1. The problem statement, all variables and given/known data Least Upper Bound (LUB) Principle: every nonempty subset S of R that is bounded above has a least upper bound. Completeness Theorem: every Cauchy sequence of real numbers converges. So R is complete. To prove that Completness Theorem implies the least upper bound property (i.e. take the copmleteness theorem as an axiom, and use it to prove the LUB principle), my textbook suggests the following: "Begin with any a1 in S and let b1 be an upper bound of S. Let I1 = [a1,b1] . Let c be the midpoint of I1 . If c is an upper bound of S, let I2 be the first half of I1, otherwise let I2 be the second half. Repeat the process to get a nested sequence In = [an,bn] , the interested reader can verify that: (i) (an) and (bn) are Cauchy sequences with a common limit L (ii) and that L = sup S. " 2. Relevant equations N/A 3. The attempt at a solution I'm struggling to prove (i). 1) By definition, an is Cauchy iff for all ε>0, there exists N s.t. if n,m≥N, then |an-am|<ε. But I have no idea how to prove that (an) and (bn) in our problem are Cauchy. Can someone help me, please? 2) [solved] Now if we have proved that both an and bn converge, why does it follow that they must converge to a COMMON limit L? I can see that the legnth of the interval = bn-an ->0 as n->∞, so it's intuitvely believable that they both converge to a common limit, but how can we actually PROVE rigorously? Thanks for any help!