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Proyecto Fin de Máster en Investigación Matemática
Facultad de Ciencias Matemáticas
Universidad Complutense de Madrid
Bornologies in Metric Spaces
Ana Soledad Meroño Moreno
Directora Prof. Mª Isabel Garrido Carballo
2009-2010
Abstract. In the same way that the Samuel Compactication, it
can be constructed the, also called, Samuel Realcompactication, but
taking all the real-valued uniformly continuous functions, not necessarily bounded. To construct the Samuel Realcompactication of a metric
space (X, d) we study the case in which all the real-valued uniformly
continuous functions f ∈ U C(X) are bounded in the closed metric balls
of (X, d). In particular, this functions are bounded in all the bounded
subsets of (X, d). Taking these metric spaces, we have that the Samuel
Realcompactication is the quotient of a subspace of the Stone-ƒech
compactication of X , which can be written as the (countable) union
of the closure in βX of the closed metric balls in (X, d) of radius n ∈ N
and center in a xed point x0 ∈ X . This problem of characterization
of the Samuel Realcompactication, for these type of metric spaces, is
equivalent to solve the following problem. Let B a subset of (X, d),
then it is said bounded in the sense of Bourbaki if and only if every
real-valued uniformly continuous function f ∈ U C(X) is bounded for
it. These types of subsets form what is called a bornology for the metric
space (X, d), that is, they form a cover of X stable by inclusions and
nite unions. We have that given two uniformly continuous metrics, d
and d0 , the bounded sets in sense of Bourbaki for (X, d) and (X, d0 )
are the same. If we solve the problem of nding a uniformly equivalent metric d0 for (X, d) such that all the bounded sets in the sense
on Bourbaki of (X, d) concide with the bounded metric sets of (X, d0 ),
then we have that all the real-valued uniformly continuous functions
f ∈ U C(X) are bounded for the bounded metric sets of (X, d0 ). In
this way (X, d0 ) has the Samuel Realcompactication developed before
and the Samuel Realcompactication of (X, d) and (X, d0 ) is the same
because both metric spaces have the same uniformity.
1
Resumen. Similarmente a la Compacticación de Samuel puede
construirse la Realcompacticación, también llamada de Samuel, pero
tomando todas las funciones reales uniformemente continuas no necesariamente acotadas. Para construir la Realcompacticación de Samuel
de un espacio métrico (X, d) tomamos el caso particular en el que todas
las funciones reales uniformemente continuas f ∈ U C(X) son acotadas
para las bolas métricas cerradas en (X, d). En particular estas funciones son acotadas para todos los subconjuntos acotados de (X, d).
Tomando estos espacios métricos tenemos que la Realcompacticación
de Samuel es el cociente de un subespacio de βX , la compacticación
de Stone-ƒech de X , que puede escribirse como la unión (numerable)
de las clausuras en βX de las bolas cerradas de radio n ∈ N y centro
un punto jo x0 ∈ X . Este problema de caracterización de la Realcompacticación de Samuel para éste tipo de espacios métricos, es
equivalente a resolver el siguiente problema. Sea B un subconjunto de
(X, d), entonces éste se dice que es acotado en el sentido de Bourbaki
si y sólo si toda función real uniformemente continua f ∈ U C(X) es
acotada para B . Éste tipo de subconjuntos forman lo que se llama
una bornología para el espacio (X, d) i. e. forman un recubrimiento
de X estable para las inclusiones y las uniones nitas. Además, dadas
dos métricas uniformemente equivalentes d y d0 los acotados en el sentido de Bourbaki para (X, d) y para (X, d0 ) coinciden. Si resolvemos el
problema de encontrar una métrica uniformemente equivalente d0 para
(X, d) tal que los subconjuntos acotados en el sentido de Bourbaki de
(X, d) coinciden con los acotados métricos de (X, d0 ), entonces tenemos
que todas las las funciones reales uniformemente continuas f ∈ U C(X)
son acotadas para los acotados métricos de (X, d0 ). De esta manera
(X, d0 ) tiene la Realcompacticación de Samuel descrita anteriormente
y la Realcompacticación de Samuel de (X, d0 ) y (X, d) es la misma ya
que los dos espacios métricos tienen la misma uniformidad.
2
Key words: Metric spaces, Bornologies, Bounded sets, Uniformly
continuous, Realcompactication, Bourbaki bounded, Totally bounded,
Samuel compactication, Heine-Borel property.
Palabras clave: Espacios métricos, Bornologías, Conjuntos acotados, Uniformemente continuo, Realcompacticación, Bourbaki acotado, Compacticación de Samuel, Propiedad de Heine-Borel.
