3The topology of C(K)
II Linear Analysis
3.2 Tietze-Urysohn extension theorem
The objective of this chapter is to show that if we have a continuous function
defined on a closed subset of a normal space, then we can extend this to the
whole of the space.
We start a special case of this theorem.
Lemma
(Urysohn’s lemma)
.
Let
X
be normal and
C
0
, C
1
be disjoint closed
subsets of
X
. Then there is a
f ∈ C
(
X
) such that
f|
C
0
= 0 and
f|
C
1
= 1, and
0 ≤ f(x) ≤ 1 for all X.
Before we prove this, let’s look at a “stupid” example. Let
X
= [
−
1
,
2]. This
is compact Hausdorff. We let
C
0
= [
−
1
,
0] and
C
1
= [1
,
2]. To construct the
function f, we do the obvious thing:
−1 0 1 2
We can define this function f (in [0, 1]) by
f(x) = inf
n
a
2
n
: a, n ∈ N, 0 ≤ a ≤ 2
n
, x ≤
a
2
n
o
.
This is obviously a rather silly way to write our function out. However, this is
what we will end up doing in the proof below. So keep this in mind for now.
Proof. In this proof, all subsets labeled C are closed, and all subsets labeled U
are open.
First note that normality is equivalent to the following: suppose
C ⊆ U ⊆ X
,
where
U
is open and
C
is closed. Then there is some
˜
C
closed,
˜
U
open such that
C ⊆
˜
U ⊆
˜
C ⊆ U .
We start by defining
U
1
=
X \C
1
. Since
C
0
and
C
1
are disjoint, we know
that C
0
⊆ U
1
. By normality, there exists C
1
2
and U
1
2
such that
C
0
⊆ U
1
2
⊆ C
1
2
⊆ U
1
.
Then we can find C
1
4
, C
3
4
, U
1
4
, U
3
4
such that
C
0
⊆ U
1
4
⊆ C
1
4
⊆ U
1
2
⊆ C
1
2
⊆ U
3
4
⊆ C
3
4
⊆ U
1
.
Iterating this, we get that for all dyadic rationals
q
=
a
2
n
,
a, n ∈ N,
0
< a <
2
n
,
there are some U
q
open, C
q
closed such that U
q
⊆ C
q
, with C
q
⊆ U
q
0
if q < q
0
.
We now define f by
f(x) = inf {q ∈ (0, 1] dyadic rational : x ∈ U
q
},
with the understanding that inf ∅ = 1. We now check the properties desired.
– By definition, we have 0 ≤ f(x) ≤ 1.
– If x ∈ C
0
, then x ∈ U
q
for all q. So f (x) = 0.
– If x ∈ C
1
, then x 6∈ U
q
for all q. So f (x) = 1.
–
To show
f
is continuous, it suffices to check that
{x
:
f
(
x
)
> α}
and
{x
:
f
(
x
)
< α}
are open for all
α ∈ R
, as this shows that the pre-images of
all open intervals in R are open. We know that
f(x) < α ⇔ inf{q ∈ (0, 1) dyadic rational : x ∈ U
q
} < α
⇔ (∃q) q < α and x ∈ U
q
⇔ x ∈
[
q<α
U
q
.
Hence we have
{x : f(x) < α} =
[
q<α
U
q
.
which is open, since each U
q
is open for all q. Similarly we know that
f(x) > α ⇔ inf{q : x ∈ U
q
} > α
⇔ (∃q > α) x 6∈ C
q
⇔ x ∈
[
q>α
X \ C
q
.
Since this is a union of complement of closed sets, this is open.
With this, we can already say that there are many continuous functions. We
can just pick some values on some
C
0
, C
1
, and then get a continuous function
out of it. However, we can make a stronger statement.
Theorem
(Tietze-Urysohn extension theorem)
.
Let
X
be a normal topological
space, and
C ⊆ X
be a closed subset. Suppose
f
:
C → R
is a continuous
function. Then there exists an extension
˜
f
:
X → R
which is continuous and
satisfies
˜
f|
C
= f and k
˜
fk
C(X)
= kfk
C(C)
.
This is in some sense similar to the Hahn-Banach theorem, which states that
we can extend linear maps to larger spaces.
Note that this implies the Urysohn’s lemma, since if
C
0
and
C
1
are closed,
then
C
0
∪ C
1
is closed. However, we cannot be lazy and not prove Urysohn’s
lemma, because the proof of this theorem relies on the Urysohn’s lemma.
Proof.
