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I am doing the exercise of Bartle and Sherbert. The question and my answer is as follow. Is my proof valid?

Let $S_3$ = { ${\frac{1}{n}}:n\in$ N}. Show that sup $S_3$ = 1 and inf $S_3$ $\geq 0 $

Here is my proof:

Part I

$\because$ $n\in N$ and $N:=$ {1, 2, 3, ... }

when n = 1, $\frac{1}{n} = \frac{1}{1} = 1$ ;

when n = 2 , $\frac{1}{n} = \frac{1}{2}$ ;

$\frac{1}{n} < 1$, $\forall n > 1 $

$\therefore$ u:= upper bound of $S_3$ = 1

Let $\epsilon = \frac{3}{4}$ , $u-\epsilon = \frac{1}{4} = \frac{1}{n}$ when n = 4

Thus, $(u-\epsilon) < \frac{1}{2} < 1 $ and $ \frac{1}{2} \in S_3 $

i.e. u = $sup S_3 $

Part II

$S_3 \ne \emptyset$ and $\frac{1}{n} \in R$ $\implies \exists n \geq \frac{1}{n} $

$\because \frac{1}{n} > 0 \implies 0 $ is one of the lower bound of $S_3$

$\therefore S_3$ is bounded below by $ 0 $

$\implies \exists inf S_3:= w$ and $ w \geq 0 $

Let $\epsilon >0 $,

By Archimedean Property, $\exists n> \frac{1}{\epsilon}$ where $n \in N$

$\therefore 0 \leq w \leq \frac{1}{n} < \epsilon$

$\because \epsilon >0 $ is arbitrary and $\epsilon \in R$ ,

$0 \leq w \leq 1/n < \epsilon $ $\forall \epsilon>0 $

Thus, $ w := inf S_3 = 0$

Problem Then how can we prove $ inf S >0$ so that $ inf S_3 \geq 0 $ ?

THANKS!

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That's not the way one would expect you to write a formal proof. To show that $\sup S_3 = 1$ you can observe that $n \ge 1$ implies $1/n \le 1$ and $\sup S_3 \le 1$, combined with the fact that $1 \in S_3$, the equality is obvious.

The part II is very similar.

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