MSC2000:
- 54C35 - Metric spaces - Metrizability;
- 54C30 - Real-valued functions;
- 54D60 - Realcompactness and realcompactication.
3
La abajo rmante, Ana Soledad Meroño Moreno, matriculada en el
Máster en Investigación Matemática de la Facultad de Ciencias Matemáticas, autoriza a la Universidad Complutense de Madrid (UCM) a difundir y utilizar con nes académicos no comerciales y mencionando expresamente a su autor el presente Trabajo Fin de Master:"Bornologies
in Metric Spaces", realizado durante el curso académico 2009-2010 bajo
la dirección de M Isabel Garrido Carballo en el Departamento de Geometría y Topología, y a la Biblioteca de la UCM a depositarlo en el
Archivo Institucional E-Prints Complutense con el objeto de incrementar la difusión, uso e impacto del trabajo en Internet y garantizar su
preservación y acceso a largo plazo.
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Contents
1 Introduction.
6
2 Some basic notions about bornologies.
8
2.1 Some examples of bornologies. . . . . . . . . . . . . . . 8
2.2 Metrization of bornological universes. . . . . . . . . . . 10
2.3 The Heine-Borel property. . . . . . . . . . . . . . . . . 15
3 Bornologies in uniform spaces.
3.1 Totally bounded sets and bounded sets in the sense of
Bourbaki. . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Characterization of the bounded sets in the sense of
Bourbaki. . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Uniform metrization of bornologies. . . . . . . . . . . .
3.4 The Heine-Borel property for uniformly equivalent metrics.
18
18
21
24
27
4 Realcompactication v.s compactication.
30
5 Samuel Realcompactication of a metric space.
45
4.1 Compactication of a topological space. . . . . . . . . . 32
4.2 Realcompactication of a topological space. . . . . . . . 35
4.3 Realcompactications as sets of lattice homomorphisms. 39
5.1 Samuel Realcompactication and Compactication. . . . 45
5.2 Conclusion. . . . . . . . . . . . . . . . . . . . . . . . . 48
5
Chapter 1
Introduction.
The notion of bounded set has been very important in the theory of
metric spaces. So that, new notions of bounded sets have been dened
in other types of spaces as vector topological spaces, uniform spaces
and more generally, topological spaces. In this way we can see that
there exists new notions of bounded sets, very dierent between them
which can be viewed as an extension of the classic notion of bounded
set. All of them have the common property of forming a bornology on
the considered space X . We say that a family of subsets of a set X is
a bornology of X if it is a cover of X and if it is stable under inclusions
and nite unions. If we take a topological space X and a bornology B
on X , then the pair (X, B) is called a bornological universe.
A rst interesting question raised in this report, is when a bornological universe (X, d) is metrizable or uniformly metrizable, that is , when
there exists an admissible metric d such that d denes the same topology or uniformity of X and the metric bounded sets for (X, d) coincide
with the bounded sets of the bornology B.
Second, as an application of these results, the theorem have been
applied to the compact bornology for a T2 topological space X , that is,
the bornology of all the subsets in X such that their closure is compact.
As an answer of this application we have a way to nd admissible proper
metrics for the topology or the uniformity of X , that is, metrics with
the Heine-Borel property.
6
Finally we can applied the theorem of uniform metrization to a metric space (X, d) with the bornology of the bounded sets in the sense
of Bourbaki, that is, the subsets of X which are bounded for every
real-valued uniformly continuous function f ∈ U C(X). Thanks to this
result we have a useful characterization of the Samuel Realcompactication of the metric space (X, d). Recall that the Samuel Realcompactication of a metric space is dened as the smallest realcompactication of (X, d) such that every real-valued uniformly continuous function
f ∈ U C(X) can be extended to a real-valued continuous function on
the realcompactication.
After all these months of work, I want to thank Prof. M Isabel
Garrido Carballo who has helped me to the understanding of this interesting theme and to the hard redaction of this report.
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Chapter 2
Some basic notions about
bornologies.
2.1 Some examples of bornologies.
In a metric space (X, d) we have the classical notion of metric
bounded set, that is, a subset B is bounded if there is a positive number
K > 0 such that for all x, y ∈ B , d(x, y) ≤ K . The collection of all the
metric bounded sets in (X, d) has the following properties: they form
a cover of X , every subset of a bounded set is bounded, and the nite
union of bounded sets is also bounded. In general, for a non empty
set X , if we have a collection of subsets of X with the same properties
than the metric bounded sets of a metric space, we have what is called
a bornology on X .
Denition 1. Let X be a non empty set, a bornology on it is a non
empty collection of subsets {Bi} of X such that:
1.
