April 21, 2006
Munkres
§
29
Ex.
29.1.
Closed intervals [
a, b
]
∩
Q
in
Q
are not compact for they are not even sequentially
compact [Thm 28.2]. It follows that all compact subsets of
Q
have empty interior (are nowhere
dense) so
Q
can not be locally compact.
To see that compact subsets of
Q
are nowhere dense we may argue as follows: If
C
⊂
Q
is
compact and
C
has an interior point then there is a whole open interval (
a, b
)
∩
Q
⊂
C
and also
[
a, b
]
∩
Q
⊂
C
for
C
is closed (as a compact subset of a Hausdorff space [Thm 26.3]). The closed
subspace [
a, b
]
∩
Q
of
C
is compact [Thm 26.2]. This contradicts that no closed intervals of
Q
are
compact.
Ex. 29.2.
(a)
.
Assume that the product
X
α
is locally compact. Projections are continuous and open [Ex
16.4], so
X
α
is locally compact for all
α
[Ex 29.3]. Furthermore, there are subspaces
U
⊂
C
such
that
U
is nonempty and open and
C
is compact. Since
π
α
(
U
) =
X
α
for all but finitely many
α
,
also
π
α
(
C
) =
X
α
for all but finitely many
α
. But
C
is compact so also
π
α
(
C
) is compact.
(b)
.
We have
X
α
=
X
1
×
X
2
where
X
1
is a finite product of locally compact spaces and
X
2
is
a product of compact spaces. It is clear that finite products of locally compact spaces are locally
compact for finite products of open sets are open and all products of compact spaces are compact
by Tychonoff. So
X
1
is locally compact.
X
2
is compact, hence locally compact. Thus the product
of
X
1
and
X
2
is locally compact.
Conclusion
:
X
α
is locally compact if and only if
X
α
is locally compact for all
α
and compact
for all but finitely many
α
.
Example
:
R
ω
and
Z
ω
+
are not locally compact.
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 Fall '08
 Brown
 Topology, Sets, Metric space, Compact space, General topology, locally compact Hausdorff, Morten Poulsen

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