§ 28. The shrinking lemma. Partition of unity.
28.1. Definition. Let ϕ : X → R be a continuous function. The open set
Uϕ =
x ∈ X : ϕ(x) 6= 0
will be called the support of the function ϕ and denoted by supp(ϕ).
28.2. Definition. Let u = {Uα : α ∈ A} be an open cover of X. A family of continuous functions ϕα : X → [0, 1], α ∈ A, is called a partition of unity subordinate to the cover
u, if
(1) Cl(supp(ϕα)) ⊂ Uα, α ∈ A;
(2) the family {supp(ϕα) : α ∈ A} is locally finite;
(3) Pϕα(x) = 1 for any x ∈ X (by virtue of condition (2), at each point only finitely many
functions ϕα are nonzero).
28.3. Lemma (shrinking lemma (case of locally finite covers)). Let X be a paracompact Hausdorff space, u = {Uα : α ∈ A} an open cover of X. Then
there exists a locally finite open cover v = {Vα : α ∈ A} such that Cl(Vα) ⊂ Uα,
α ∈ A.
Proof. Let Ω be the family of open subsets O of X such that Cl(O) is contained in some element of u. By virtue of the regularity of X, Ω is an open cover of X.
Let v
0 be a locally finite open cover of X refining Ω. For any V ∈ v
0
fix a Uα such that Cl(V ) ⊂ Uα ∈ u (Cl(V ) may be contained in various Uα, but
one of them is fixed). Set
Vα =
[
{V ∈ v
0
: Cl(V ) ⊂ Uα}, α ∈ A.
If the set Uα is not fixed for any element V ∈ v
0
, then set Vα = ∅.
v = {Vα : α ∈ A} is an open cover of X.
Cl(Vα) ⊂ Uα, α ∈ A, by virtue of the local finiteness of the family v
0
(Proposition 27.3 of Lecture
8). For any point x ∈ X there exists its neighborhood Ox intersecting only finitely
many elements of v
0
. The neighborhood Ox will intersect only those elements of v to which
the sets V ∈ v
0
belong. Hence v is locally finite.
28.4. Proposition. For every open cover u = {Uα : α ∈ A} of a paracompact
Hausdorff space X there exists a partition of unity ϕα : X → [0, 1], α ∈ A, subordinate to the cover u.
Proof. By Lemma 28.3 there exist locally finite covers w = {Wα : α ∈ A}
and v = {Vα : α ∈ A} such that
Cl(Wα) ⊂ Vα ⊂ Cl(Vα) ⊂ Uα, α ∈ A.
By Urysohn's lemma (Lemma 23.1 of Lecture 6) there exist continuous functions ψα : X → [0, 1],
α ∈ A such that
ψα
Cl(Wα)
= 1, ψα(X \ Vα) = 0.
P
By virtue of the local finiteness of the cover v, a continuous function ψ(x) =
{ψα(x) : α ∈ A} is well defined. Since w is a cover of X, ψ(x) > 0 for any point x ∈ X. Then
the family of functions
ϕα(x) = ψα(x)
ψ(x)
, α ∈ A,
will be the required partition of unity.
28.5. Remark. A partition of unity subordinate to an arbitrary finite
open cover of a normal space is defined.
28.6. Example. Let X be a paracompact Hausdorff space, C a locally finite family of subsets of X, εC , C ∈ C, a family of positive numbers. Then there exists a
positive continuous function f : X → R such that f(x) ≤ εC for x ∈ C.
50
Consider an open cover u = {Uα : α ∈ A} of the space X each element of which
intersects no more than finitely many elements of C. Let ϕα : X → [0, 1], α ∈ A, be
a partition of unity subordinate to the cover u. Set for α ∈ A
δα = min{εC : C ∩ Cl(supp(ϕα)) 6= ∅, C ∈ C}.
If the set {C ∈ C : C∩Cl(supp(ϕα)) 6= ∅} is empty, then set δα = 1. A continuous function f(x) = P{δαϕα(x) : α ∈ A} is well defined. Fix C ∈ C. To complete
the proof it suffices to show that for any α ∈ A
δαϕα(x) ≤ εC ϕα(x), for x ∈ C. (28.1)
For x 6∈ Cl(supp(ϕα)) inequality (28.1) is obvious (ϕα(x) = 0). If x ∈ Cl(supp(ϕα)), then δα ≤ εC
by construction.
§ 29. Compact spaces.
29.1. Definition. A family of subsets of a set X is called centered,
if the intersection of any finite number of its elements is nonempty.
29.2. Theorem. A space X is compact if and only if every centered system of closed subsets of X has nonempty intersection.