S
Bi = X ;
2. every subset of a bounded set is bounded;
3. nite unions of bounded sets are bounded.
Remark 1. It follows immediately that:
8
• ∅ is a bounded set;
• every x ∈ X is a bounded set;
• nite or not intersections of bounded sets are bounded sets.
With this denition we have an axiomatic extension of the concept
of bounded subset on a generic set X . Obviously the metric bounded
sets of a metric space (X, d) form a bornology on X called trivially the
metric bornology for d. We are going to denote it by Bd(X).
Remark 2. Let B and B0 be two bornologies on a set X . Then, B and
B0 coincide, that is, B = B0 i for every B ∈ B there exists B 0 ∈ B0
such that B ⊆ B 0 and for every B 0 ∈ B0 there exists B ∈ B such that
B0 ⊆ B.
Example 1. For a non empty set X the collection of all the subsets in
X , P(X), is a bornology on X called the trivial bornology. Obviously
if we have a bornology B on a set X and X ∈ B, that is, X is bounded,
then B is the trivial bornology.
Example 2. For a metric space (X, d) the metric bornology for d coincides with the trivial bornology i d is a bounded metric. If d is not
a bounded metric then the trivial bornology conicides with the metric
bornology for d∗, where d∗ = min{1, d} is a bounded metric uniform
equivalent to d.
Example 3. Let F be the collection of all the nite subsets on X .
Then it is a bornology on X called the nite bornology.
Example 4. We can have dierent bornologies generated in the next
way. Let A be a family of subsets in a set X such that it forms a
cover of X and let B be the family of subsets of X which consists of
the totality of the subsets of the nite unions of the family A. Then B
is the weakest bornology generated by the cover A.
Example 5. As in metric spaces it should be natural in topological
spaces that the compact sets be bounded. Let A be the collection of all
9
the compact sets of a topological space X , then it generates a bornology
as in the previous example. This is the compact bornology denoted
by CB. Recall that B ∈ CB i B is compact.
2.2 Metrization of bornological universes.
Example 6. Let CBd(X) denote the compact bornology on a metric
space (X, d). It coincides with the metric bornology of d i (X, d) has
the Heine-Borel Property that is every closed bounded set in (X, d)
is compact. In this case d is said to be a proper metric. If X is a
compact space then trivially the compact bornology coincides with the
trivial bornology.
As in the last example it is an interesting problem to know when, in a
metric space (X, d) with a given bornology B, the bornology B coincide
with the metric bornology for d, Bd (X). A more general problem has
been solved by Hu in [7] and [8], for a metrizable topological space.
So we are going to present this result in the same way as Hu did it.
Consider the next denition:
Denition 2. A bornological universe is a pair (X, B) where X is
a topological space and B is a given bornology on X .
We want to know when there exists a metric d on (X, B) such that
X is metrizable by d and the bornology B concide with the metric
bornology Bd (X) on (X, d). In this case we say that the bornological
universe is metrizable.
Denition 3. A base for a bornology B is a subfamily A of B such
that every element in B is contained in some element of A.
Denition 4. A characteristic function of a bornological universe
(X, B) is a real-valued non negative continuous function
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χ : X → [0, ∞) such that a subset E ⊂ X is bounded i χ(E) is
bounded in [0, ∞), that is,
B = {E ⊂ X : ∃K > 0, χ(E) ≤ K}.
Denition 5. A bornological universe (X, B) is normal i its topological space is normal, that is, every pair of disjoint closed sets in X
has disjoint neighborhoods.
Lemma 1. A normal bornological universe (X, B) admits a characteristic function χ i B has a countable base and for each B1 ∈ B there
◦
is B2 ∈ B such that B1 ⊆ B 2.
Proof. ⇒) Assume that (X, B) is a bornological universe which admits
a characteristic function χ. Now, for each positive integer n, let Bn
denote the subset of X which consists of all points x ∈ X satisfying
χ(x) ≤ n. Then A = {Bn : n ∈ N} is a countable sequence of bounded
sets. Let B denote an arbitrary bounded set of X and choose an integer
n satisfying n ≥ χ(B), then we have B ⊂ Bn . Hence A is a countable
base of the bornology B.
Next let B denote an arbitrary bounded set. Then there exists a
real number K > 0 with K ≥ χ(B). By continuity of χ, we have that
χ(B) ⊂ χ(B) ≤ K < K + 1. Then again by continuity, BK+1 = {x ∈
X : χ(x) < K + 1} is a bounded open set and B ⊂ BK+1 .
