3The topology of C(K)
II Linear Analysis
3.1 Normality of compact Hausdorff spaces
At this point, it is helpful to introduce a new class of topological spaces.
Definition
(Normal space)
.
A topological space
X
is normal if for every disjoint
pair of closed subsets
C
1
, C
2
of
X
, there exists
U
1
, U
2
⊆ X
disjoint open such
that C
1
⊆ U
1
, C
2
⊆ U
2
.
This is similar to being Hausdorff, except that instead of requiring the ability
to separate points, we want the ability to separate closed subsets.
In general, one makes the following definition:
Definition
(
T
i
space)
.
A topological space
X
has the
T
1
property if for all
x, y ∈ X, where x 6= y, there exists U ⊆ X open such that x ∈ U and y 6∈ U.
A topological space X has the T
2
property if X is Hausdorff.
A topological space
X
has the
T
3
property if for any
x ∈ X
,
C ⊆ X
closed
with
x 6∈ C
, then there are
U
x
, U
C
disjoint open such that
x ∈ U
x
, C ⊆ U
C
.
These spaces are called regular.
A topological space X has the T
4
property if X is normal.
Note here that
T
4
and
T
1
together imply
T
2
. It suffices to notice that
T
1
implies that
x
is a closed set for all
x
— for all
y
, let
U
y
be such that
y ∈ U
y
and x 6∈ U
y
. Then we can write
X \ {x} =
[
y6=x
U
y
,
which is open since it is a union of open sets.
More importantly, we have the following theorem:
Theorem.
Let
X
be a Hausdorff space. If
C
1
, C
2
⊆ X
are compact disjoint
subsets, then there are some
U
1
, U
2
⊆ X
disjoint open such that
C
1
⊆ U
1
, C
2
, ⊆
U
2
.
In particular, if
X
is a compact Hausdorff space, then
X
is normal (since
closed subsets of compact spaces are compact).
Proof.
Since
C
1
and
C
2
are disjoint, by the Hausdorff property, for every
p ∈ C
1
and q ∈ C
2
, there is some U
p,q
, V
p,q
⊆ X disjoint open with p ∈ U
p,q
, q ∈ V
p,q
.
Now fix a
p
. Then
S
q∈C
2
V
p,q
⊇ C
2
is an open cover. Since
C
2
is compact,
there is a finite subcover, say
C
2
⊆
n
[
i=1
V
p,q
i
for some {q
1
, ··· , q
n
} ⊆ C
2
.
Note that n and q
i
depends on which p we picked at the beginning.
Define
U
p
=
n
\
i=1
U
p,q
i
, V
p
=
n
[
i=1
V
p,q
i
.
Since these are finite intersections and unions,
U
p
and
V
p
are open. Also,
U
p
and V
p
are disjoint. We also know that C
2
⊆ V
p
.
Now note that
S
p∈C
1
U
p
⊇ C
1
is an open cover. By compactness of
C
1
, there
is a finite subcover, say
C
1
⊆
m
[
j=1
U
p
j
for some {p
1
, ··· , p
m
} ⊆ C
1
.
Now define
U =
m
[
j=1
U
p
j
, V =
m
\
j=1
V
p
j
.
Then U and V are disjoint open with C
1
⊆ U, C
2
⊆ V . So done.
Why do we care about this? It turns out it is easier to discuss continuous
functions on normal spaces rather than Hausdorff spaces. Hence, if we are given
that a space is compact Hausdorff (e.g. [0, 1]), then we know it is normal.