Equivalence of Completeness Axioms of Real Numbers There are many equivalent versions of completeness in the real number system:
i) LUB/supremum property
ii) Monotone Convergence property
iii) Nested Interval property
iv) Bolzano Weierstrass property
v) Cauchy Criterion property

I've been able to prove: (i)$\implies$(ii)$\implies$(iii)$\implies$(iv)$\implies$(v)
I need help with
a) (v)$\implies$(i)
b) (iii)$\implies$(i)
P.S. In proving (v)$\implies$(i), we use the construction of 2 sequences by using mid-points. I am having problem with showing that they are Cauchy sequences
 A: Given a non-empty set $T\subset \mathbb{R}$ with an upper bound, $y_0$.  Let $x_0$ be any elment of $T$.
Given $x_n$ and $y_n$, define $z_n=\frac{x_n+y_n}{2}$.
If $z_n$ is an upper bound for $T$, then let $x_{n+1}=x_n$ and $y_{n+1}=z_n$.
If $z_n$ is not an upper bound for $T$, let $x_{n+1}>z_n$ be a member of $T$ greater than $z_n$, and let $y_{n+1}=y_n$.
Lemma: For every $n$, $|x_n-y_n|\leq \frac{|x_0-y_0|}{2^n}$
Lemma: If $m>n$, then $|y_m-y_n|\leq \frac{|x_0-y_0|}{2^n}$
So $\{y_n\}$ is Cauchy.  Now you just have to prove the limit of $\{y_n\}$ is the LUB of $T$. (Hint: By the same reasoning, $\{x_n\}$ is Cauchy, and the two limits must be equal.)
A: You don't need to prove that (iii) $\implies$ (i) if you prove that (v) $\implies$ (i). But here is a sketch.
Let $S \subset \mathbb R$ be non-empty and bounded above. Let $a \in S$ and $b$ be an upper bound for $S$. Start with the interval $I_0=[a,b]$. Consider the midpoint $m=(a+b)/2$. If $m \in S$, let $I_1=[m,b]$. If $m$ is an upper bound for $S$, let $I_1=[a,m]$. Otherwise, there is $a'\in S$ such that $m<a'$. Let $I_1=[a',b]$. Repeat. You get a nested sequence of intervals that converge to $\sup S$.