⇐) Conversely, assume that X is a normal universe satisfying the
two conditions of the theorem. If X is bounded, then the constant zero
function on X is a characteristic function of X . So, suppose that X is
not bounded. So this implies that X cannot has a maximal bounded
set, that is, it doesn't exist a bounded set B ∗ ∈ B such that for every
B ∈ B, B ⊆ B ∗ because every point in X is bounded and X is not
bounded.
First we are going to nd a base G for the bornology which consists
of a strictly increasing sequence of open bounded sets G1 , G2 , ..., Gn , ...
satisfying Gn ⊂ Gn+1 for every n = 1, 2, ... . Indeed let D
= {D1 , D2 , ..., Dn , ...} be a countable base for B. By hypothesis there
11
◦
exists an open bounded set G1 = B, B ∈ B such that D1 ⊆ D1 ⊆ G1 .
Assume that bounded open sets G1 , G2 , ..., Gn have been choosen in
such a way that Gi ⊃ Gi−1 ∪ Di and Gi 6= Gi−1 for each i = 2, ..., n.
By hypothesis Gn is a bounded set because there exists a B ∈ B such
◦
that Gn ⊆ B . As B has no maximal bounded set then there exists a set
B0 ∈ B which is not a subset of Gn . Again by hypothesis there exists an
open bounded set Gn+1 ∈ B such that Gn ∪ Dn+1 ∪ B0 ⊂ Gn+1 . Hence
Gn+1 ⊃ Gn ∪ Dn+1 , and Gn+1 6= Gn . This completes the inductive
construction of a strictly increasing sequence G = {G1 , G2 , ..., Gn , ...}
of bounded open sets, satisfying Gn ⊂ Gn+1 for every n = 1, 2, ....
Finally, let B a bounded set, since D is a base of B, there exists a
positive integer n satisfying B ⊂ Dn ⊂ Gn . Hence G is a base of B.
Since Gn and X −Gn+1 are two disjoint closed sets of a normal space
X , it follows from Urysohn's lemma that there exists a continuous
function φn : X → I satisfying φn (x) = 0 for every x ∈ Gn and
φn (x) = 1 for every x ∈ X − Gn+1 . Now dene a function χ : X → R
by taking :
1
χ(x) = φ1 (x), ∀x ∈ G1 ,
χ(x) = (n − 1) + φn (x), ∀x ∈ Gn+1 − Gn , ∀n > 1.
The function χ is obviously real-valued and nonnegative. The continuity of χ follows from the fact that φn−1 (x) = 1 and φn (x) = 0 hold
for every point x ∈ Gn − Gn and every n > 1. Now let E denote an
arbritary subset of X . If E is bounded, then there exists a positive
integer n satisfying E ⊂ Gn+1 . This implies χ(x) ≤ n for every x ∈ E .
Conversely, assume that χ(E) < K . Choose a positive integer n so
large that χ(x) < n holds for every x ∈ E . Then we have E ⊂ Gn+1
and hence E is bounded. Therefore χ is a characteristic function of
X.
1 Urysohn's Lemma
If
A and B
are two disjoint closed sets in a normal space
a continuous function
φ:X→I
such that
φ(A) = 0
and
φ(B) = 1.
12
X , there exists
Theorem 2. Let (X, B) a bornological universe. Then B is metrizable
i the following three conditions are satised:
1. the topological space X is metrizable;
2. the bornology B has a countable base;
◦
3. for each B1 ∈ B there is B2 ∈ B such that B1 ⊂ B 2.
Proof. ⇒) Condition 1. is clear. For 2., choose a point z ∈ X . For
each positive integer n the open balls Bn (z) = {x ∈ X : d(z, x) < n}
form a countable base for B. As for every B ∈ B there exists a Bn (z)
such that B ⊂ Bn (z) so B ⊂ Bn (z) ⊂ Bn+1 (z) and we have 3..
⇐) Assume that (X, B) satises the conditions 1., 2., 3.. Since X
is metrizable then there exists a bounded metric e : X 2 → R which
denes the topology of X . Again by 1. X is normal and by 2. and 3.
and the last theroem we have a characteristic function χ : X → R on
the universe (X, B).
Now dene the function d : X 2 → R by taking
d(x, y) = e(x, y) + |χ(x) − χ(y)|
for every x, y ∈ X . In fact d is a metric on X .
To prove that d denes the same topology of X , it is enough to
show that d and e are equivalent. For this purpose, consider any point
p ∈ X and an arbitrary positive real number r. Since χ is continuous
and since e denes the same topology of X , there exists a positive real
number s < r/2 such that
|χ(p) − χ(x)| < r/2
holds for every point x ∈ X satisfying e(p, x) < s. It follows that, for
every x ∈ X , e(p, x) < s implies
d(p, x) = e(p, x) + |χ(p) − χ(x)| < s + r/2 < r/2 + r/2 = r.
