2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Possible duplicate: math.stackexchange.com/questions/354673/… $\endgroup$ – Adrian Keister Aug 7 '13 at 2:06
  • 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 Solomon Aug 7 '13 at 2:09
  • $\begingroup$ 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 Nicolas Aug 7 '13 at 2:25
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/22873/…. $\endgroup$ – lhf Aug 7 '13 at 2:42
5
$\begingroup$

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".

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.