ROYDEN, REAL ANALYSIS 3RD ED.
CHAPTER 8
Problem 810 a.
Let us assume
A
1
\
A
2
=
Á
. For
x
2
A
1
[
A
2
;x
2
A
1
or
x
2
A
2
.
Enough to consider the case
x
2
A
1
. Then
x =
2
A
2
. Or
x
2
A
n
A
2
which is open in
A
.
Hence there exists an open set
O
, such that
(1)
x
2
O
1
½
A
n
A
2
;
which means
O
1
\
A
2
=
Á:
Since
f
j
A
1
is continuous, there exists, for any open set
U
of
f
(
x
), an open set
O
2
in
A
such that
(2)
O
2
\
A
1
μ
f
¡
1
(
U
)
:
Then
x
2
O
1
\
O
2
and by (1)
(
O
1
\
O
2
)
\
A
= (
O
1
\
O
2
)
\
A
1
μ
f
¡
1
(
U
)
:
That is
f
j
A
is continuous at
x
.
b.
A
1
= (
¡ 1
;
0] and
A
2
= (0
;
1)
: f
1
(
x
)
´
0 on
A
1
and
f
2
(
x
) = 1
=x
on
A
2
.
c.
For
x
2
A
1
n
A
2
we can ¯nd an open set
O
1
such that
x
2
O
1
and
O
1
\
(
A
2
¡
A
1
) =
Á:
Together with (2), we have
x
2
O
1
\
O
2
and
O
1
\
O
2
\
(
A
1
[
A
2
) =
O
1
\
O
2
\
A
1
μ
f
¡
1
(
U
)
:
Hence
f
A
1
[
A
2
is continuous at
x
. Similar arguments work for
x
2
A
2
n
A
1
.
Typesetby
A
M
S
T
E
X
1
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2
Since
F
μ
U
r
½
O
for any
r
=
p
¢
2
¡
m
, we immediately get
f
j
F
= 0 and
f
j
e
O
= 1
:
Because
U
1
´
X;
0
·
f
·
1. It remains to show the continuity of
f
.
Suppose
f
(
x
) =
®
. By de¯nition
x
2
U
®
+
"
and
x =
2
U
®
¡
"
μ
¹
U
®
¡
"
for any
" >
0 with
®
§
"
=
p
¢
2
¡
n
. Since
U
®
+
"
is open, there is an open set
N
1
(
x
)
μ
U
®
+
"
which implies
(1)
f
(
y
)
·
®
+
"
for
y
2
N
1
(
x
)
:
As
x
2
(
¹
U
®
¡
"
)
c
which is open, there is an open set
N
2
(
x
) such that
N
2
(
x
)
½
¹
U
®
¡
"
)
c
or
equivalently
N
2
(
x
)
\
¹
U
®
¡
"
=
Á
. Since
U
®
μ
U
s
for
® < s
, we have
N
2
(
x
)
½
U
®
=
;
and
thus
(2)
f
(
y
)
¸
®
¡
"
for
y
2
N
2
(
x
)
:
Let
N
(
x
) =
N
1
(
x
)
\
N
2
(
x
)
:
Combining (1) and (2)
j
f
(
y
)
¡
®
j ·
"
for
y
2
N
(
x
)
:
d.
(
)
) Apply
c.
with
O
c
=
B
.
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 Spring '08
 Staff
 Topology, Metric space, Open set, Topological space, General topology

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