Proof. Let X be a compact space and Φ a centered system of its
closed subsets. If the intersection of all elements of Φ is empty, then the set {X\F : F ∈ Φ}
is an open cover of the space X. From it one can extract a finite subcover
{X \ Fi
, i = 1, . . . , k}. Then F1 ∩ . . . ∩ Fk = ∅, which contradicts the centeredness of the family Φ.
Conversely, suppose that in the space X every centered system of closed sets has
nonempty intersection. Take an arbitrary open cover u of the space X. Suppose that the cover u does not contain a finite subcover. Then the family
{X \ (U1 ∪ . . . ∪ Uk) : Ui ∈ u}
is centered and has empty intersection. This contradiction
completes the proof.
29.3. Proposition. If Y is a compact subspace of a Hausdorff space
X, then Y is closed in X.
Proof. Take an arbitrary point x ∈ X \Y . For any point y ∈ Y there exist disjoint neighborhoods Oy and Oyx of the points y and x respectively. From the family {Oy : y ∈ Y },
which covers Y , one can select a finite subfamily {Oy1, . . . , Oyk} covering Y . Then OY =
Sk
i=1 Oyi and Ox =
Tk
i=1 Oyix are disjoint neighborhoods of the subset Y
and the point x respectively.
Thus x is an interior point of the set X \ Y , which is open. Hence Y
is closed.
29.4. Proposition. The continuous image of a compact space is compact.
Proof. Let X be compact and f : X → Y a continuous surjective map. Let us show that Y is compact. Let u be an open cover of Y. By virtue of the continuity of f,
the family f
−1
(u) =
f
−1
(U) : U ∈ u
is an open cover of the space X. Let us choose
in it a finite subcover f
−1
(U1), . . . , f −1
(Uk). Then the family {U1, . . . , Uk} will be a finite
subcover of u.
29.5. Proposition. A continuous bijective map f of a compact space X onto a Hausdorff space Y is a homeomorphism.
Proof. To prove the continuity of the map f
−1 it suffices to show
that the image of an arbitrary closed subset F of X is closed in Y.
By Proposition 27.11 of Lecture 8 the subset F is compact in X. From Proposition 29.4
the compactness of the set f(F) follows, and from Proposition 29.3 the closedness of the set
f(F) follows.
§ 30. Alexander's lemma. The First Tychonoff Theorem.
Denote the family of open covers of X by cov(X).
30.1. Alexander's Lemma. If in the space X there exists a subbase B such that
from any cover of the space X by its elements one can extract a finite subcover,
then X is compact.
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Proof. Suppose that there exists a cover u0 ∈ cov(X) from which one cannot
extract a finite subcover. Set
U = {u ∈ cov(X) : u0 ⊂ u, u contains no finite u
0 ∈ cov(X)
, (30.1)
considering all elements of the covers under consideration to be pairwise distinct. The set U is ordered by the inclusion relation (u1 ≤ u2, if u1 ⊂ u2).
Let U0 ⊂ U be a linearly ordered subset. If U0 has a maximal element,
then it is an upper bound of the set U0 in U. If U0 does not have a maximal element,
then set
v =
[
u∈U0
u. (30.2)
Take an arbitrary finite set v
0 = {V1, . . . , Vn} ⊂ v. From (30.2) it follows that each Vi ∈ v
0 is an element of some ui ∈ U0. Hence, v
0 ⊂ u1 ∪ . . . ∪ un ∈ U0 ⊂ U.
From definition (30.1) it follows that v
0 is not a cover. Hence, v ∈ U is an upper bound of the set U0 in U. Thus, the set U satisfies the conditions of the Kuratowski–Zorn
Lemma (§ 7 of Lecture 1) and, consequently, has a maximal element v0.
Let us show that v0 ∩ B ∈ cov(X) (i.e. v0 contains a subcover consisting of elements of the subbase B).
Suppose the contrary. Then there exists a point x such that no set from B containing it belongs to v0.
For the point x ∈ X there exists an element V of the cover v0 containing it. There exist
elements G1, . . . , Gn of the subbase B such that
x ∈ G1 ∩ . . . ∩ Gn ⊂ V. (30.3)
By virtue of the maximality of the family v0 in U,
v0 ∪ {Gi} ∈ U / , i = 1, . . . , n.
Hence, v0 ∪ {Gi} contains a finite subcover. Consequently, for each i = 1, . . . , n
there exist sets
V
i
1
, . . . , V i
j(i) ∈ v0,
such that
Gi ∪
j
[
(i)
k=1
V
i
k
= X. (30.4)
From (30.4) it follows
\n
i=1
Gi
∪
[
i,k
V
i
k
= X. (30.5)
Hence, by virtue of (30.3) and (30.5), the family v0 contains a finite subcover {V, V i
k
: i =
1, . . . , n, k = 1, . . . , j(i)}. This contradiction shows that v0 ∩ B ∈ cov(X). From this, by the property
of the subbase B, by Alexander's lemma a finite subcover can be extracted. All the more, a finite
subcover can be extracted from the cover v0. This contradiction concludes the proof.
