In baby Rudin, it is proved that if an ordered set that includes the integers Z has the Least Upper Bound Property, then it is archimedean. That is, if every nonempty set that is bounded above has a least upper bound, then for every x $\in$ R, $\exists$ n $\in$ Z s.t. n > x.

Now, the extended real line, that is R $\cup$ {+$\infty , -\infty$}, clearly is not archimedean, so it should not have the LUB property. However, every set is bounded above in the extended real line, and every set seems to have a least upper bound in the extended reals. For example, if E is a subset of R with no upper bound, then $\infty$ would be the sup of E in the extended reals. Any ideas on how to resolve these apparent contradiction?

  • 1
    $\begingroup$ What is the exact statement of the theorem in Rudin? $\endgroup$ – Brian M. Scott Sep 26 '16 at 19:05
  • $\begingroup$ What exactly does "archimedean" mean in the context of the extended real line? I doubt it is defined. $\endgroup$ – arkeet Sep 26 '16 at 19:07
  • $\begingroup$ If x,y in R, x > 0, then there is a positive integer n s.t. nx > y. This is from page 9 chpt.1 in my version, and I'm pretty sure it implies the statement in my question. $\endgroup$ – David Warren Katz Sep 26 '16 at 19:08
  • 1
    $\begingroup$ @Auburn: But what is the statement of the theorem that you paraphrase in your first sentence. $\endgroup$ – Brian M. Scott Sep 26 '16 at 19:10
  • 1
    $\begingroup$ There's a hidden hypothesis in there: Perhaps the right statement is "If an ordered subset of the reals that includes the integers has the LUB property, then it is archimidean." Or maybe "If an ordered field that includes the integers has the LUB property, then..." Or maybe any of a number of other possibilities. $\endgroup$ – John Hughes Sep 26 '16 at 19:24

Your Answer

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

Browse other questions tagged or ask your own question.