Econ 4111
Professor: John Nachbar
9/25/08
Compactness and Completeness in
R
N
.
1
R
is complete.
Theorem 6, the HeineBorel theorem, states that a set in
R
N
is compact iff it is
closed and bounded. Theorem 6 is immediate if I can show (a) that
R
N
is complete
and (b) that bounded sets in
R
N
are totally bounded. I demonstrate the latter in
Theorem 5. As for completeness, I start by showing here that
R
is complete, then
show that this implies that
R
N
is complete.
Theorem 1.
R
is complete.
Proof.
Let
{
x
t
}
be a Cauchy sequence in
R
. I claim that
{
x
t
}
is bounded. Since
{
x
t
}
is Cauchy, there is a
T
such that for any
s > T
,
d
(
x
T
+1
, x
s
)
<
1. Set
r
= max
{
d
(
x
1
, x
T
+1
)
, . . . , d
(
x
T
, x
T
+1
)
}
+ 1
.
Then for any
t
,
d
(
x
t
, x
T
+1
)
< r
, as was to be shown.
Therefore, for each
t
, the set
{
x
t
, x
t
+1
, . . .
}
is bounded above and hence has a
supremum
b
t
. Moreover, the set
{
x
1
, x
2
, . . .
}
is bounded below and hence has an
infimum. Therefore, for every
t
,
b
t
≥
inf
{
x
1
, x
2
, . . .
}
.
Therefore, the set
{
b
1
, b
2
, . . .
}
is bounded below and so has an infimum. Call this
infimum
b
.
1
I claim that
{
x
t
}
has a subsequence converging to
b
.
This follows from two
properties of
b
.
1.
Claim.
For any
ε >
0, there is a
T
such that for all
t > T
,
x
t
< b
+
ε
.
Proof.
By contraposition. Consider any
x
such that
x
t
≥
x
+
ε
for infinitely many
t
. Then
b
t
≥
x
+
ε
for all
t
, hence
b
≥
x
+
ε
, hence
b > x
, and in particular,
b
6
=
x
.
2.
Claim.
For any
ε >
0,
x
t
> b

ε
for infinitely many
t
.
Proof.
By contra
position. Consider any
x
such that for any
ε >
0 there is a
T
such that, for
all
t > T
,
x
t
≤
x

ε
. Then for all such
t
,
b
t
≤
x

ε
, hence
b
≤
x

ε
, hence
b < x
and, in particular,
b
6
=
x
.
1
For example, suppose that
x
t
= 1 + 1
/t
if
t
is odd and
x
t
=

1
/t
if
t
is even. Then for each
t
,
a
t
=

1
/t
,
b
t
= 1 + 1
/t
and
b
= 1. The point
b
is called the lim sup of
{
x
t
}
.
1
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From these two claims it follows that for any
ε >
0, there are infinitely many
t
such
that
x
t
∈
N
ε
(
b
). Choose
t
1
such that
x
t
1
∈
N
1
(
b
),
t
2
> t
1
such that
x
t
2
∈
N
1
/
2
(
b
),
and so on. By construction,
x
t
k
converges to
b
.
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 Fall '08
 JohnNachbar
 Topology, Metric space, Compact space, Xt, RN Tychonoff

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