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$$S=\left\{\frac{n-1}{n+1} | n\geq 1\right\}$$

I can see that $0$ is a lower bound but I want to prove that is an infimum. So I claim that there is $\epsilon>0$ where the $\inf(S)=\epsilon$.

How do I go about it from here?

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$0$ is a lower bound and this value is attained when $n=1$. So $0$ is the infimum and also the minimum of the set.

$1$ is an upper bound and no number $x$ less than $1$ can be an upper bound : $\frac {n-1} {n+1 } >x$ when $n >\frac {1+x} {1-x}$. Hence $1$ is the suprmum of the set but the set does not have a maximum.

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  • $\begingroup$ is there a way to prove that $0$ is the infimum algebraically like what was done with the supremum? or is saying that $0\in S$ when $n=1$ enough? $\endgroup$ – PLC Mar 24 at 9:06
  • $\begingroup$ Saying that $0=\frac {n-1} {n+1} \in S$ when $n=1$ is enough. @PLC $\endgroup$ – Kavi Rama Murthy Mar 24 at 9:09
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Writting $$ \frac{n+1-1-1}{n+1} = \frac{n+1 - 2}{n+1} = 1 - \frac{2}{n+1} $$ You see that the sequence is increasing so the inf is the firt term $0$ and the sup is the last term $1$.

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We have $ 0 \le s$ for all $s \in S$ and $0 \in S.$

Can you proceed ?

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