MAT 325: Topology
Professor Zoltan Szabo
Problem Set 7
Rik Sengupta
rsengupt@princeton.edu
April 7, 2010
1.
Munkres, p. 218, problem 1
Give an example showing that a Hausdorﬀ space with a countable basis need not be metrizable.
Solution.
R
K
suﬃces as an example. By Example 1 on p. 197,
R
K
[as deﬁned on p. 82] is
a space that is nonregular and Hausdorﬀ. Indeed,
R
K
is Hausdorﬀ, for the topology is ﬁner
than the standard topology [see Lemma 13.4]. Furthermore,
R
K
is 2nd countable, for the sets
(
a,b
) and (
a,b
)

K
, where the intervals have
rational
endpoints, constitute a countable basis.
However,
R
K
is not metrizable, for it is not even regular [by Example 1, p. 197]. Here, recall
that a metrizable space is necessarily
T
1
and regular, which gives rise to the contradiction.
±
2.
Munkres, p. 218, problem 3
Let
X
be a compact Hausdorﬀ space. Show that
X
is metrizable if and only if
X
has a
countable basis.
Solution.
We present two solutions:
Solution 1.
(=
⇒
:) Assume
X
is metrizable. Now,
X
is Lindel¨of because it is compact, and
metrizable Lindel¨of spaces are second countable.
(
⇐
=:) Assume
X
is second countable. Since
X
is compact and Hausdorﬀ, it is normal, and
thus regular as well. Urysohn’s metrization theorem then shows that
X
is metrizable.
±
Solution 2.
We ﬁrst prove a simple lemma.
Lemma.
Every compact metrizable space is secondcountable.
Proof.
Let
X
be a compact metrizable space, and let
d
be a metric on
X
that induces the
topology on
X
.
For each
n
∈
Z
+
, let
A
n
be an open covering of
X
with 1
/n
balls. By the compactness of
X
,
there exists a ﬁnite subcovering
A
n
.
So,
B
=
S
n
∈
Z
+
A
n
is countable, being a countable union of ﬁnite sets.
Notice that
B
is a basis. Let
U
be an open set in
X
and
x
∈
U
. By deﬁnition of the metric
topology, there exists some
± >
0 such that
B
d
(
x,±
)
⊂
U
. Choose
N
∈
Z
+
such that 2
/N < ±
.
Since
A
n
covers
X
, this means there exists some
B
d
(
y,
1
/N
) containing
x
. If
z
∈
B
d
(
y,
1
/N
),
then
1