Proving the last part of Nested interval property implying Axiom of completeness I took a non-empty set A that is bounded above. And I went on with the regular algorithm, which either gave us a LUB or gave us an infinite chain of nested intervals
$I_1$ $\supseteq$ $I_2$ $\supseteq$ $I_3$ $\dots$
Now I used the nested interval property to say that there is at least a member b $\in$ $\bigcap_n$ $I_n$. Now I am having problem showing that this element b is the upper bound of my original set A. I am trying proof by contradiction:
If b wasn't upper bound for A then, $\exists$a$\in$A such that a>b. How does this give me a contradiction? If this method, doesn't work, which way should I go?
 A: Let's apply the bisection method.
Since $A$ is bounded above, there exists an upper bound of $A$, say $b_1$; also, since $A$ is non void, choose $a_1=a-1$ where $a \in A$. So $a_1$ is not an upper bound of $A$.
If $m$ is the midpoint of $[a_1,b_1]$, let $a_2=a_1$ and $b_2=m$ if $m$ is an upper bound of $A$, let $a_2=m$ and $b_2=b_1$ otherwise.
Continuing in the same way, you obtain that every right end and none left end of the successive intervals is an upper bound of $A$.
Also the nested intervals property assures the existence of a number $b$ contained in every interval:
$b$ is unique because the length of the intervals converges to zero.
You can prove by contradiction that


*

*$b$ is an upper bound of $A$

*no number less tha $b$ is an upper bound of $A$


finding every time a number different from $b$ which is contained in every interval.
In fact


*

*if $c \in A$ with $c>b$, then $c$ is contained in every interval

*if $d\,$ is an upper bound of $A$ with $d<b$, then $d$ is contained in every interval.


Done !
A: Presumably $I_1$ is an interval such that the lower limit is less than some element of $A$ and the upper limit is greater than every element of $A$.  Now you bisect $I_1$ into two intervals with the midpoint $c$.  You ask if $c$ is an upper bound for $A$.  If so, $I_2$ is the lower subinterval and $b$ will be in it.  If not, $I_2$ is the upper subinterval and $b$ will be in it.  Now you have two things to prove.  First, you have to argue that $b$ is an upper bound for $A$ because the upper end of each interval is an upper bound for $A$ and the limit of the upper bounds converges to $b$ because the length goes to zero.  The second is that there is no lower number than $b$ that is an upper bound for $A$.  For this, argue that any lower number that is an upper bound for $A$ would force a different selection of intervals-the lower limit of each interval is less than some member of $A$.
