M
453
S
A
8
N
6, 2004
Exercise 2 from Section 26, page 171
(a) Let
X
be a subspace of
R
in the ﬁnite complement topology, and let
{
U
α
}
be an open
cover for
X
. Pick a particular
U
α
0
=
R

A
α
0
, where
A
α
0
is ﬁnite. Then
X

U
α
0
⊂
A
α
0
is ﬁnite, say
X

U
α
0
=
{
a
1
,
a
2
, . . . ,
a
n
}
. For each
a
i
, we can ﬁnd an
U
α
i
that contains
it, so
{
U
α
0
,
U
α
i
, . . . ,
U
α
n
}
is a ﬁnite subcover.
(b) [0
,
1] is not compact in this “countable complement” topology. For each
n
∈
Z
+
, let
U
n
=
[0
,
1]
 {
1
i

i
>
n
}
. Then
{
U
n
}
n
∈
Z
+
is an open cover for [0
,
1] that has no ﬁnite
subcover.
±
Exercise 3 from Section 26, page 171
Let
A
1
, . . . ,
A
n
be compact subspaces of
X
; let
A
=
S
n
i
=
1
A
i
. Suppose
C
is an open cover
for
A
. Then
C
is also an open cover for each
A
i
, so we can ﬁnd ﬁnite subcovers
C
i
. Their
union is the required ﬁnite subcover for
A
. Hence
A
is compact.
±
Exercise 5 from Section 26, page 171
By Lemma 26.4, for each
a
∈
A
, we can ﬁnd disjoint open sets
U
a
and
V
a
such that
a
∈
U
a
and
B
⊂
V
a
. The collection
{
U
a

a
∈
A
}
is an open cover for
A
, so by compactness, we can
ﬁnd a ﬁnite subcover
{
U
a
1
,
U
a
2
, . . . ,
U
a
n
}
. Let
U
=
S
n
i
=
1
U
a
i
and
V
=
T
n
i
=
1
V
a
i
. Then
U
and
V
are disjoint, and
A
⊂
U
and
B
⊂
V
.
±
Exercise 11 from Section 26, page 171
Suppose that
Y
is not connected, so there is a separation
C
∪
D
. Then
C
and
D
are both
closed in
Y
and hence in
X
since
Y
is closed in
X
. Therefore
C
and
D
are compact and by
problem 26.5 we can ﬁnd disjoint open sets
U
and
V
containing
C
and
D
respectively.