MAT 325: Topology
Professor Zoltan Szabo
Problem Set 6
Rik Sengupta
[email protected]
March 30, 2010
1.
Munkres, p. 199, problem 3
Show that every order topology is regular.
Solution.
We first prove that every order topology is Hausdorff. So let (
X,
≤
) be a simply
ordered set. Let
X
be equipped with the order topology induced by the simple order. Fur
thermore, let
a
and
b
be two distinct points in
X
, and suppose without loss of generality that
a < b
. Let
A
=
{
x
∈
X
:
a < x < b
}
,
i.e. the set of elements between
a
and
b
. So now, if
A
is empty, then
a
∈
(
∞
, b
),
b
∈
(
a,
∞
),
and (
∞
, b
)
∩
(
a,
∞
) =
∅
, and so
X
is Hausdorff.
If
A
is nonempty, then
a
∈
(
∞
, x
),
b
∈
(
x,
∞
), and (
∞
, x
)
∩
(
x,
∞
) =
∅
for any
x
∈
A
, and therefore,
X
is Hausdorff.
So in particular, single points in
X
are closed. Suppose now that
x
∈
X
, and
A
is a closed set,
disjoint from
x
. Then, there exists a basis element (
a, b
) containing
x
which is disjoint from
A
. So pick any
a
0
∈
(
a, x
), and let
U
1
= (
∞
, a
0
),
V
1
= (
a
0
,
∞
). If no such
a
0
exists, then
let
U
1
= (
∞
, x
),
V
1
= (
a,
∞
). Exactly as before, in both cases, the pair of sets is disjoint.
Similarly, try to find
b
0
∈
(
x, b
), and if that exists, let
U
2
= (
b
0
,
∞
),
V
2
= (
∞
, b
0
), and if
not, let
U
2
= (
x,
∞
),
V
2
= (
∞
, b
).
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 Fall '09
 ZoltánSzabó
 Topology, Topological space, Compact space, disjoint open sets, Munkres, Hausdorff space

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