Math 245C
Homework 1
Brett Hemenway
April 18, 2007
Folland Chapter 4
10. Let
X
be a topological space.
a. Suppose
X
is disconnected, then there exist
U, V
nonempty open
subsets of
X
such that
U
∩
V
=
∅
and
U
∪
V
=
X
. Thus
U
c
=
V
in
X
.
V
is open by assumption, and
U
c
is closed because
U
is
open, so
V
is both open and closed. If we have that if
∅
, X
are
the only sets in
X
that are closed and open, then
X
must be
connected.
Conversely, suppose there is an open set
∅
=
U
X
that is
both open and closed in
X
, then
U
c
is open,
U
∩
U
c
=
∅
, and
U
∪
U
c
=
X
.
So we have that
X
is connected iff the only sets that are closed
and open in
X
are
∅
and
X
.
b. Let
{
E
α
}
α
∈
A
be a collection of connected subsets of
X
with
α
∈
A
E
α
=
∅
.
Suppose
E
=
α
∈
A
E
α
is disconnected, then
we can find open sets
U, V
in
E
with
U
∩
V
=
∅
,
U
∩
V
=
E
.
With
U, V
=
∅
. Then, since each
E
α
is connected, we have that
U
∩
E
α
=
∅
and
V
∩
E
α
=
E
α
or viceversa, for each
α
∈
A
.
Let
x
∈
α
∈
A
E
α
, then
x
∈
U
∪
V
=
α
∈
A
E
α
. Without loss of
generality, we may assume
x
∈
U
. Thus
U
∩
E
α
=
∅
for all
α
.
This gives us that
V
∩
E
α
=
∅
for all
α
∈
A
. This gives us that
V
=
∅
, thus
E
is connected.
c. Suppose
A
is disconnected. This means that we can find open sets
U, V
in
X
such that (
U
∩
A
)
∩
(
V
∩
A
) =
U
∩
V
∩
A
=
∅
,
A
⊂
U
∪
V
,
and
U
∩
A
=
∅
=
V
∩
A
. Now
U
∩
V
∩
A
⊂
U
∩
V
∩
A
=
∅
, and
A
⊂
A
⊂
U
∪
V
. So to show that
A
is disconnected, it remains
1
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to show that
U
∩
A
=
∅
=
V
∩
A
. Since
U
∩
A
and
V
∩
A
are
nonempty, let
u
∈
U
∩
A
and
v
∈
V
∩
A
.
Then since
U, V
are
neighborhoods of
u
and
v
, and
u
and
v
are in
A
, we have that
U
∩
A
=
∅
and
V
∩
A
=
∅
.
Thus if
A
is connected, so is
A
.
d. Let
x
∈
X
. Consider the set
A
=
{
E
:
E
⊂
X, E
connected,
x
∈
E
}
Then by part (b), the set
C
x
=
E
∈
A
E
is connected. The set
C
x
is clearly maximal, and by part (c),
C
x
is closed.
13. Let
X
be a topological space,
U
⊂
X
open, and
A
⊂
X
dense in
X
.
Then since
U
∩
A
⊂
U
, we have that
U
∩
A
⊂
U
To show the other inclusion, let
x
∈
U
.
Then for any open neigh
borhood
V
of
x
, we have that
V
∩
U
=
∅
. Since
U
is open, we have
that
U
∩
V
is open.
Since
A
is dense we have that (
A
∩
U
)
∩
V
=
A
∩
(
U
∩
V
) =
∅
. If
x
∈
A
∩
U
then we are done. If
x
∈
A
∩
U
, then
since (
A
∩
U
)
∩
(
V
\{
x
}
) = (
A
∩
U
)
∩
V
, we have
x
∈
acc(
A
∩
U
). Thus
x
∈
U
∩
A
. So
U
⊂
U
∩
A
and hence
U
=
U
∩
A
26.
a. If
f
is continuous on
X
, and
f
(
X
) =
Y
is disconnected, then we
can find nonempty open sets
U, V
⊂
Y
such that
U
∩
V
=
∅
, and
U
∪
V
=
Y
. Since
f
is continuous,
f

1
(
U
), and
f

1
(
V
) are open
in
X
, they are nonempty since
U
, and
V
are nonempty, and
X
=
f

1
(
Y
) =
f

1
(
U
∪
V
) =
f

1
(
U
)
∪
f

1
(
V
)
∅
=
f

1
(
∅
) =
f

1
(
U
∩
V
) =
f

1
(
U
)
∩
f

1
(
V
)
.
Thus
X
is disconnected.
We conclude that if
X
is connected,
f
(
X
) must also be connected.
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 Spring '10
 tao/analysis
 Logic, Topology, Sets, Metric space, Topological space, General topology, open sets

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