Preliminary draft only: please check for final version
ARE211, Fall 2007
LECTURE #2: THU, AUG 30, 2007
PRINT DATE: AUGUST 21, 2007
(ANALYSIS2)
Contents
1.
Analysis (cont)
1
1.4.
Convergence of Sequences
1
1.4.1.
Boundedness of Sequences
2
1.4.2.
Boundedness of Sequences (cont)
3
1.5.
Least Upper bounds and Greatest Lower Bounds
4
1.
Analysis (cont)
1.4.
Convergence of Sequences
Definition:
We say that a sequence
{
x
n
}
in
X
converges
to an element
x
∈
X
in the metric
d
if
∀
epsilon1 >
0
∃
N
∈
N
such that for all
n > N
,
d
(
x
n
, x
)
< epsilon1
.
A sequence that converges to a point is called a convergent sequence. The point that it converges
to is called a limit of the sequence. (It’s also
the
limit of the sequence, i.e., there’s only one, but
that’s a result not a definition.)
Example:
{
x
n
}
defined by
x
n
= 1
/n
converges to zero in the Pythagorian metric, but not in the
discrete metric.
1
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LECTURE #2: THU, AUG 30, 2007
PRINT DATE: AUGUST 21, 2007
(ANALYSIS2)
Proof:
We need to show that for all
epsilon1
, there exists
N
such that for all
n > N
,
d
(
x
n
,
0)
< epsilon1
. Now
in the Pythagorian metric,
d
(
x
n
,
0) =
radicalbig
(
x
n

0)
2
=

x
n

, i.e., we just need to show that for
n
sufficiently large,

x
n

< epsilon1
. Pick
N >
1
/epsilon1
. Then 1
/N < epsilon1
and for
n > N
, 1
/n <
1
/N < epsilon1
. Done
On the other hand, consider the discrete metric. To show that
x
n
doesn’t
converge to zero, we need
to show that there exists
epsilon1 >
0 such that for all
N
, there exists
n > N
such that
d
(
x
n
,
0)
> epsilon1
. Pick
epsilon1
= 1
/
2. Let
N
be arbitrary and observe that
d
(
x
N
+1
,
0) = 1
> epsilon1
. Done
Note that what we did here was an example of a general proof technique: to prove that an assertion
involving
∀
’s and
∃
’s is false, go through and change the
∀
’s to
∃
and the
∃
’s to
∀
.
Theorem:
A sequence
{
x
n
}
converges to
α
∈
R
in the discrete metric if and only if there exists
N
such that for all
n > N
,
x
n
=
α
.
1.4.1.
Boundedness of Sequences.
Definition:
A sequence is
bounded above
by
b
∈
R
if
x
n
≤
b
, for all
n
∈
N
. A sequence is
bounded
above
if it is bounded above by
some
b
∈
R
. Defn of bounded below is parallel.
Example:
The sequence
{
x
n
}
defined by
x
n
= (

1)
n
is bounded above by 1 and below by 1. (It’s
bounded above by lots of other things too, e.g, 2,
π
, etc, and similarly bounded below by lots of
other things. The point is that in order to be bounded above, a sequence just has to be bounded
above by
something
).
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 Fall '07
 Simon
 Supremum, Order theory, lower bound

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