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.
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$\begingroup$ Possible duplicate: math.stackexchange.com/questions/354673/… $\endgroup$– Adrian KeisterAug 7, 2013 at 2:06
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1$\begingroup$ 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. $\endgroup$– Elchanan SolomonAug 7, 2013 at 2:09
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$\begingroup$ 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. $\endgroup$– André NicolasAug 7, 2013 at 2:25
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$\begingroup$ Possible duplicate of math.stackexchange.com/questions/22873/…. $\endgroup$– lhfAug 7, 2013 at 2:42
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2 Answers
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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!
As usual for these questions, my source is James Propp's "Real Analysis in Reverse".
answered Aug 7, 2013 at 2:30
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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.
