Real numbers via equivalence classes of Cauchy sequences and the Completeness Axiom Hi so there's a question in my elementary real analysis course I'm a little bugged about. The question goes: Use the definition of the real numbers via equivalence classes of Cauchy sequences to prove the Completeness Axiom. I tried using proof by contradiction by first stating Completeness is false. Then every nonempty subset S of R that is bounded above has no least upper bound. I wanted to construct a Cauchy sequence of descending upperbounds and show that these upperbounds converge to real number which would be the least upperbound, resulting in a contradiction. Is this a valid method? If so, I'm a little stuck on the details of such a Cauchy sequence. Any help is much appreciated.
 A: A similar argument to this that follows is used in my favorite book of Analysis. I think could help you.
Suppose for the sake of contradiction that exists a non-empty set of the real numbers which is bounded but not have a least upper bound. 
Let $E$ be such a set, $n$ be a positive integer, $M$ an upper bound of $E$ and $x_0$ an element of $E$, using the Archimedean property we know that there is a natural number $K$ such that $x_0+K/n >M$. So $x_0+K/n$ is an upper bound of $E$ and clearly $x_0-1/n$ is not an upper bound. 
Then there exists a unique natural number $0\le i \le K$ such that $x_0+i/n$ is an upper bound, but $x_0+(i-1)/n$ is not (why? argue by contradiction). So  $x_0+(i-1)/n <x_0+i/n$. By the denseness of $\mathbb{Q}$ there is a $q_n$ such that $x_0+(i-1)/n < q_n <x_0+i/n$. Then $q_n-1/n$ is not an upper bound and $q_n+1/n$ is an upper bound (why?). We define the sequence $(q_n)_{n=1}^\infty$.  
Let $M$ be a positive integer and $n,m\ge M$ then $|q_n-q_m|<1/n+1/m\le 2/M$. Thus is a cauchy sequence and so, it must be the formal limit of a real number. Let $S$ be such a real numbers.
(1) We shall show that $S$ is an upper bound. Let $x\in E$. Then $x\le q_n + 1/n$ and so $x\le S$, which shows that $S$ is an upper bound.
(2) Let $T$ be an upper bound. Then $q_n-1/n<T$ and so $S\le T$. In other words $S$ is at most $T$. 
Then by (1) and (2) we can conclude that $S$ is the least upper bound, a contradiction. 
