Solution.
ABNS
(c)
X
=
R
and
C
consists of all sets that are either countable or have a countable complement. (Finite sets count as
countable!)
Solution.
SA
(d)
X
=
R
and
C
is the family of all open subsets of
R
.
Solution.
N
(e)
X
=
R
and
C
is the family of all sets that are either open or closed.
Solution.
N
(f)
X
=
R
and
C
is the family consisting of
R
, and all sets of one of the following forms:
i.
n
[
k
=1
(
a
k
,b
k
],
ii. (
a
0
,
∞
]
∪
n
[
k
=1
(
a
k
,b
k
],
iii. (
∞
,b
0
]
∪
n
[
k
=1
(
a
k
,b
k
].
(The number of sets
n
in the unions above varies, all
n
∈
N
are involved.)
Solution.
This was supposed to be ABNS, but is actually N. See Homework 7 for a corrected version.
3. The smallest
σ
algebra containing all open subsets of a metric space
X
is known as the
Borel
σ
algebra of
X
, or the
family of Borel subsets of
X
. We will write
B
(
X
) to denote the
σ
algebra of Borel subsets of
X
. To repeat,
B
(
X
) is
the
σ
algebra generated by the family of open subsets of
X
. Prove that
B
(
X
) is also the
σ
algebra generated by all
closed sets of
X
.
Solution.
Let
O
be the family of all open subsets of
X
,
C
the family of all closed subsets of
X
. Let
A
∈ O
. Then
A
c
is closed, hence
A
c
∈
σ
(
C
), hence also
A
∈
σ
(
C
). This proves that
σ
(
C
) contains all open sets, being a
σ
algebra
containing all open sets, it must contain
σ
(
O
) =
B
(
X
). Similarly one sees that
B
(
X
) =
σ
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 Spring '11
 Speinklo
 Topology, Empty set, Metric space, Open set, Topological space, Closed set

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