On the other hand, d(p, x) < s obviusly implies e(p, x) < s. So we
have the topological equivalence between the metrics e and d.
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It remains to verify that the metric d denes the bornology B. For
this purpose, choose a positive real number k such that e(x, y) < k
holds for every point (x, y) of X 2 . Let E denote any subset of X . If E
is a member of B, then it follows from the denition of a characteristic
function that χ(E) < K . Hence we obtain
d(x, y) = e(x, y) + |χ(x) − χ(y)| < k + K
for all x, y ∈ E . This proves that E is bounded for the metric d. Conversely, assume that E is a member of the metric bornology generated
by d. Then there exists a positive real number r such that d(x, y) < r
holds for all points x, y ∈ E . Without loss of generality, assume E is
not empty. Choose a point p ∈ E . Then we have
χ(x) ≤ χ(p) + |χ(p) − χ(x)| ≤ χ(p) + d(p, x) < χ(p) + r
for all x ∈ E . This implies E ∈ B. Hence the bornologies are equivalent, that is, d denes the given bornology B of X .
Now we give an example of application of this theorem.
Corollary 3. Let (X, F) a bornological universe where X is a T2 and
F is the nite bornology, then we have the following equivalence:
1. (X, F) is metrizable;
2. X is countable and discrete.
Proof. 1. ⇒ 2.) If (X, F) is metrizableSthen by the previous theorem F
has a coumerable base such that X = n∈N Bn where Bn are nite sets.
So that X is countable. Also let x ∈ X then, as (X, F) is metrizable
◦
◦
then there exists F ∈ F such that {x} = {x} ⊆ F and obviously F is
a nite set. As X is T2 then there exists an open neigborhood G of {x}
◦
such that G ∩ F = {x}. Then we have that {x} is open and closed,
because X is T2 , ∀x ∈ X .
◦
1. ⇐ 2.) If X is discrete then for every F ∈ F we have F = F = F ,
so we have that for every F1 ∈ F there exists a F2 ∈ F such that
14
◦
F1 ⊆ B . Now if X is countable then let x0 ∈ X be xed, and dene
B0 = {x0 }, B1 = B0 ∪ {x1 } with x1 ∈
/ B0 and Bn = Bn−1 ∪ {xn },
xn ∈
/ Bn−1 . This is possible because X is numerable. In this way we
have a numerable base of F because every F is nite.
2.3 The Heine-Borel property.
Let (X, d) a metric space. If C is a compact subset of it then it is
closed and bounded. Now we ask for the converse, that is, when (X, d)
has the Heine-Borel property: every closed and bounded subsets is
compact. This problem can be solved more generally for a Hausdor
space X , that is, we can nd, under some hypothesis, and admissible
metric d for X such that (X, d) has the Heine-Borel property. In this
case we say that d is a proper metric.
Theorem 4. Let X be a T2 topological space. The next cases are
equivalent:
1. X is metrizable by a proper metric d;
2. (X, CB) is metrizable;
3. the topology of X is metrizable, CB has a countable base and for
◦
every B1 ∈ CB there exists a B2 ∈ CB such that B1 ⊆ B 2;
4. X is locally compact and second countable.
Proof. 1. ⇔ 2. If X is metrizable by a proper metric d, that is, (X, d)
has the Heine-Borel property, then we have that by Example 6. CB =
Bd(X). Conversely if CB = Bd(X) by an admisible metric d, then
obviously X is metrizable, and for every closed B ∈ Bd (X) there exists
a C ∈ CB such that B ⊂ C so B is compact.
2. ⇔ 3. This equivalence follows from the last Theorem 2..
3. ⇒ 4. As every point in (X, d) is a bounded set of the compact
bornology CB, then there exists a bounded set B ∈ CB such that x ∈
15
◦
◦
B and as B is compact we can concluded that (X, d) is a locally compact space. Now, we take a countable base of CB, {B1 , B2 , ..., Bn , ...}.
Obviously, by denition of CB, for every j ∈ N, Bj is a compact subset of X . Given an open cover {Gi }i∈I of X , every Bj is covered by
a nite subcover of {Gi }i∈I by compactness. For every j ∈ N take
its respective nite cover {Gik j }k=1,...,nk , Then {{Gik j }k=1,...,nk }j∈N is
an open countable subcover of X . So (X, d) is a Lindelöf space, and
because it is metrizable then it is second countable.
