Preliminary draft only: please check for final version
ARE211, Fall 2007
LECTURE #3: THU, SEP 6, 2007
PRINT DATE: AUGUST 21, 2007
(ANALYSIS3)
Contents
1.
Analysis (cont)
1
1.6.
Two Preliminary Results
1
1.7.
Cauchy Sequences
2
1.
Analysis (cont)
1.6.
Two Preliminary Results
Theorem:
(Theorem A for future ref): Given two subsets,
P
and
Q
, of the real line, if
P
⊂
Q
then
sup(
P
)
≤
sup(
Q
) and inf(
P
)
≥
inf(
Q
).
Obvious just by looking at a picture: if
P
is smaller than
Q
then then every upper bound to
Q
must also be an upper bound to
P
and there may be upper bounds to
P
that aren’t upper bounds
to
Q
. Doing the proof is just a matter of saying the obvious in symbols.
Proof:
We’ll just prove the first half about sups. Pick
P, Q
⊂
R
such that
P
⊂
Q
. There are two
cases to consider:
Case A
: sup(
Q
) =
∞
; in this case, there’s nothing to prove because
any
possible sup for
P
has to
be less than or equal to infinity.
Case B
: sup(
Q
) =
¯
b
. We’ll consider
b >
¯
b
and prove that
b
cannot be a least upper bound for
P
.
This will establish that sup(
P
)
≤
sup(
Q
). Well,
¯
b
≥
q
, for every
q
∈
Q
. Since
P
⊂
Q
,
¯
b
≥
p
, for
1
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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
.
square
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 iff
b
is an upper bound for
S
and if for every
epsilon1 >
0, there exists
s
∈
S
such that
b

s < epsilon1
.
Proof:
: Need to prove this in both directions, i.e., (a) if
b
satisfies 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
epsilon1 >
0, there exists
s
∈
S
such that
b

s < epsilon1
. We need to show
that
b
≤
b
prime
for every upper bound
b
prime
for
S
. We’ll do this by showing that if
b
prime
< b
then
b
prime
cannot be
an upper bound for
S
. Pick
b
prime
=
b

epsilon1
, for some
epsilon1 >
0. By assumption,
∃
s
∈
S
s.t.
b

s < epsilon1
=
b

b
prime
,
i.e.,

s <

b
prime
or
s > b
prime
proving that
b
prime
isn’t an upper bound for
S
.
Proof of (b): Now assume that one of the requirements of the theorem isn’t satisfied.
The first
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
∀
epsilon1
condition fails, i.e.,
∃
epsilon1 >
0 such that
∀
s
∈
S
,
b

s
≥
epsilon1
(see Fig. 1).
We’ll show that
b
cannot be a
least
upper bound for
S
, i.e., we’ll show that there
exists
b
prime
< b
such that
b
prime
is an upper bound for
S
. Pick
b
prime
=
b

epsilon1
. Since
b

s
≥
epsilon1
, for all
s
∈
S
, or,
equivalently,
b

epsilon1
≥
s
, for all
s
∈
S
, then
b
prime
=
b

epsilon1
≥
s
, for all
s
∈
S
. Hence
b
prime
is an upper bound
for
S
and is also smaller than
b
. So we’ve proved that
b
isn’t a least upper bound.
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 Fall '07
 Simon
 Metric space, Limit of a sequence, Cauchy sequence, Cauchy

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