Necessity of LUB property to prove the nested interval theorem

Is least upper bound property of the set of real numbers necessary to prove the nested interval theorem ? I know that it is sufficient, but I doubt whether it is necessary.

• Possible duplicate: math.stackexchange.com/questions/354673/… – Adrian Keister Aug 7 '13 at 2:06
• What do you mean by "is it necessary"? The rational numbers $\mathbb{Q}$ are similar to the reals in certain ways, but lack the LUB property and also the nested intervals theorem. – Elchanan Solomon Aug 7 '13 at 2:09
• What other properties does one want to keep? If we want simply an ordered set in which the intersection of a nested sequence of closed intervals is non-empty, there are many examples that do not have the least upper bound property, Take for example the reals, with a copy of the non-negative reals appended on the right. – André Nicolas Aug 7 '13 at 2:25
• Possible duplicate of math.stackexchange.com/questions/22873/…. – lhf Aug 7 '13 at 2:42

It is not necessary, in the following sense. There exists an ordered field $S$ where the nested interval theorem holds, but the least upper bound property does not. To prove the nested interval theorem for $S$, one need not assume LUB; indeed one cannot assume LUB, since it isn't true!
It is necessary because the nested interval theorem is not true in $\mathbb Q$. Consider for instance the bisection method for solving $x^2=2$ starting from the interval $[0,2]$. You get a sequence of nested closed intervals with rational endpoints that has empty intersection.