The idea is to repeatedly use Urysohn’s lemma to get better and better
approximations. We can assume wlog that 0
≤ f
(
x
)
≤
1 for all
x ∈ C
. Otherwise,
we just translate and rescale our function. Moreover, we can assume that the
sup
x∈C
f
(
x
) = 1. It suffices to find
˜
f
:
X → R
with
˜
f|
C
=
f
with 0
≤
˜
f
(
x
)
≤
1 for
all x ∈ X.
We define the sequences of continuous functions
f
i
:
C → R
and
g
i
:
X → R
for
i ∈ N
. We want to think of the sum
P
n
i=0
g
i
to be the approximations, and
f
n+1
the error on C.
Let
f
0
=
f
. This is the error we have when we approximate with the zero
function.
We first define g
0
on a subset of X by
g
0
(x) =
(
0 x ∈ f
−1
0
0,
1
3
1
3
x ∈ f
−1
0
2
3
, 1
.
We can then extend this to the whole of
X
with 0
≤ g
0
(
x
)
≤
1
3
for all
x
by
Urysohn’s lemma.
1
3
2
3
1
g
0
(x)
f(x)
We define
f
1
= f
0
− g
0
|
C
.
By construction, we know that 0
≤ f
1
≤
2
3
. This is our first approximation.
Note that we have now lowered our maximum error from 1 to
2
3
. We now repeat
this.
Given f
i
: C → R with 0 ≤ f
i
≤
2
3
i
, we define g
i
by requiring
g
i
(x) =
0 x ∈ f
−1
i
h
0,
1
3
2
3
i
i
1
3
2
3
i
x ∈ f
−1
i
h
2
3
i+1
,
2
3
i
i
,
and then extending to the whole of
X
with 0
≤ g
i
≤
1
3
2
3
i
and
g
i
continuous.
Again, this exists by Urysohn’s lemma. We then define f
i+1
= f
i
− g
i
|
C
.
We then have
n
X
i=0
g
i
= (f
0
− f
1
) + (f
1
− f
2
) + ··· + (f
n
− f
n+1
) = f − f
n+1
.
We also know that
0 ≤ f
i+1
≤
2
3
i+1
.
We conclude by letting
˜
f =
∞
X
i=0
g
i
.
This exists because we have the bounds
0 ≤ g
i
≤
1
3
2
3
i
,
and hence
P
n
i=0
g
i
is Cauchy. So the limit exists and is continuous by the
completeness of C(X).
Now we check that
n
X
i=0
g
i
|
C
− f = −f
n+1
.
Since we know that kf
n+1
k
C
(C)
→ 0. Therefore, we know that
∞
X
i=0
g
i
C
=
˜
f|
C
= f.
Finally, we check the bounds. We need to show that 0
≤
˜
f
(
x
)
≤
1. This is true
since g
i
≥ 0 for all i, and also
|
˜
f(x)| ≤
∞
X
i=0
g
i
(x) ≤
n
X
i=0
1
3
2
3
i
= 1.
So done.
We can already show what was stated last time — if
K
is compact Hausdorff,
{p
1
, ··· , p
n
} ⊆ K
a finite set of points, and
{y
1
, ··· , y
n
} ⊆ R
, then there exists
f
:
K → R
continuous such that
f
(
p
i
) =
y
i
. This is since compact Hausdorff
spaces are normal, and singleton points are closed sets in Hausdorff spaces. In
fact, we can prove this directly with the Urysohn’s lemma, by, say, requesting
functions
f
i
such that
f
i
(
p
i
) =
y
i
,
f
i
(
p
j
) = 0 for
i 6
=
j
. Then we just sum all
f
i
.
Note that normality is necessary for Urysohn’s lemma. Since Urysohn’s
lemma is a special case of the Tietze-Urysohn extension theorem, normality is
also necessary for the Tietze-Urysohn extension theorem. In fact, the lemma
is equivalent to normality. Let
C
0
, C
1
be disjoint closed sets of
X
. If there is
some
f
:
X → R
such that
f|
C
0
= 0,
f|
C
1
= 1, then
U
0
=
f
−1
−∞,
1
3
and
U
1
= f
−1
2
3
, ∞
are open disjoint sets such that C
0
⊆ U
0
, C
1
⊆ U
1
.
Closedness of
C
0
and
C
1
is also necessary in Urysohn’s lemma. For example,
we cannot extend f : [0,
1
2
) ∪ (
1
2
, 1] to [0, 1] continuously, where f is defined as
f(x) =
(
0 x <
1
2
1 x >
1
2
.