4. ⇒ 3. Conversely suppose that (X, d) is locally compact and second countable. Then X is metrizable because it is regular and second
countable . As X is locally compact, for every x ∈ X there exists an
open neighborhood V x such that V x is compact. Let B1 ∈ CB then
its closure is compact so B1 is covered by a nite subcover V xk . If we
◦
S
put B2 = k V xk then it is an open set in CB, such that B1 ⊂ B 2 .
Now suppose that X is bounded, then obviously CB has a countable
base given by X itself. So suppose that X is not bounded. Then recall
that CB doesn't have a maximal bounded set. Remember that X is
metrizable and second countable, so it is Lindelöf. Take the open cover
of X given by V x , then there exists a countable subcover of X , {V xj }.
Let B1 ∈ CB. As B1 is compact, then
exists a nite subcover
Sn1 there
xjk
xjk
{V }k=1,...,n1 of B1 . Maybe B1 = k=1 V
but as {V x } is a cover
of X and X is not bounded then there S
exists a V x1∗ such that it is
1
not contained in B1 . So if we pick B2 = nk=1
V xjk ∪ V x1∗ ∪ V x1 then
B1 is strictly contained in B2 and B2 ∈ CB. Repeating this process
we have a strictly increasing sequence of sets {B1 , ..., Bn , ...} such that
V xj−1 ⊂ Bj , j ≥ 2. Then let B ∈ CB then its closure is compact so
B is covered by a nite subcover {V xjk }k=1,...n . By construction of the
Bj0 s, we can take the biggest BjK , K ∈ {1, ..., n} such that V xjk ⊆ BjK ,
∀k = 1, ..., n. So {B1 , ..., Bn , ...} is countable base of (X, CB).
2
3
2X
is a regular space if every point
x∈X
3 Urysohn's metrization theorem.
a) X is regular and second countable;
b) X is separable and metrizable;
c) X can be embedded as subspace of
has a nhood base consisting of closed sets.
The following are equivalent for a
the Hilbert cube I
16
N
.
T1
space:
Remark 3. Regarding in the proof of Theorem 2, and applying it to a
unbounded metric space (X, d) and to its compact bornology CBd(X)
we have a way to construct a proper metric from d. As in the proof it
is given by
d0 (x, y) = e(x, y) + |χ(x) − χ(y)|
where e = min{d, 1} and χ : X → [0, ∞) is the characteristic function
associated to a countable base of CBd(X).
To have more information about this theme please consult the articles [13] and [10].
17
Chapter 3
Bornologies in uniform spaces.
3.1 Totally bounded sets and bounded sets in the
sense of Bourbaki.
Now, we take a uniform space (X, U) and we look for some natural
denition of bounded sets on it. To make easier the work we are going
to suppose that U generates a T2 topology on X .
Denition 6. A subset B of a uniform space (X, U) is called bounded
in the sense of Bourbaki if for each U ∈ U, we can nd an
n ∈ N and a nite set F ⊆ X such that B ⊆ U n (K) where U n =
n
z
}|
{
U ◦ U ◦ ... ◦ U . If n = 1 then we say that B is
precompact) .
totally bounded (or
Remark 4. In these denitions we can suppose that our entourages
are from a base of U and that they are symmetric, that is, they satisfy
U = U −1 .
Obviously the bounded sets in the sense of Bourbaki and the totally bounded sets form a bornology which we are going to denote by
BB (X) and TB (X), respectively.
U
U
Remark 5. If a space X has two equivalent uniformities then the
Bourbaki bounded sets and the totally bounded sets of both uniform
18
spaces coincide respectively and they dene the same bornology because
the entourages, which are needed to be bounded in the sense of Bourbaki
or totally bounded, are the same in both uniformities.
Remark 6. To be totally bounded implies to be bounded in the sense
of Bourbaki, but the converse is not true in general, as we will see in
the next example.
Example 7. Let (X, k.k) an innite dimensional Banach space. The
closed ball of radius 1, B1(x), is bounded in the sense of Bourbaki because, as we will see later, its image under every real-valued uniformly
continuous function f ∈ U C(X), f (B1(x)) is bounded in R.
If BBk.k(X) = TBk.k(X), that is, BBk.k(X) ⊂ TBk.k(X), this implies that B1(x) is totally bounded and then compact because it was
already complete. But this is impossible because (X, k.k) is innite
dimensional.
