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