Royden, Real Analysis 3rd ed.
Chapter 7
Problem 73
b.
Let
X
x
=
f
y
2
X
:
½
(
y;x
)
<
1g
:
Then
(i)
X
x
is open.
For each
y
2
X
x
;
we have
½
(
y;x
)
<
1
:
By using the
triangle inequality it can be shown easily that
B
(
y; r
)
μ
X
x
for any
r >
0
:
(ii)
X
c
x
is open. Similarly, it can be shown that
B
(
y; r
)
μ
X
c
x
if
y
2
X
c
x
i.e.
½
(
y;x
) =
1
:
Hence
X
c
x
is open which is the same to say
X
x
is closed.
Problem 77
L
1
(
¡ 1
;
1
)
and
L
1
[0
;
1]
are separable. Take
L
1
[0
;
1]
for
instance. Any integrable function can be approximated by continuous func
tions, and then by polynomials (StoneWeierstrass Theorem).
Finally we
consider the class of all polynomials with rational coe¢ cients, which is count
able.
L
1
(0
;
1)
is not separable. Consider
f
c
(
x
) =
Â
(0
;c
)
(
x
)
for
c
2
(0
;
1)
:
Then
k
f
c
¡
f
d
k
1
= 1
for
c
6
=
d:
Problem 711
a.
If
½ < ";
then
¾
=
½
1 +
½
< ½ < ":
If
½ <
"
2
<
1
2
;
then
½
=
¾
1
¡
¾
<
2
¾ < ":
Hence
¾
is equivalent to
½:
Problem 717
d.
Let
*
x
n
= (
x
n
1
;x
n
2
; :: :
)
be a Cauchy sequence in
(
X
¤
; ½
¤
)
:
By taking a proper subsequence we may assume
½
¤
(
*
x
n
;
*
x
n
+1
)
<
2
¡
n
:
Since
f
x
nk
:
k
¸
1
g
is a Cauchy sequence, we can …nd
N
n
such that
½
(
x
nm
;x
nl
)
<
2
¡
n
and
½
(
x
n;m
;x
n
+1
;m
)
<
2
¡
n
for
m
and
l
¸
N
n
:
(1)
WLOG we may assume
N
1
< N
2
< N
3
<
¢¢¢
:
De…ne
*
x
= (
x
1
N
1
;x
2
N
2
;x
3
N
3
; : : :
)
:
We claim
½
¤
(
*
x
n
;
*
x
)
!
0
:
First we need
to show
f
x
n;N
n
g
is a Cauchy sequence. For
k
¸
1
we have from
(1)
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 Spring '08
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 Metric space, Baire Category Theorem

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