Recall the
Axiom of Completeness
, which states that every nonempty subset of
R
that is bounded above
has a least upper bound in
R.
In this problem, we consider some of the consequences of the
completeness of
R
.
Theorem (Archimedean Property):
For every real number
x
, there exists a positive integer
n
such that .
Theorem (Greatest Lower Bound Property):
Every nonempty subset of
R
that is bounded
below has a greatest lower bound.
Exercise:
Define what it means for
to be the least upper bound of a set
S
. Prove that if
is the least upper bound
of
S
, then for every , there is some
x
in
S
such that .
Proof (Archimedean Property)
Suppose not for contradiction. Then there is some
x
in
R
such that
for all
positive integers
n
. It follows that
N
is bounded above. Put , which we know exists by the axiom of
completeness. Then
is not an upper bound of
N
.
Exercise:
Use this and the exercise above to find a contradiction. Conclude that
N
is not bounded above.
This completes the proof of the first theorem.
Proof (Greatest Lower Bound Property):
Let
S
be a nonempty subset of real numbers that is bounded above,
and suppose . Define the set
T
by
Exercise:
Show that
is a lower bound for the set
T
.
Exercise:
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 Spring '08
 PLOTKIN
 Continuous function, Metric space, greatest lower bound

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