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1st December 2004
Munkres
§
31
Ex. 31.1 (Morten Poulsen).
Let
a
and
b
be distinct points of
X
. Note that
X
is Hausdorﬀ,
since
X
is regular. Thus there exists disjoint open sets
A
and
B
such that
a
∈
A
and
b
∈
B
. By
lemma 31.1(a) there exists open sets
U
and
V
such that
a
∈
U
⊂
U
⊂
A
and
b
∈
V
⊂
V
⊂
B.
Clearly
U
∩
V
=
∅
.
Ex. 31.2 (Morten Poulsen).
Let
A
and
B
be disjoint closed subsets of
X
. Since
X
normal
there exists disjoint open sets
U
0
and
U
1
such that
A
⊂
U
0
and
B
⊂
U
1
. By lemma 31.1(b) there
exists open sets
V
0
and
V
1
such that
A
⊂
V
0
⊂
V
0
⊂
U
0
and
B
⊂
V
1
⊂
V
1
⊂
U
1
Clearly
U
∩
V
=
∅
.
Ex. 31.3 (Morten Poulsen).
Theorem 1.
Every order topology is regular.
Proof.
Let
X
be an ordered set. Let
x
∈
X
and let
U
be a neighborhood of
x
, may assume
U
= (
a, b
),
∞ ≤
a < b
≤ ∞
. Set
A
= (
a, x
) and
B
= (
x, b
). Using the criterion for regularity in
lemma 31.1(b) there are four cases:
(1) If
u
∈
A
and
v
∈
B
then
x
∈
(
u, v
)
⊂
(
u, v
)
⊂
[
u, v
]
⊂
(
a, b
).
(2) If
A
=
B
=
∅
then (
a, b
) =
{
x
}
is open and closed, since
X
Hausdorﬀ, c.f. Ex. 17.10.
(3) If
A
=
∅
and
v
∈
B
then
x
∈
(
a, v
)
⊂
= [
x, v
)
⊂
[
x, v
)
⊂
[
x, v
]
⊂
(
a, b
).
(4) If
u
∈
A
and
B
=
∅
then
x
∈
(
u, b
)
⊂
(
u, x
]
⊂
(
u, x
]
⊂
[
u, x
]
⊂
(
a, b
).
Thus
X
is regular.
±
Ex. 31.5.
The diagonal Δ
⊂
Y
×
Y
is closed as
Y
is Hausdorﬀ [Ex 17.13]. The map (
f, g
) :
X
→
Y
×
Y
is continuous [Thm 18.4, Thm 19.6] so
{
x
∈
X

f
(
x
) =
g
(
x
)
}
= (
f, g
)

1
(Δ)
is closed.
Ex. 31.6.
Let
p
:
X
→
Y
be closed continuous surjective map. Then
X
normal
⇒
Y
normal.
For this exercise and the next we shall use the following lemma from [Ex 26.12].
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This note was uploaded on 01/12/2011 for the course MATH 110 taught by Professor Brown during the Fall '08 term at Arizona Western College.
 Fall '08
 Brown
 Topology, Sets

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