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?