2
LECTURE #3: THU, SEP 6, 2007
PRINT DATE: AUGUST 21, 2007
(ANALYSIS3)
every
p
∈
P
. Hence
¯
b
is an upper bound for
P
. Since
b >
¯
b
,
b
cannot be a
least
upper bound for
P
.
s
Notice that if the statement of Theorem A had omitted any mention of the real line, then the
theorem would have been false. Why?
Theorem:
(Theorem B for future ref): Given
S
⊂
R
,
b
∈
R
is the least upper bound (supremum) of
S iF
b
is an upper bound for
S
and if for every
e >
0, there exists
s
∈
S
such that
b

s < e
.
Proof:
: Need to prove this in both directions, i.e., (a) if
b
satis±es the requirements of the theorem
then it is a supremum; (b) if it
doesn’t
satisfy the requirements of the theorem, then it
isn’t
a
supremum.
Proof of (a): Assume that for every
e >
0, there exists
s
∈
S
such that
b

s < e
. We need to show
that
b
≤
b
p
for every upper bound
b
p
for
S
. We’ll do this by showing that if
b
p
< b
then
b
p
cannot be
an upper bound for
S
. Pick
b
p
=
b

e
, for some
e >
0. By assumption,
∃
s
∈
S
s.t.
b

s < e
=
b

b
p
,
i.e.,

s <

b
p
or
s > b
p
proving that
b
p
isn’t an upper bound for
S
.
Proof of (b): Now assume that one of the requirements of the theorem isn’t satis±ed. The ±rst
requirement is that
b
is an upper bound for
S
. If this requirement is violated, then trivally
b
can’t
be a supremum; Now suppose that the
∀
e
condition fails, i.e.,
∃
e >
0 such that
∀
s
∈
S
,
b

s
≥
e
(see ²ig. 1). We’ll show that
b
cannot be a
least
upper bound for
S
, i.e., we’ll show that there
exists
b
p
< b
such that
b
p
is an upper bound for
S
. Pick
b
p
=
b

e
. Since
b

s
≥
e
, for all
s
∈
S
, or,
equivalently,
b

e
≥
s
, for all
s
∈
S
, then
b
p
=
b

e
≥
s
, for all
s
∈
S
. Hence
b
p
is an upper bound
for
S
and is also smaller than
b
. So we’ve proved that
b
isn’t a least upper bound.