set 6

3. Partitions of unity.
We define the sum of the non-negative real numbers ri , i ∈ I, by the formula
X
i∈I
and we note that if
P
i∈I
ri = sup
X
j∈J
rj : J ⊂ I and J is finite .
ri < ∞, then ri 6= 0 for at most countably many i ∈ I.
A partition of unity of a topological space X is a collection F = {fi : i ∈ I} of
P
continuous functions X → [0, 1] such that i∈I fi (x) = 1 for every x ∈ X. The partition
of unity F is subordinated to a cover U of X provided that the family {Supp(f ) : f ∈ F }
is a refinement of U .
1 Theorem Every open cover of a pseudometrizable space has a subordinated partition
of unity.
Proof. Let G be an open cover of a pseudometric space (X, d). By the well-ordering
theorem, we can write G = {Gα : α < λ}, where λ is an ordinal number. We set sup ∅ = 0.
For every α ≤ λ, the formula gα (x) = min 1, supβ<α d(x, X r Gβ ) defines a function
gα : X → [0, 1]; this fuction is continuous, because for all x, y ∈ X we have that
|gα (x) − gα (y)| ≤ sup |d(x, X r Gβ ) − d(y, X r Gβ )| ≤ d(x, y) .
β<α
For every α < λ, we set fα = gα+1 − gα , and we note that fα is continuous X → [0, 1]. If
a point x ∈ X satisfies the inequality fα (x) > 0, then we have that gα+1 (x) > gα (x) and
hence that supβ≤α d x, X r Gβ > supβ<α d x, X r Gβ ; as a consequence, we have that
d x, X r Gα > 0. By the foregoing, we see that the support of fα is contained in Gα .
P
We use transfinite induction to show that
β<α fβ = gα for every α ≤ λ. The
P
equation holds for α = 0, because
β<0 fβ ≡ 0 ≡ g0 . Let 0 < α ≤ λ be such that
the required equation holds for every γ < α. Note that gα = supγ<α gγ+1 . For each
γ < α we have that fγ = gγ+1 − gγ and hence it follows from the inductive assumption
P
P
that gγ+1 = fγ + gγ = fγ + β<γ fβ =
β≤γ fβ . By the foregoing, we have that
P
P
gα = supγ<α β≤γ fβ = β<α fβ . This completes the induction.
P
By the foregoing, we have that α<λ fα = gλ . Since G is an open cover, all values of
gλ are strictly positive. It follows that the collection F = {fα /gλ : α < λ} is a partition of
unity. Moreover, F is subordinated to G.
39
The following result indicates the rôle of partitions of unity in the theory of pseudometrizable spaces.
2 Theorem A space is pseudometrizable iff the space has the weak topology induced by
some partition of unity.
Proof. Necessity. Let (X, d) be a pseudometric space. By Theorem 1 there exists, for
every n ∈ N, a partition of unity {fi : i ∈ In } of X subordinated to the open cover
S
{Bd (x, n1 ) : x ∈ X}. We may assume that In ∩ Ik = ∅ for n 6= k. We set I = n∈N In ,
and we define functions gi , i ∈ I, by setting gi = 2−n fi for i ∈ In . Then the collection
G = {gi : i ∈ I} is a partition of unity of X.
We denote by τ the weak topology induced on X by G. Since G consists of τd -
continuous functions, we have that τ ⊂ τd . To show that τd ⊂ τ , let x ∈ X and > 0.
Choose n ∈ N so that
2
n
< and choose i ∈ In so that fi (x) > 0. Let z ∈ X be such that
Supp(fi ) ⊂ Bd (z, n1 ). Then we have that x ∈ Bd (z, n1 ) and hence that Bd (z, n1 ) ⊂ Bd (x, n2 ).
It follows from the foregoing that we habve x ∈ Supp(gi ) ⊂ Bd (x, ) for the the τ -open set
Supp(gi ). We have shown that τd ⊂ τ .
