Question

Suppose that is nonempty and bounded above. Then has a supremum.

Note: Show that there is a least element $k_{1}\in\mathbb{Z}$ such that $k_{1}$ is an upper bound for . if $k_{1}$ is not a least upper bound for , show there is at least $k_{2}\in\mathbb{N}$ such that $k_{1} - \frac{1}{2^{k_{2}}}$ is an upper bound for . Proceed in this way to find the supremum.

Answer 1

gsk a least upper TS 'S Here, we ale asked to Prove the completenes Anion, which states that every nen-empty set q A that bounded from above has bound. In the book, Introduction to Analysis 54) Edition by Edward D. Gaughan, it gives a Proof On page 23, It s is a non empty set q real numbers that bounded from below, then had a greatest lower bound. Proof let T=&t: t = -5 for some sesy. It M. is any lower bound for s then Mss for all s ins; hence -sa M for all ses, S0-M bound for T. Similarly, it K is an upper bound for T, then -k is since bounded S bounded from below, Tis from above, and the completeness Aniom, T has a least copper bound, call it B. we claim that for s. the greatest lower bound Certainly -B is a lower bound for so let C be any other cooee bound fors. Then -c is hence bound for T.; В. С. socе в is the But least upper bound for T. Thus -B is the greatest lowes bound an uppee a lower bound for s. is Is an uppes this implies cc - B for si CS Scanned with Scanner