Math 4200
Suppose every non empty subset S of R that is bounded above has a least upper bound.
Prove that every non empty subset T of R that is bounded below has a greatest lower
bound.
Proof:
Suppose T is non empty and is bounded below by the number w. Then, x  w x  T .
Let S  { x : x  T } . Since x  w x  T ,  x   w  x  T . Hence S is bounded above
by w . By our initial assumption, S has a least upper bound, call it sup S . Let
m   sup S . We will show that m is the greatest lower bound of S.
Since sup S is an upper bound of S,  x  sup S ,  x  T . So, x   sup S ,  x  T ,
and x  m ,  x  T . We have shown that m is a lower bound for T. We must now
show that m is the greatest lower bound of T. Suppose that u is a lower bound of T and
u  m . Then,  sup S  u  x , x  T . It now follows that sup S  u   x , x  T
Then, u is an upper bound for S and u is less that the least upper bound of S, a logical
contradiction. Hence, m is the greatest lower bound of S.