For a metric space (X, d) ina easy way we can say that a set B is
totally bounded if ∀ > 0, can be covered by a nite union of disks
of
, that is, ∀ > 0 there exist x1 , x2 , ..., xn such that B ⊆
Snradius
i=1 B (xi ), where the B (xi ) are the open metric balls of radius and
center xi of (X, d). If B is bounded in the sense of Bourbaki in (X, d)
then its visualization it's a little more dicult. Let > 0 and let U ∈ U
such that U [x] = {y ∈ X : d(x, y) < }. Then
U 2 [x] = {y ∈ X : ∃z ∈ X, d(x, z) < , d(z, y) < }
and
U 3 [x] = {y ∈ X : ∃z ∈ X, d(x, z) < , z ∈ U 2 [y]} =
= {y ∈ X : ∃z, w ∈ X, d(x, z) < , d(y, w) < , d(w, z) < }.
More generally
U n [x] = {y ∈ X : ∃z ∈ Xd(x, z) < , z ∈ U n−1 [y]}.
So, if B bounded in the sense of Bourbaki then ∀ > 0 ∃n ∈ N and
there exist x1 , x2 , ..., xm ∈ X such that
B⊆
m
[
U n [xi ],
i=1
19
where we can imagine U n [xi ] as a sequence of at most n balls in X of
radius at most each one, needed to join xi and another point y ∈ X
in a path of length at most n.
Also for a metric space (X, d) we have that each bounded set in
the sense of Bourbaki for the uniformity generated by the metric d is
bounded for the metric, and so that each totally bounded set for d is
also bounded for d. The converses are not true in general.
Example 8. Let R with the Euclidean metric d2. We know that it is
uniformly equivalent to the bounded metric d∗ = min{1, d2}, so they
dene the same uniformity and then
BBd (X) = BBd (X)
TBd (X) = TBd (X).
∗
2
∗
2
Also we know that to be totally bounded in (R, d2) is equivalent to be
bounded for the metric by completeness of it , so we have that to be
bounded in the sense of Bourbaki for the uniformity generated by d2 is
equivalent to be bounded and totally bounded for d2. That is,
Bd (X) = TBd (X) = BBd (X).
2
2
2
Now, as d∗ is a bounded metric we have that every set is bounded by
d∗ in R, but this doesn't happen to d2 , so the bornology generated by
d∗ is bigger but not equal to the bornology generated by d2 although if
they generate the same uniformity. So
Bd (X) 6= Bd (X)
∗
2
and
BBd (X) = TBd (X) = Bd (X) ⊂ Bd (X)
∗
∗
2
20
∗
3.2 Characterization of the bounded sets in the
sense of Bourbaki.
We have the following results for a uniform space (X, U).
Lemma 5. [9] Metrization Lemma.
Let {Un}n∈N be a sequence of subsets of X × X such that U0 =
X × X , each Un contains the diagonal, and Un+1 ◦ Un+1 ◦ Un+1 ⊂ Un
for each n. Then there is a non-negative real-valued function d on
X × X such that
1. d(x, y) + d(y, z) ≥ d(x, z) for all x, y and z ; and
2. Un ⊂ {(x, y) : d(x, y) < 1/2n} ⊂ Un−1 for each positive integer n.
If each Un is symmetric, then there is a pseudometric d satisfying condition 2..
Denition 7. We say that a pseudometric d is uniformly continuous in a uniform space (X, U) if it denes a uniformity contained in
U.
Theorem 6. [6] Let (X, U) be a uniform space. The following statements are equivalent:
1. B is bounded in the sense of Bourbaki;
2. each real valued uniformly continuous map is bounded on B ;
3. B is bounded for each pseudometric d uniformly continuous on
(X, U).
Proof. 1. ⇒ 2.) Let f : X → R be an uniformly continuous function,
then let > 0 there exists a U ∈ U such that for every x, y ∈ X
if (x, y) ∈ U then |f (x) − f (y)| < . If B ⊂ X is bounded in the
sense of Bourbaki, there exists a nite set K and a natural number n
such that U n (K) ⊃ B . If z ∈ f (B), there exists y ∈ B such that
f (y) = z . As y ∈ U n (K), there exist y = y1 , ..., yn ∈ X , yn+1 ∈ K
21
so that (yi , yi+1 ) ∈ U for i = 1, ..., n. Then |f (yi ) − f (yi+1 )| < and
z ∈ Bn (f (K)).
2. ⇒ 3.) If d is a uniformly continuous pseudometric on X and
b ∈ B , dene, for all x ∈ X f (x) = d(x, b). Then f : X → R is
uniformly continuous and B is bounded for d.
3. ⇒ 1.)Let B ⊂ X be a set given. Let U ∈ U be a symmetric
entourage. We dene a relation on X as x ≡ y i there exists a
n ∈ N such that x ∈ U n (y). The relation ≡ is an equivalent
relation
S
on X . Therefore it denes a decomposition X = α∈A Xα . By the
Metrization Lemma there exists a uniformly continuous pseudometric
d on X such that
1
{(x, y) : d(x, y) < 1} ⊂ U.
