Math 202A Homework 8
Roman Vaisberg
Problem 1.
(a) Let
X
α
be a Hausdorff topological space for each
α
∈
I
.
Prove that if an infinite number of the
coordinate spaces
X
α
are noncompact, then each compact subset of the product
X
=
α
∈
I
X
α
is nowhere
dense.
Proof.
Let
K
be an arbitrary compact subset of
X
and suppose
x
= (
x
α
) is an interior point of
K
. Then
there exists an open set
U
in
X
containing
x
and contained in
K
and thus in turn
x
∈
α
∈
I
U
α
⊂
U
⊂
K
,
where each
U
α
is open in
X
α
and for all but finitely many
α
,
U
α
=
X
α
. From the statement of the problem
there must exist an
i
∈
I
such that
X
α
is noncompact and
U
α
=
X
α
. Letting
π
i
be the projection onto the
i
th coordinate we see that
X
i
=
π
i
(
α
∈
I
U
α
)
⊂
π
i
(
K
). We also clearly have
π
i
(
K
)
⊂
X
i
. However,
π
i
(
K
)
must be compact and
X
i
is not, which is a contradiction.
Thus
int
(
K
) =
int
(
K
) =
∅
, where the first
equality comes from the fact that
X
is a product of Hausdorff spaces and thus Hausdorff, and
K
being a
compact subset must then already be closed. Thus
K
is nowhere dense.
(b) Show that a continuous map need not map a locally compact space to a locally compact space.
Proof.
Consider the identity map from the space
R
with the discrete topology to
R
with the lower limit
topology. It is obvious that a space with the discrete topology is locally compact as every singleton is open,
closed, and compact. Furthermore the identity map is continuous. To prove that
R
with the lower limit
topology is not compact we first show that [0
,
1) is not compact. This is easy since
{
[0
,
1

1
2
n
)
}
n
∈
N
is a
cover for [0
,
1) which does not have a finite subcover. Thus [
a, b
) is not compact for any real
a
,
b
. Next we
show that 0 does not have an open neighbourhood whose closure is compact. Let
U
be a neighbourhood
of 0. Then there must exist a basic set [
a, b
) such that 0
∈
[
a, b
)
⊂
U
. But since [
a, b
) is also closed in
the lower limit topology (it is easy to see that its complement is open), it would have to be compact if the
closure of
U
were compact. Thus we obtain that the closure of
U
cannot be compact and
R
with the lower
limit topology is not locally compact.
(c) Suppose
f
:
X
→
Y
is continuous, open, and surjective.
X
is locally compact and
Y
is Hausdorff.
Prove that
Y
is locally compact.
Proof.
Let
y
∈
Y
and select
x
∈
X
such that
f
(
x
) =
y
. Since
X
is locally compact there exists an open
set
U
containing
x
such that
U
is compact. let
V
=
f
(
U
). Then
V
is open. Now observe that
f
(
U
) is
compact and since
Y
is Hausdorff,
f
(
U
) is closed. Thus
V
⊂
f
(
U
) implies
V
⊂
f
(
U
), which implies that
V
is a closed subset of a compact space and thus compact.
This concludes the proof that
Y
is locally
compact.
(d) Prove that if a product of Hausdorff spaces is locally compact, then each coordinate space is locally
compact.
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 Spring '10
 tao/analysis
 Logic, Topology, Metric space, Topological space, General topology, CLX

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