1st December 2004
Munkres
§
16
Ex. 16.1 (Morten Poulsen).
Let (
X,
T
) be a topological space, (
Y,
T
Y
) be a subspace and let
A
⊂
Y
.
Let
T
Y
A
be the subspace topology on
A
as a subset of
Y
and let
T
X
A
be the subspace topology
on
A
as a subset of
X
. Since
U
∈ T
Y
A
⇔ ∃
U
Y
∈ T
Y
:
U
=
A
∩
U
Y
⇔ ∃
U
X
∈ T
:
U
=
A
∩
(
Y
∩
U
X
)
⇔ ∃
U
X
∈ T
:
U
=
A
∩
U
X
⇔
U
∈ T
X
A
it follows that
T
Y
A
=
T
X
A
.
Ex.
16.3 (Morten Poulsen).
Consider
Y
= [

1
,
1] as a subspace of
R
with the standard
topology. By lemma 16.1 a basis for the subspace topology on
Y
is sets of the form:
Y
∩
(
a, b
) =
(
a, b
)
,
a, b
∈
Y
[

1
, b
)
,
a /
∈
Y, b
∈
Y
(
a,
1]
,
a
∈
Y, b /
∈
Y
Y,
∅
,
a, b /
∈
Y.
Note that intervals of the form [
a, b
) are not open in
R
, since there are no basis element (
c, d
)
such that
a
∈
(
c, d
)
⊂
[
a, b
). Similarly are intervals of the form (
a, b
] and [
a, b
] not open in
R
.
A
=
{
x

1
2
<

x

<
1
}
:
A
= (

1
,

1
2
)
∪
(
1
2
,
1), hence
A
open in
R
. Since
A
=
Y
∩
A
it follows
that
A
open in
Y
.
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 Fall '08
 Brown
 Logic, Topology, Topological space, subspace topology, Morten Poulsen, basis element

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