{\frac{{{1}}}{{{m}}}}:{n},{m}\in{N}\right\rbrace},{f}\in{d}\in{f}{S}{\quad\text{and}\quad}\supset{S}.{e}{n}{d}{\left\lbrace{t}{a}{b}\underline{{a}}{r}\right\rbrace}\)

Answers (1)

2021-02-14

Let \(S={1−\frac{1}{n}:n\in N} Since 1−\frac{1}{n}<1\), 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<1. Since 1−a>01−a>0 by Archimedean Property there exist a natural number N0 such that
N0 \((1−a)>1 \Rightarrow 1-a>1/N0 \Rightarrow 1-1/N0>a\).
This contradicts that aa is not an uper bound of SS. Hence 11 is the least upper bound of SS consequently
\(\sup{1-1/n:n\in N}=1.\) (2) Let \(S={\frac{1}{n} −1/m:m,n\in N}\). Since \(\frac{1}{n}-\frac{1}{m}>\frac{1}{n}-1>1-1\), therefore −1 is a lower bound of S. We have to prove that 1 is the gratest upper bound. Let \(\epsilon >0\), by Archimedean Property there exist a natural number N0 such that ​
\(N0\epsilon >1 \Rightarrow \frac{1}{N0}<\epsilon \Rightarrow \frac{1}{N0}-1<\epsilon -1=-1+\epsilon\)
Now since \(\frac{1}{N0}−1<−1+\epsilon\) for every \(\epsilon >0 \ and \ \frac{1}{N0}-1\in S\), thus
\(inf{\frac{1}{n}-1/m:m,n\in N}=-1\)
Similarly we can prove that
\(\sup{\frac{1}{n}-\frac{1}{m}:m,n\in N}=1\)

0

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