# A question about Archimedean property

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!

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.