# Section 2.8: Problem 7 Solution

Working problems is a crucial part of learning mathematics. No one can learn... 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 . Munkres

Let
be a subset of
having no greatest member. Then as a subset of
,
will have upper bounds (with respect to the ordering
) in
. But show that
does not have a least such bound.

We, first, argue that
is an upper bound of
in
iff
is positive and infinite. Indeed, if
is an upper bound of
in
, then, since there is
such that
,
. Further, suppose that
is finite. Then
and there is
such that
,
, implying (by the definition of infinitesimal numbers)
, and
. Therefore,
must be infinite. Now, if
is positive and infinite, then, by definition,
for every
, in particular,
.

Next, we show that if
is an upper bound of
, then there is a strictly lower upper bound of
. This follows from the fact that if
is positive and infinite, then
is positive and infinite. Indeed, for every
,
,
, so that
.

We conclude that there is no least upper bound of
.