Sufficiency. We assume that the topology τ of a space X is the weak topology induced by
a partition of unity G = {gi : i ∈ I}. We define a pseudometric d of X by the formula
d(x, y) = sup |gi (x) − gi (y)| .
i∈I
We show that τd = τ . To show that τd ⊂ τ , it suffices to show that, for all x ∈ X and
P
r > 0, we have that Bd (x, r) ∈ ηx (τ ). Let x ∈ X and r > 0. Since i∈I gi (x) = 1, there
P
exists a finite set J ⊂ I such that i∈J gi (x) > 1 − r3 . Set n = |J |. The functions gi , i ∈ J ,
are τ -continuous, and hence there exists V ∈ ηx (τ ) such that we have |gi (z) − gi (x)| <
r
3n
for all z ∈ V and i ∈ J . We show that V ⊂ Bd (x, r). Let v ∈ V . Then we have that
X
gi (v) >
i∈J
and it follows, since
P
X
(gi (x) −
r
3n )
i∈J
i∈I
r
3
r
3
−
r
3
gi (v) = 1, that we have gi (v) <
the foregoing, we have that |gi (v) − gi (x)| <
|gi (v) − gi (x)| <
>1−
2r
3
=1−
2r
3
2r
3
for every i ∈ I r J . By
for every i ∈ I r J . It follows, since
for every i ∈ J , that the inequality d(v, x) = supi∈I |gi (v) − gi (x)| ≤
2r
3
holds. As a consequence, we have that V ⊂ Bd (x, r). Hence Bd (x, r) ∈ ηx (τ ).
The family C = {gi−1 (O) : i ∈ I ja O ⊂◦ R} is a subbase of the topology τ . To prove
the inclusion τ ⊂ τd , it suffices to show that C ⊂ τd . Let i ∈ I, O ⊂◦ R and x ∈ gi−1 (O).
40
Since gi (x) ∈ O ⊂◦ R, there exists r > 0 such that (g(x) − r, g(x) + r) ⊂ O. Now we
have that Bd (x, r) ⊂ gi−1 (O), because if d(y, x) < r, then |gi (y) − gi (x)| < r and hence
gi (y) ∈ (g(x) − r, g(x) + r) ⊂ O, which implies that y ∈ gi−1 (O). By the foregoing, we have
that gi−1 (O) ∈ τd .
We know (from Topology II) that a metric space (X, d) can be isometrically imbedded
in the Banach space `∞
X (this is the linear space of all bounded functions X → R, equipped
with the supremum-norm). We can obtain an isometrism as follows. We fix a point x0 of
X, and for x ∈ X, we define fx : X → R be the formula fx (y) = d(y, x) − d(y, x0 ). Then
x 7→ fx is the required isometrism X → `∞
X.
As we mentioned in Example II.4.19, the Banach space `∞
X is isometric with the
linear space C(βD) (equipped with the supremum-norm), where D is a discrete space with
|D| = |X|. However, if we note that the functions fx above are continuous on X, then
we can embed X isometrically into a smaller Banach space: every fx can be extended to
continuous function f¯x on βX, and the mapping x 7→ f¯x is an isometrism X → C(βX).
We can use Theorem 2 to show that every metrizable space X can be topologically
embedded in a Banach space, which is in many ways much simpler than C(βX). Before
we define these Banach spaces, we extend the sum notation to cases where some of the
P
summands may be negative. Let ri ∈ R, for i ∈ I, satisfy
i∈I |ri | < ∞. We set
I+ = {i ∈ I : ri ≥ 0} and I− = {i ∈ I : ri < 0}, and we define
X
X
X
ri =
ri −
−ri .
i∈I
i∈I+
i∈I−
This generalized sum fulfills the usual rules for addition and gives a finite sum the same
value as the ordinary sum.
Let A be a set. We define
n
o
X
`2A = (xα )α∈A ∈ RA :
x2α < ∞ .
α∈A
By the Schwarz Inequality, we have, for all (xα ), (yα ) ∈ `2A , that
X
2
X
X
|xα yα | ≤
x2α ·
yα2
α∈E
for every finite E ⊂ A. Hence we have that
α∈E
P
α∈A
α∈E
|xα yα | < ∞. As a consequence, we can
define a function (, . , ) : `2A × `2A → R by the formula
X
(xα ), (yα ) =
xα y α .
α∈A
41
By the foregoing, we have for all (xα ), (yα ) ∈ `2A that
X
(xα + yα )2 =
α∈A
X
(x2α + 2xα yα + yα2 ) =
α∈A
X
α∈A
X 2
x2α + 2 (xα ), (yα ) +
yα < ∞ .
α∈A
Hence we can make `2A into a linear space by equipping it with pointwise operations:
(xα ) + (yα ) = (xα + yα ) and
r(xα ) = (rxα ) .
