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453
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7, 2004
Exercise 2 from Section 17, page 100
Using Theorem 17.2,
A
is closed in
Y
implies that it is the intersection of
Y
and a closed
subset
C
of
X
:
A
=
C
∩
Y
. Since
Y
is given to be closed in
X
,
A
is the intersection of two
closed sets in
X
, and hence is closed in
X
.
±
Exercise 6 from Section 17, page 101
(a)
¯
B
is a closed set that contains
B
, so it contains
A
as well. Hence
¯
A
⊂
¯
B
.
(b)
¯
A
∪
¯
B
is closed, being the ﬁnite union of closed sets, and contains
A
∪
B
, so it must
contain the closure
A
∪
B
as well. The other inclusion is a special case of (c).
(c) For each
α
,
A
α
⊂
S
A
α
, so
¯
A
α
⊂
S
A
α
. Consequently,
S
¯
A
α
⊂
S
A
α
. An example
where equality fails is given by
A
n
=
[1
/
n
,
1],
n
∈
Z
+
in
R
. Each
A
n
is closed, so
S
¯
A
n
=
S
A
n
=
(0
,
1]. However,
S
A
n
=
(0
,
1]
=
[0
,
1].
±
Exercise 7 from Section 17, page 101
The given “proof” fails because, while every neighborhood
U
of
x
must intersect some
A
α
,
each such neighborhood may intersect a di
ﬀ
erent
A
α
, leaving
x
in the closure of none. For