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Serge Ballif
MATH 527 Homework 4
September 28, 2007
Problem 1.
Show that every locally compact Hausdorff space is completely regular.
Let
X
be locally compact Hausdorﬀ. Then
X
satisﬁes the conditions necessary
for a onepoint compactiﬁcation. That is, there exists a compact Hausdorﬀ set
Y
containing
X
as a subspace. In particular,
Y
is normal. Therefore,
Y
is com
pletely regular. The space
X
is completely regular as a subspace of a completely
regular space.
Problem 2.
A space
X
is locally metrizable if each point
x
∈
X
has a neighborhood that is metrizable in the subspace
topology. Show that a compact Hausdorff space
X
is metrizable if and only if it is locally metrizable.
(
⇒
) Any metrizable space is locally metrizable, since a subspace of a metric space
is a metric space.
(
⇐
) Let
X
be compact, Hausdorﬀ, and locally metrizable.
X
is normal (and
hence, regular) as a compact Hausdorﬀ space. If
X
has a countable basis, then
we can invoke the Urysohn metrization theorem to show that
X
is metrizable.
To show that
X
has a countable basis, we note that local metrizability of
X
means that for each
x
in
X
, there is an open neighborhood
U
x
of
x
which is
metrizable. Then the set
U
c
x
is closed in
X
and is disjoint from
x
. Therefore, by
regularity of
X
there are disjoint open sets
V
x
⊃ {
x
}
and
W
x
⊃
U
c
x
. The set
V
x
is a closed compact subset that is contained in
U
x
. We can repeat this procedure
for each
x
in
X
. Then since
X
is compact, there exists a ﬁnite subcover of sets
V
x
1
,V
x
2
,...,V
x
k
. Each space
V
x
i
is compact as a closed subspace of a compact
space, and is metrizable as a subset of
U
x
i
. Hence,
V
x
i
is 2nd countable with basis
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This note was uploaded on 04/01/2008 for the course MATH 527 taught by Professor Rotman,regina during the Fall '07 term at Pennsylvania State University, University Park.
 Fall '07
 ROTMAN,REGINA
 Math, Topology

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