It is easy to see that the function (· , ·) is an inner product of the linear space `2A (i.e., it is
symmetric, bilinear and satisfies the condition (x, x) > 0 for each x 6= 0̄). In this situation
we can define a norm for the linear space `2A by the formula
||(xα )|| =
in other words, by the formula ||(xα )||2 =
p
((xα ), (xα )) ,
P
α∈A
x2α .
The foregoing gives us the following equation between the norm and the inner product:
||(xα ) + (yα )||2 = ||(xα )||2 + 2 (xα ), (yα ) + ||(yα )||2 .
The inner product space `2A is a generalized Hilbert space; it is quite easy to show that
the norm metric of `2A is complete; hence `2A , as a normed space, is a Banach space. The
unit sphere of the space `2A is the subset S`2A = {x̄ ∈ `2A : ||x̄|| = 1}.
3 Theorem let X be a metrizable space. Then there exists a set A and an embedding
X → S`2A .
Proof. It follows from Theorem 2 that X has a partition of unity F = {fα : α ∈ A} such
that the weak topology induced by F coincides with the topology of X. We can define a
p
mapping ϕ : X → S`2A by the formula ϕ(x)α = fα (x). We show that ϕ is an embedding.
Since X is T1 and F induces the topology of X, we see that ϕ is one-to-one. To show
that ϕ is continuous, let xn → x in X. We show that ϕ(xn ) → ϕ(x), in other words, that
√
||ϕ(xn ) − ϕ(x)|| → 0. For all a, b ≥ 0 we have that ab ≤ 21 (a + b), and it follows that we
have, for every n ∈ N that
Xp
X
1 X
fα (xn )fα (x) ≤
fα (xn ) +
fα (x) = 1 < ∞
2
α∈A
α∈A
and hence that
42
α∈A
||ϕ(xn ) − ϕ(x)||2 =
X
fα (xn ) +
α∈A
X
α∈A
2
X p
p
fα (xn ) − fα (x) =
α∈A
Xp
X p
fα (xn )fα (x) .
fα (x) −
2 fα (xn )fα (x) = 2 − 2
α∈A
α∈A
As a consequence, to show that ϕ(xn ) → ϕ(x), it suffices to show that
P
α∈A
p
fα (xn )fα (x) →
1 when n → ∞. Let > 0. Then there exists a finite set ∅ 6= B ⊂ A such that
P
α∈B fα (x) > 1 − 2 and fα (x) > 0 for every α ∈ B. We denote by m the number |B|
and by δ the minimum of the numbers
2m
and fα (x), α ∈ B. As the functions fα , α ∈ B,
are continuous, there exists k ∈ N such that fα (xn ) > fα (x) − δ 2 for all n ≥ k and α ∈ B.
Now we have, for all n ≥ k and α ∈ B, that
q
q
q
2
fα (xn )fα (x) >
fα (x) − δ fα (x) ≥ fα (x)2 − δ 2 ≥ fα (x) − δ ≥ fα (x) −
2m
.
By the foregoing, we have, for every n ≥ k, that
Xp
Xp
X
fα (xn )fα (x) ≥
fα (xn )fα (x) >
fα (x) −
α∈A
α∈B
=
X
i∈B
2m
α∈B
fα (x) −
2
≥1−
2
−
2
= 1− .
p
P
fα (x` )fα (x) ≤ 1, we have shown that
Since we have, for every ` ∈ N, that
α∈A
p
P
fα (xn )fα (x) → 1 and hence that ϕ(xn ) → ϕ(x).
α∈A
Finally, let us note that also the mapping ϕ−1 : ϕ(X) → X is continuous: if x and
xn , n ∈ N, are points of X such that ϕ(xn ) → ϕ(x), then we have for every α ∈ A that
ϕ(xn )i → ϕ(x)i , i.e., that fα (xn ) → fα (x), and it follows, since X has the weak topology
induced by {fα : α ∈ A}, that we have xn → x.
The proofs of Theorems 2 and 3 show that, if a T1 -space X has the weak topology
induced by a partition of unity {fα : α ∈ A}, then the mapping x 7→ hfα (x)iα∈A gives an
p
embedding of X in `∞
A and the mapping x 7→ h fα (x)iα∈A gives an embedding of X in
`2A . We mention (without proof) that the first mapping also gives an embedding of X into
(the unit sphere of) the Banach space `1A , where `1A is the linear space of all “summable”
P
functions A → R, equipped with norm ||h|| = α∈A |h(α)|.
We have seen above that pseudometrizability of a space can be characterized in terms
of the existence of a certain partition of unity. Next we note that also full normality of a
space can be characterized in terms of the existence of partitions of unity.
