Archimedian Property and Axiom of completeness

Yesterday we discussed Archimedian Property. It says that Natural numbers are not bounded above i.e there is no maximum value of natural number. On discussion with Andy he gave me a very good way to visualize this property. Think of a Real number line starting from zero to infinity. Now on this number line you choose any number, how so big, I can find a Natural number bigger than that. Similarly choose a number on this Real line, how so small, I can find a Natural number whose reciprocal will be smaller than this number.

Formally Archimedian property is stated as
(i) Given any number x belongs to R, there exists an n belonging to N satisfying n > x
(ii) Given any number y > 0, there exists an n belonging to N satisfying 1/n < y.

Axiom of completness says that any bounded above set of real number has a sup, i.e if we have a set of real numbers which is bounded above we can always assume that it has a sup. This axiom has a wide implications and one of the important application is in proving "Nested Interval Property".

The nested interval property says that the intersection of nested closed intervals is NON EMPTY. How this comes ? Well its easy to see that Axiom of completeness says that whenever we have a bounded above set we will have a sup. In case of nested interval we have upper bounds, which in turn implies that we have a sup (recall that all bn's are upper bound