30.2. The First Tychonoff Theorem. The product of compact spaces is compact.
Proof. Let X =
Q
α∈A Xα. By Alexander's Lemma it suffices to show
that from a cover u of the space X, consisting of sets of the form pr−1
α (V ), where V is open in Xα,
α ∈ A, (a subbase of the Tychonoff topology on the product) one can extract a finite subcover. For this it suffices to find an index α
0 and a cover v ∈ cov(Xα0 ), such that pr−1
α0 (V ) ∈ u,
V ∈ v. Indeed, then pr−1
α0 (v) = {pr−1
α0 (V ) : V ∈ v} is a subcover of u. By virtue of the compactness of Xα0 , from the cover v one can extract a finite subcover. Thus, from the subcover
pr−1
α0 (v) (and hence from the cover u) one can extract a finite subcover.
Suppose that no such α
0 exists. Then for every α ∈ A there is a point xα ∈ Xα,
such that xα 6∈ V for any element of the cover u of the form pr−1
α (V ), where V is an open subset
of Xα. Hence the point x = (xα) does not lie in any element of the cover u. Indeed,
if x ∈ pr−1
β
(V ) ∈ u, where V is open in Xβ, then prβ(x) = xβ ∈ V . This contradiction
concludes the proof.
30.3. Theorem. A subset X of Euclidean space R
n is compact if and only
if it is closed and bounded.
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Proof. Necessity. The closedness of X follows from Proposition 29.3. If
X were unbounded, then from the cover of X by open disks Dn
m, m ∈ N, of radius m centered
at the origin, one could not extract a finite subcover.
Sufficiency. Let X be closed and bounded. From the boundedness of X follows the existence of a closed ball Bn
m of radius m centered at the origin, such that X ⊂ Bn
m. But
Bn
m lies in the product of segments [−m, m]
n, which is compact by the Tychonoff Theorem. Then
X is compact as a closed subset of the compact cube [−m, m]
n (see Proposition 27.11
of Lecture 8).
30.4. Example. The sphere S
n−1 = {(x1, . . . , xn) ∈ R
n : x
2
1 + . . . + x
2
n = 1} and the closed ball
Bn = {(x1, . . . , xn) ∈ R
n : x
2
1 + . . . + x
2
n ≤ 1} are compact spaces.
From Proposition 29.4 and Theorem 30.3 it follows
30.5. Corollary. Every continuous function on a compact space is bounded and
attains its greatest and least values.
§ 31. Compactifications. The Stone-Cech compactification.
31.1. Definition. A compactification of a space X is a pair (Y, i), where Y is a
compact Hausdorff space, i : X → Y an embedding such that Cl(i(X)) = Y . Here
two compactifications (Y1, i1) and (Y2, i2) of the space X are equivalent, if there exists
a homeomorphism h : Y1 → Y2 such that h ◦ i1 = i2.
From Proposition 25.2 and Theorem 25.4 of Lecture 7 it follows that a space X has a compactification if and only if it is Tychonoff. If it is clear which embedding
of the space into a compact Hausdorff space is meant, then in the notation for the compactification the symbol for the embedding will be omitted.
31.2. Examples. 1. The circle S
1 is a compactification of the interval (0, 1) (the embedding
t → (cos(2πt),sin(2πt)).
2. The segment [0, 1] is a compactification of the interval (0, 1).
3. The compactum X ⊂ R
2
(called the “sin 1
x
” space), being the union of the vertical segment
X1 = {(0, y) : −1 ≤ y ≤ 1} and the graph X2 = {(x,sin 1
x
) : 0 < x ≤ 1} of the function y = sin 1
x
, is
a compactification of the interval (0, 1) (the embedding t → (t,sin 1
t
)).
Assignment N 9
1. Partition of unity. Prove that for every finite open cover u = {U1, . . . , Uk}
of a normal space X there exists a partition of unity {ϕ1, . . . , ϕk}, subordinate to the cover u.
2. Which topologies on the line R from Example 8.5 of Lecture 2 give a compact topology?
Which subsets of the Sorgenfrey line are compact?
Which subsets of the line with the Zariski topology are compact?
Give an example of a T1 space and its non-closed compact subset.
3. Let X be a linearly ordered space with the order topology. Prove that any segment of X is compact if and only if every bounded
subset of X has a least upper bound in X.
4. Prove that the lexicographically ordered square is compact, satisfies the first axiom of countability, and is not separable.
Prove that the subset (0, 1]×{0}∪[0, 1)×{1} of the lexicographically ordered square (the "Alexandrov double arrow" space) is compact, separable, and does not satisfy
the second axiom of countability (and hence is not metrizable).
