Math 202A
Fall 2005
David Penneys
Problems by Sarason.
(11) Let
B
be a Baire space. Let
U ⊆ B
be an open subset. Let
M ⊂ U
be a meager set in
U
.
Since
M
is cleary meager in
B
,
B \ M
is residual in
B
and therefore dense in
B
since
B
is
Baire. Let
V ⊆ U
be an open subset in the relative topology for
U
, i.e.,
V
=
U ∩W
for some
open
W ⊂ B
. Since
V
is open in
B
as the intersection of two open sets in
B
, (
B \ M
)
∩ V
is nonempty, and since
V ⊆ U
, (
U \ M
)
∩ V
is nonempty. Thus
U \ M
is dense, and
U
is a
Baire space.
(12) Let
X
be a
T
1
Baire space with no isolated points. Since
X
is
T
1
, all singletons are closed
in
X
, and since
X
has no isolated points, all singletons are not open in
X
.
Case 1
:
X
is counbtable. Let
x
∈
X
.
{
x
}
is meager, so
X
\ {
x
}
is residual. But
X
\ {
x
}
is the
countable union of singletons, which are nowhere dense sets, so
X
\ {
x
}
is meager and
thus
{
x
}
is residual. But
{
x
}
is not dense since
X
\ {
x
}
is open, and (
X
\ {
x
}
)
∩ {
x
}
is empty. Thus
X
is not a Baire space, a contradiction.
Case 2
:
X
is uncountable. Suppose
R
⊂
X
is residual. Suppose
R
is countable and has one
point. Then
X
\
R
is open, and
R
is not dense since (
X
\
R
)
∩
R
is empty. Thus
X
is not a
Baire space, a contradiction. Suppose
R
is countable and has more than one point. Let
x
∈
R
. Then
R
\{
x
}
is meager as the countable union of singletons, which are nowhere
dense. Thus (
X
\
R
)
∪
(
R
\ {
x
}
) is meager as the union of two meager sets (which is
still the countable union of nowhere dense sets). So
{
x
}
=
X
\
((
X
\
R
)
∪
(
R
\ {
x
}
))
is residual, but
{
x
}
is not dense as before, a contradiction. Thus, any residual subset
must be uncountable.
(13) Let
X
be a Baire space,
f
:
X
→
Y
a continuous, open surjection. Let
R
⊂
Y
be residual.
We need to show
R
is dense in
Y
. We have that
Y
\
R
is meager, so
Y
\
R
=
∞
n
=1
C
n
where
C
n
is nowhere dense
∀
n
∈
N
.
First, we show
f

1
(
∞
n
=1
C
n
) is meager in
X
. To show this, it suffices to show that
f

1
(
C
n
)
is meager since
f

1
(
∞
n
=1
C
n
) =
∞
n
=1
f

1
(
C
n
) and a countable union of meager sets is meager
as the countable union of a countable union of nowhere dense sets, which is still a countable
union of nowhere dense sets. Let
U ⊆
f

1
(
C
n
) for some
n
∈
N
be an open subset. Then
f
(
U
)
⊆
f
(
f

1
(
C
n
))
⊆
f
(
f

1
(
C
n
)) =
C
n
where the first inclusion follows from the continuous image of a subset, the second inclusion
follows from the defintion of continuity (a problem from the last homewrok set), and the
equality follows from the fact that
f
is a surjection.
But since
f
is an open map,
f
(
U
)
is open, so
int
(
C
n
) cannot be empty, a contradiction. Thus, there does not exist an open
U ⊆
f

1
(
C
n
), so
int
(
f

1
(
C
n
)) is empty
∀
n
∈
N
, and thus
f

1
(
C
n
) is meager
∀
n
∈
N
, and
f

1
(
∞
n
=1
C
n
) is meager in
X
. Thus
f

1
(
R
) is dense in
X
.
It remains to show that for all open
V ∈
Y
,
V ∩
R
is nonempty. It suffices to show that
for any open
V
such that
V
contains a point in
Y
\
R
,
V ∩
R
is nonempty. Let
y
∈
Y
\
R
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
and let
V
be open in
Y
such that
y
∈ V
. Then
f

1
(
V
)
∩
f

1
(
R
) is nonempty since
f

1
(
R
)
is dense in
X
, and
f

1
(
V
) is open in
X
by continuity. Let
x
∈
f

1
(
V
)
∩
f

1
(
R
). Then
f
(
x
)
∈ V ∩
R
, so
V ∩
R
is nonempty. Thus R is dense and we are done.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 tao/analysis
 Math, Topology, Metric space, CN, General topology, lim µ, countable union

Click to edit the document details