43
4 Theorem A space is fully normal iff every open cover of the space has a subordinated
partition of unity.
Proof. Necessity. Let X be fully normal space, and let U be an open cover of X. By
Theorem 1.8, there exists a continuous pseudometric d of X and a τd -open cover V which
refines U . By Theorem 1, the pseudometric space (X, d) has a partition of unity F which
is subordinated to V. Since d is continuous on X, the collection F is also a partition of
unity of the space X. Moreover, since V refines U , the partition of unity F is subordinate
to U .
Sufficiency. Assume that every open cover of X has a subordinated partition of unity.
To show that X is fully normal, let U be an open cover of X. Let F be a partition of
unity of X which is subordinate to U . By Theorem 2, there exists a pseudometric d of
X such that the weak topology induced on X by F coincides with τd . Since every f ∈ F
is continuous on X, the weak topology induced by F is coarser than the topology of X.
As a consequence, d is a continuous pseudometric of X. Moreover, the τd -open cover
{Supp(f ) : f ∈ F } of X refines U .
We have shown that every open cover of X has a τd -open refinement for some conti-
nuous pseudometric of X. By Theorem 1.8, the space X is fully normal.
Before we turn to consider the next topic of “continuous selections”, we observe that
partitions of unity have certain properties of locally finitely supported families which can
be used to construct partitions of unity which actually are locally finitely supported.
5 Lemma Let {fα : α ∈ A} be a partition of unity of X. For all x ∈ X and > 0, there
P
exists V ∈ ηx and a finite E ⊂ A such that α∈ArE fα (z) < for every z ∈ V .
Proof. Exercise.
6 Lemma Let {fα : α ∈ A} be a partition of unity of X and let B ⊂ A. Then the
P
functions α∈B fα and supα∈B fα are continuous.
Proof. Exercise.
7 Proposition Let {fα : α ∈ A} be a partition of unity of X. Then there exists a
partition of unity {gα : α ∈ A} of X such that we have Supp(gα ) ⊂ Supp(fα ) for every
α ∈ A and the family {Supp(gα ) : α ∈ A} is locally finite (as an indexed family).
Proof. By Lemma 6, the function h = supα∈A fα is continuous. It follows that, for
every α ∈ A, the set Uα = {x ∈ X : fα (x) >
44
1
2 h(x)}
is open. Note that the family
U = {Uα : α ∈ A} covers X. We show that U is locally finite. Let x ∈ X. Then there
exists αx ∈ A such that fαx (x) > 43 h(x). Denote by V the nbhd {z ∈ X : fαx (z) > 34 h(x)}.
By Lemma 5, there exists W ∈ ηx and a finite B ⊂ A such that we have fα (z) < 41 h(x)
for all z ∈ W and α ∈ A r B. We show that {α ∈ A : Uα ∩ V ∩ W 6= ∅} ⊂ B. Assume
on the contrary that there exist α ∈ A r B and z ∈ Uα ∩ V ∩ W . Then we have that
fα (z) >
1
4 h(x)
1
2 h(z),
fαx (z) >
> fα (z) >
1
2 h(z)
3
4 h(x)
and fα (z) <
and h(z) ≥ fαx (z) >
1
4 h(x).
3
4 h(x),
As a consequence, we have that
but this is a a contradiction. It
follows from the foregoing that U is locally finite (as an indexed family).
For every α ∈ A, the function kα = 0 ∨ (fα − 21 h) is continuous and Supp(kα ) = Uα .
Hence the collection {kα : α ∈ A} is locally finitely supported, and it follows that the
P
function k = α∈A kα is continuous. If we set gα = kα /k for each α ∈ A, then we get a
partition of unity {gα : α ∈ A} with the required properties.
Next we shall use partitions of unity to prove a fundamental result in the “theory of
selections”. We need some definitions before can state the result.
Let ϕ be a mapping from a space X into the family of all non-empty subsets of a
space Y . We say that ϕ is a carrier. A selection for the carrier ϕ is a mapping f : X → Y
such that we have f (x) ∈ ϕ(x) for every x ∈ X; if f is continuous, then we say that f is a
continuous selection for ϕ.
8 Example Extensions of mappings can be considered as selections. Let A ⊂ X and
g : A → Y . If we define ϕ : X → P(Y ) by setting
g{x} , for x ∈ A
ϕ(x) =
Y
, for x ∈ X r A,
then a (continuous) selection for ϕ is a (continuous) extension of g.
