# open and closed intervals and real numbers

by reb659
Tags: intervals, numbers, real
 P: 64 1. The problem statement, all variables and given/known data Show that: Let S be a subset of the real numbers such that S is bounded above and below and if some x and y are in S with x not equal to y, then all numbers between x and y are in S. then there exist unique numbers a and b in R with a
 PF Patron HW Helper Sci Advisor Thanks P: 25,499 hi reb659! i think i'd start by proving that there must be a greatest lower bound and a least upper bound, and then call them a and b, and carry on from there.
 P: 64 Good idea. Isn't it an axiom that if a nonempty subset of R has an upper bound, then it has a least upper bound/sup(S)?
PF Patron
HW Helper
Thanks
P: 25,499

## open and closed intervals and real numbers

I don't think it's an axiom, I think it's the Dedekind-completeness theorem:

A bounded real-valued function has a least upper bound and a greatest lower bound.

(See Rolle's theorem in the PF Library )
 P: 64 So far: Since S is bounded above and below, by Dedekind completeness there exists a supremum of S. Call it b. Again by dedekind completeness we can say there exists an infimum of S. Call it b. By definition of sup and inf, a
 PF Patron HW Helper Sci Advisor Thanks P: 25,499 Yup, that's the proof!
 P: 64 Yay! How exactly does uniqueness follow though? It seems like its trivial to prove.
 PF Patron HW Helper Sci Advisor Thanks P: 25,499 as you said, "by Dedekind completeness there exists a supremum of S" … there can't be two supremums, can there?

 Related Discussions Set Theory, Logic, Probability, Statistics 1 Calculus & Beyond Homework 2 Calculus & Beyond Homework 4 Calculus & Beyond Homework 7 Calculus & Beyond Homework 6