Why are the least upper bound principle and the fundamental axiom of analysis equiva

The least upper bound principle states that every non-empty subset of R that is bounded above has a least upper bound. The fundamental axiom of analysis (this has several names) states that every increasing sequence bounded above converges. They are equivalent.

Assuming the least upper bound principle, I can derive the fundamental axiom--by arguing, if an is an increasing sequence bounded above, then {an} is a set of real numbers that has a least upper bound. Therefore, the increasing sequence "gets arbitrarily close to the least upper bound and does not get further away afterwards" and so converges.

How to prove the fundamental axiom implies the least upper bound principle?

Reply 1

simba_

Show that An+1 > An as its an increasing sequence

This will mean A1 will equal the least upper bound.

Prove by contradiction... assume that A1 is not the greatest lower bound

This gives two cases that, the greatest lower bound is greater than A1 or that it is less than A1

if a number is greater than A1 e.g A1 + k then it is not a lower bound for An, show why!!

If a number is less than A1 e.g. A1 -k then there is also a number greater than A1-k that is a lower bound so A1-k is not a least lower bound, show why!!

Reply 2

boromir9111

17

Original post by simba_

Show that An+1 > An as its an increasing sequence

This will mean A1 will equal the least upper bound.

Prove by contradiction... assume that A1 is not the greatest lower bound

This gives two cases that, the greatest lower bound is greater than A1 or that it is less than A1

if a number is greater than A1 e.g A1 + k then it is not a lower bound for An, show why!!

If a number is less than A1 e.g. A1 -k then there is also a number greater than A1-k that is a lower bound so A1-k is not a least lower bound, show why!!