prove of supremum and infimum principle using Cauchy convergence criterium

prove: suppose S is a non-empty set having upperbound number set
by Archimedes property, for any positive number $\alpha$,exists a
integer,it makes $\lambda_{\alpha}\alpha$ a S's upperbound, while
$\lambda$ - a ,being not a upperbound to S.
so,
$\lambda_{\alpha}=k_{\alpha}\alpha$ , a upperbound to S
$\lambda_{\alpha}-\alpha=(k_{\alpha}-1)\alpha$ , not a upperbound to S
set $\alpha=\frac{1}{n}$,n=1,2,3....
so, for each positive n ,exists a corresponding $\lambda_{n}$,
it makes a upperbound to S, but not $\lambda_{n}-\frac{1}{n}$
they are as belows:
$\lambda_{\alpha}=k_{\alpha}\alpha$
$\lambda_{\alpha}-\alpha=(k_{\alpha}-1)\alpha$
so, there must be a $\alpha'\in S$,with:
$\alpha'>\lambda_{n}-\frac{1}{n}$
so,get p$\in {1,2,3,....}$
$\lambda_{p}=k_{p}\frac{1}{p}$
there must be a $\alpha_{p}$
$\alpha_{p}>\lambda_{p}-\frac{1}{p}$
choose q $\in${1,2,3,...},we have:
$\lambda_{q}=k_{q}\frac{1}{q}$ is an upperbound .
so $\lambda_{q}>\alpha_{p}>\lambda_{p}-\frac{1}{p}$
so $\lambda_{p}-\lambda_{q}<\frac{1}{p}$
similarly,we get $\lambda_{q}-\lambda_{p}<\frac{1}{q}$

so,we get |$\lambda_{q}-\lambda_{p}|<MAX\{\frac{1}{p},\frac{1}{q}\}$
$\forall \epsilon>0$,we choose N>$\frac{1}{\epsilon}$
then we get $\frac{1}{N}<\epsilon$
so when p,q >N,then
$\frac{1}{p},\frac{1}{q}<\frac{1}{N}<\epsilon$
so |$\lambda_{p}-\lambda_{q}|<MAX\{\frac{1}{p},\frac{1}{q}\}<\epsilon$
so , by Cauchy convergence criterium,$\{\lambda_{n}\}$ converges
set $lim_{n\to\infty}\lambda_{n}=\lambda$
$\forall \alpha \in S,and \forall \lambda_{n},\alpha \leqslant \lambda_{n}
so ,by 数列极限的保不等式性， $\forall \alpha \in S, \alpha <lim_{n\to \infty}\lambda_{n}=\lambda$
so, $\lambda$ is a upperbound to S.
we need to prove :$\forall \epsilon>0$,there exists a $\alpha_{n}$,with $\alpha_{n}>\lambda-\epsilon$
we know $\forall \frac{1}{n},$there is a $\alpha_{n}>\lambda_{n}-\frac{1}{n}$
if $\lambda_{n}-\frac{1}{n}>\lambda-\epsilon$
ie, $\lambda-\lambda_{n}+\frac{1}{n}<\epsilon$
we need to have $\frac{1}{n}<\frac{\epsilon}{2}$,and $\lambda-\lambda_{n}<\frac{\epsilon}{2}$
ie: $n>\frac{2}{\epsilon}$,$\lambda-\lambda_{n}<\frac{\epsilon}{2}$
cause $lim_{n\to\infty}\lambda_{n}=\lambda$,
so, $\exists$N_{0}$\in N^+$,when n>$N_{0}$,$|\lambda-\lambda_{n}|<\frac{\epsilon}{2}$
ie,$-\frac{\epsilon}{2}<\lambda-\lambda_{n}<\frac{\epsilon}{2}$
so, if n>MAX{$N_{0},\frac{2}{\epsilon}\}$
the theorem is proven
likely to prove the infimum