Solutions for HW 4
Problem 55: 7.
For each
n
pick
x
n
∈
F
n
. We first show that (
x
n
) is a Cauchy sequence.
Let
>
0.
Then there exists
N
such that
ρ
(
F
n
)
<
for all
n
≥
N
.
Let
n, m
≥
N
and
assume
n < m
.
Then
ρ
(
x
n
, x
m
)
≤
ρ
(
F
n
)
<
.
Hence (
x
n
) is a Cauchy sequence.
As
X
is complete there exists
x
0
such that
x
n
→
x
0
. Now
x
m
∈
F
n
for all
m
≥
n
implies that
x
0
∈
F
n
(as
F
n
is closed) for all
n
≥
1. Hence
x
0
∈ ∩
∞
n
=1
F
n
. Assume that also
y
0
∈ ∩
∞
n
=1
F
n
,
then
ρ
(
x
0
, y
0
)
≤
ρ
(
F
n
)
→
0 implies that
ρ
(
x
0
, y
0
) = 0, i.e.,
x
0
=
y
0
. Hence
∩
∞
n
=1
F
n
=
{
x
0
}
.
Counterexamples: (1) Take
X
=
Q
, and
F
n
= [
√
2
,
√
2 +
1
n
]
∩
X
. Then (ii) and (iii) hold,
but
∩
∞
n
=1
F
n
=
∅
. (2) Take
X
=
R
, and
F
n
= (0
,
1
n
). Then (i) and (iii) hold, but
∩
∞
n
=1
F
n
=
∅
.
(3) Let
X
=
R
, and
F
n
= [
n,
∞
). Then (i) and (ii) hold, but
∩
∞
n
=1
F
n
=
∅
.
Problem 55: 13.
(1) Ω(
x
)

x >
0, so Ω has no fix point.
(2) One can see using calculus that Ω
0
(
x
) = 1

e
x
(1+
e
x
)
2
. We claim that 0
<
Ω
0
(
x
)
<
1
for all
x
. The right hand side of this inequality is obvious. For the left hand side we
need that
e
x
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 Fall '11
 Schep
 Topology, Metric space, Compact space, Closed set, FN

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