Section 3: Problem 13 Solution
Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Prove the following:
Theorem. If an ordered set
has the least upper bound property, then it has the greatest lower bound property.
Let
be a non-empty subset bounded from below. Let
be the set of all lower bounds of
.
is non-empty, as
is bounded from below and there is at least one lower bound, and bounded from above by any element in
. Therefore,
has the lowest upper bound
such that for any upper bound
of
:
. We show that
is the greatest lower bound of
. First, it is a lower bound of
, as any element
in
is an upper bound of
, implying
for all
in
. Besides, for any other lower bound
of
,
, therefore,
.