M
453
S
A
10
N
23, 2004
Exercise 5 from Section 30, page 194
(a) Let
D
be a countable dense subset of the metrizable space
X
. I claim that
B
=
{
B
(
x
,
1
/
n
)

x
∈
D
and
n
∈
Z
+
}
is a countable basis. First,
B
is countable, since both
D
and
Z
+
are. To show that it
is a basis for the topology on
X
, let
U
be any open set, and let
y
∈
U
. Then there is
an open ball,
B
(
y
, ε
), around
y
that is contained in
U
. Choose
n
to be larger that 2
/ε
,
and pick an
x
∈
D
such that
x
∈
B
(
y
,
1
/
n
). This gives
y
∈
B
(
x
,
1
/
n
)
⊂
B
(
y
, ε
)
⊂
U
.
Hence
B
is a basis.
(b) Let
X
be a metrizable Lindel¨of space. For each
n
∈
Z
+
, the collection of open sets
{
B
(
x
,
1
/
n
)

x
∈
X
}
covers
X
. Pick a countable subcover and call it
B
n
. Then the
collection
B
=
S
∞
n
=
1
B
n
is a countable basis. Countability is immediate, and the
proof that it is a basis is analogous to that in part (a).
±
Exercise 6 from Section 30, page 194
We know from Example 3 (p. 192) that
R
`
is not second countable, but has a countable
dense subset. Thus part (a) above implies that
R
`
cannot be metrizable.
Suppose
I
2
o
were metrizable. Since
I
2
o
is also compact (and hence Lindel¨of), part (b)
above implies that it is second countable. This in turn shows that the subspace
A
=
I
×
(0
,
1)
is second countable, and hence Lindel¨of. However, Example 5 (p. 193) shows that
A
is not
Lindel¨of, so
I
2
o
cannot be metrizable.
±
Exercise 12 from Section 30, page 194
I’ll show the case for second countability; the only di
ﬀ
erence for ﬁrst countability is to start
with a countable basis at a point instead. Suppose
B
is a countable basis for
X
. Since
f
is
an open map,
B
0
=
{
f
(
B
)

B
∈
B
}
is a collection of open sets in
f
(
X
). To show that it is a basis for