5. Prove that in a regular space for any disjoint closed and compact subsets there exist disjoint neighborhoods.
6. Let a Hausdorff topology T1 and a compact topology T2 be given on a set X. Show that
if T1 ⊂ T2, then T1 = T2.
Show that any two Hausdorff compact topologies on a set X either coincide, or are not comparable.
53
7 "Tube lemma". In the product X × Y let the factor Y be compact. Prove that
for any neighborhood O of the "slice" {x} × Y , where x is an arbitrary point of X, there exists a neighborhood
Ox of the point x such that Ox × Y ⊂ O.
Give an example showing that the requirement of compactness of the factor Y is essential.
Prove the First Tychonoff Theorem in the case of a finite number of factors, using
the "tube lemma".
8. Prove that a map f : X → Y , where Y is a compact Hausdorff space,
is continuous if and only if the graph of the map is closed.
9. Let X be a Hausdorff space, Kα, α ∈ A, a family of compact subsets of
X, U a neighborhood of T
{Kα : α ∈ A}. Prove that then there exists a finite subset
AF in ⊂ A such that T
{Kα : α ∈ AF in} ⊂ U.
10. Prove that in a Hausdorff compact space the union of a countable number of
closed subsets with empty interior has empty interior.
11. Which of the subsets of the space of matrices Mat(n × n, R) ⊂ R
n
2
are compact:
(a) GL(n, R) = {A ∈ Mat(n × n, R) : det A 6= 0};
(b) SL(n, R) = {A ∈ Mat(n × n, R) : det A = 1};
(c) O(n, R) = {A ∈ Mat(n × n, R) : AAT = E}?
12. Give an example of a metric space and its closed, bounded, non-compact subset.
13. Prove that for any compact subsets A and B of a metric space
(X, ρ) there exist points a ∈ A, b ∈ B such that ρ(a, b) = inf{ρ(x, y) : x ∈ A, y ∈ B}.
14. Will the compactification X of the interval (0, 1) from item 3 of Example 31.2 of Lecture 9 be equivalent to the compactification X0 ⊂ R
2
, which is the union of the vertical segment X1 =
{(0, y) : −1 ≤ y ≤ 1} and the graph X0
2 of the function y = sin 1
1−x
; 0 ≤ x < 1 (the embedding t → (1 − t,sin 1
1−t
))?
15. (The Second Tychonoff Theorem). The weight of a topological space is called the least of the cardinalities of its bases. Prove that a Tychonoff space X has a compactification
bX, whose weight equals the weight of X.
Additional problems of Assignment N 9
16 "Compact hedgehog". Let Λ be some infinite set. Let us assign to
each element λ ∈ Λ a segment [0, 1], which we denote by [0, 1]λ and assume that all
these segments are pairwise disjoint except for the point 0, which is assumed
to belong to all segments. Set X = ∪{[0, 1]λ : λ ∈ Λ} and define a topology: on
the half-intervals (0, 1]λ, λ ∈ Λ, the ordinary interval topology, basic neighborhoods of {0}
being all possible finite unions of half-intervals [0, a(λ))λ, 0 < a(λ), λ ∈ ΛF in ⊂
Λ, and all segments [0, 1]λ, λ ∈ Λ \ ΛF in. Show that the topology is well defined, and that the compact
hedgehog is compact.
"Hedgehog". Let Λ be an infinite set. Let us assign to each element λ ∈ Λ
a segment [0, 1], which we denote by [0, 1]λ. On the sum ⊕
λ∈Λ
[0, 1]λ let us define a partition R into the one-point subsets ⊕
λ∈Λ
(0, 1]λ and the set ⊕
λ∈Λ
{0}λ. The quotient space of the sum ⊕
λ∈Λ
[0, 1]λ
by the partition R we call the "hedgehog".
Prove that the "hedgehog", the "compact hedgehog" and the "metrizable hedgehog" (Problem 21 of Assignment 2) are pairwise non-
homeomorphic.
17. Prove that a nonempty Hausdorff compact space X without isolated points is uncountable.
Prove that |X| ≥ c.
18. Prove that any two norms on R
n are equivalent (define the same topology).
19. Prove that any map f : X → Y of a Tychonoff space X into a compact
space Y extends to a map of its Stone-Cech compactification βX.
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Establish the uniqueness (up to equivalence) of the Stone-Cech compactification βX of a Tychonoff space X.
20. For compactifications (Y1, i1) and (Y2, i2) we say Y1 ≤ Y2, if there exists a map
f : Y2 → Y1 such that f ◦ i2 = i1. Prove that f(Y2 \ i2(X)) = (Y1 \ i2(X)).
Prove that ≤ is an ordering on the set of compactifications, for which βX is the greatest element.
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