Let S be a subset of Z. By the axiom of completeness there exists an x in R s.t. x is an upper bound of S.
By the Archimedean principle there exists an n >x. Then n is an upper bound of S.
Do I need to show that S has a Sup in Z to show it is complete? Confused.
By the Archimedean principle there exists an n >x. Then n is an upper bound of S.
Do I need to show that S has a Sup in Z to show it is complete? Confused.