Let ϕ : X → P(Y ) be a carrier. We say that ϕ is lower semi-continuous (lsc) provided
that, for every G ⊂◦ Y , we have that {x ∈ X : ϕ(x) ∩ G 6= ∅} ⊂◦ X.
9 Examples (a) To every mapping f : X → Y we can associate the carrier x 7→ f {x}.
We denote also this carrier by f . The carrier f is lsc iff we have f −1 (G) ⊂◦ X for each
G ⊂◦ Y , in other words, iff the mapping f is continuous.
(b) Let f be an open and continuous mapping X → Y . Then the formula ϕ(x) = f −1 f {x}
defines an lsc carrier ϕ : X → P(X). To see this, let G ⊂◦ X. Since f is open and
continuous, we have that f −1 (f (G)) ⊂◦ X. Moreover, we have that f −1 (f (G)) = {x ∈ X :
ϕ(x) ∩ G 6= ∅}.
45
For the proof of the basic selection theorem, we need two auxiliary results. The first
of these is a technical result, but the second one is already a “near selection theorem”.
10 Lemma Let ϕ and ψ be lsc carriers X → P(Y ), let d be a continuous pseudometric
of Y and let r > 0 be such that we have d(ϕ(x), ψ(x)) < r for every x ∈ X. Then the
formula θ(x) = Bd (ϕ(x), r) ∩ ψ(x) defines an lsc carrier θ : X → P(Y ).
Proof. To show that θ is lsc, let G ⊂◦ Y and let x be a point of the set {z ∈ X : θ(z) ∩ G 6=
∅}. Let u ∈ θ(x) ∩ G. Then u ∈ Bd (ϕ(x), r) and hence there exists δ > 0 such that
ϕ(x) ∩ Bd (u, r − δ) 6= ∅. Since ϕ is lsc, the set V = {z ∈ X : ϕ(z) ∩ Bd (u, r − δ) 6= ∅}
is a nbhd of x. We also have that u ∈ ψ(x), and it follows, since ψ is lsc, that the set
W = {z ∈ X : ψ(z) ∩ G ∩ Bd (u, δ)} is a nbhd of x. We show that V ∩ W ⊂ {z ∈
X : θ(z) ∩ G 6= ∅}. Let z ∈ V ∩ W . Since z ∈ V , there exists v ∈ ψ(z) ∩ G such that
d(v, u) < δ. Since z ∈ W , there exists w ∈ ϕ(z) such that d(w, u) < r − δ. Now we have
that d(v, ϕ(z)) ≤ d(v, w) ≤ d(v, u)+d(u, w) < δ+r−δ = r and hence we have that v ∈ θ(z),
and further, that θ(z) ∩ G 6= ∅. We have shown that V ∩ W ⊂ {z ∈ X : θ(z) ∩ G 6= ∅}. By
the foregoing, we have that {z ∈ X : θ(z) ∩ G 6= ∅} ⊂◦ X.
11 Lemma Let X be a fully normal space, Y a normed linear space and ϕ : X → P(Y )
an lsc carrier such that every ϕ(x) is convex. For every > 0, there exists a continuous
mapping f : X → Y such that we have, for every x ∈ X, that d(f (x), ϕ(x)) < , where d
denotes the norm distance in Y .
Proof. For every y ∈ Y , let Uy = {x ∈ X : ϕ(x) ∩ Bd (y, ) 6= ∅}, and note that we have
Uy ⊂◦ X because ϕ is lsc. By Theorem 4 and Proposition 7, the open cover U = {Uy : y ∈ Y }
of X has a subordinated locally finitely supported partition of unity {gα : α ∈ A}. For
every α ∈ A, let yα ∈ Y be such that Supp(gα ) ⊂ Uyα . Define f : X → Y by setting
f (x) =
X
gα (x)yα .
α∈A
Since {gα : α ∈ A} is locally finitely supported, the function f is continuous. To show that
f has the required property, let x ∈ X. Denote by B the finite set {α ∈ A : gα (x) 6= 0}.
For every α ∈ B, we have that x ∈ Supp(gα ) ⊂ Uyα and hence that d(yα , ϕ(x)) < . As a
P
consequence, the point f (x) = α∈B gα (x)yα of Y is a convex combination of points from
the set Bd (ϕ(x), ). Since the set ϕ(x) is convex, also the set Bd (ϕ(x), ) is convex. By
the foregoing, we have that f (x) ∈ Bd (ϕ(x), ).
46