I need to prove Completeness property and Cauchy criterion in Real numbers are equal, and I think its true. to prove this equality I think should use Bolzano-Weierstrass Theorem. I be so thankful if anyone could help me.
Equality of Completeness property (Supremum and Infimum property) and Cauchy criterion in Real numbers
analysisreal-analysis
Best Answer
supremum property $\Rightarrow$ Cauchy criterion
The supremum property (which is what you call completeness) also implies the infimum property: every subset $S\subset\mathbb R$ which is bounded below has an infimum. For proof, just apply the supremum property to $-S:=\{-s~\vert~s\in S\}$. These two imply the existence of limit superior and limit inferior for all bounded sequences, since they are defined as
$$\limsup a_n:=\inf_{k\in\mathbb N}\sup_{n\geq k} a_n,\\ \liminf a_n:=\sup_{k\in\mathbb N}\inf_{n\geq k} a_n,$$
both of which exist for bounded $a_n$ due to the supremum and infimum properties. If limit inferior and superior of a sequence are equal, then the sequence converges. Now the limits superior and inferior of a Cauchy sequence can't be different if they exist (I leave the proof to you). And since Cauchy sequences are bounded (again, I leave the proof to you), the limits superior and inferior do in fact exist. They are then equal, and thus the Cauchy sequence has a limit.
Cauchy criterion $\Rightarrow$ supremum property
Let $\emptyset\neq S\subset\mathbb R$ be a set which is bounded above. Let $b_0$ be an upper bound of $S$ and let $a_0\in S$. Now define recursively that if $\frac{a_n+b_n}{2}$ is an upper bound of $S$, then $a_{n+1}:=a_n$ and $b_{n+1}:=\frac{a_n+b_n}{2}$, and otherwise $a_{n+1}:=\frac{a_n+b_n}{2},~~b_{n+1}:=b_n$.
With these definitions, both $a_n$ and $b_n$ are Cauchy sequences since $\vert a_m-a_n\vert\leq\frac{1}{2^{\vert m-n\vert}}\vert b_0-a_0\vert$, same for $\vert b_m-b_n\vert$. Because of the Cauchy criterion, both sequences converge. Even more, both converge to the same number, since the sequence $b_n-a_n=\frac{1}{2^n}(b_n-a_n)$ converges to $0$. The limit of these sequences must be an upper bound of $S$ because all the $b_n$ are upper bounds. There also can't be a lower upper bound because none of the $a_n$ are upper bounds (if $b$ is an upper bound, then all $c>b$ are upper bounds, too). So this limit is a least upper bound, in other words, a supremum.