Now we want to construct another pseudometric, dening it rst
on each set Xα . Recall that for every x, y ∈ Xα there exists a nite sequence or a chain of points xi ∈ Xα , i = 1, ..., n + 1 such
that
Pn x1 = x, xn+1 = y, (xi , xi+1 ) ∈ U for i = 1, ..., n. The number
i , di+1 ) will be called the length of this chain. Put φ(x, y) =
i=1
Pd(x
n
inf i=1 d(xi , di+1 ) where the inmun is taken over all the chains on
U connectingSx and y. It is easy to see that φ is a nite non negative
function on α∈A (Xα × Xα )and that in fact φ is a pseudometric on
each Xα . It is uniformly continuous because it is equal to d on U .
Now we shall extend the pseudometric onto the hole space X . First
we dene the function µ on the set M . Put
AB = {α ∈ A : Xα ∩ B 6= Φ}.
If AB is nite then we put µ(α) = 1 for each α ∈ AB . If AB is innite,
there exists an innite sequence {αn }∞
n=1 where αn ∈ AB , αm 6= αn if
m 6= n and we put µ(αn ) = n, µ(α) = 1 for α ∈ AB , α 6= αn . Choose
wα ∈ Xα for each α ∈ A. If x ∈ Xα , y ∈ Xβ we put
ρ(x, y) = φ(x, y) if α = β ,
1 The
same demostration can be done for an uniformly continuous function
any uniform space
(Y, V).
If
n=1
f
from
(X, U)
then we have the same result for a totally bounded set.
22
to
ρ(x, y) = φ(x, wα ) + φ(y, wβ ) + µ(α) + µ(β) if α 6= β.
It easy to prove that ρ is a pseudo-metric on X . We are going
to prove the triangle inequality. Suppose that x ∈ Xα , y ∈ Xβ and
x ∈ Xγ . There are only three possible cases:
1. if α = β = γ then ρ = φ and this is clear;
2. if α 6= β 6= γ then
ρ(x, y) ≤ φ(x, wα ) + φ(z, wγ ) + µ(α) + µ(γ) <
φ(x, wα ) + φ(y, wβ ) + µ(α) + µ(β) + φ(y, wβ ) + φ(z, wγ ) + µ(β) +
µ(γ) =
ρ(x, y) + ρ(y, z);
3. if α 6= β = γ then wβ = wγ and we have
φ(wβ , y) + φ(y, z) ≥ φ(wβ , z),
φ(x, wα ) + φ(wβ,y ) + µ(α) + µ(β) + φ(y, z) ≥
φ(x, wα ) + φ(wβ , z) + µ(α) + µ(β),
ρ(x, y) + ρ(y, z) ≥ ρ(y, z).
If ρ(x, y) < 1, then x and y belong to the same class Xα and
ρ(x, y) = φ(x, y), therefore ρ is also uniformly continuous.
Now let B be bounded for the pseudometric ρ. Then AB must be
nite and φ(x, wα ) is a bounded function for x ∈ B ∩ Xα . Write
b = maxα∈A (supx∈B∩Xα φ(x, wα )) < ∞.
Let m > b, m ∈ N. Consider an element α ∈ AB . For each point
x ∈ B ∩ Xα there exists an irreducible chain on U conecting x with
wα . Therefore, therePexists a sequence x = x1 , x2 , ..., xn+1 = wα such
that (xi , xi+1 ) ∈ U , ni=1 d(xi , xi+1 ) < m. As this chain is irreducible,
the inequality d(xi , xi+1 ) < 21 cannot hold for two consecutive indices
i. It means that d(xi , xi+1 ) < 12 holds at most for 21 (n + 1) indices i
and therefore d(xi , xi+1 ) ≥ 21 holds at least for 12 (n − 1) indices. Then
the length of the chain is at least 41 (n − 1), which implies n < 4m + 1.
23
therefore B ⊂ U 4m+1 (wα , α ∈ AB ) and the boundedness in sense of
Bourbaki of B is proved.
Corollary 7. [6] Let (X, U) be a uniform metrizable space. Then
B ⊆ X is bounded in the sense of Bourbaki i B is bounded for each
metric on X dening the same uniformity as U (uniformly equivalent).
3.3 Uniform metrization of bornologies.
As in the rst chapter, now we are interested in the study of the
problem of when a uniform space (X, U) with a bornology B is metrizable, that is, there exists an admissible metric d such that it denes the
same uniformity than U and the bornology B coincide with the metric
bornology Bd (X). If we suppose that the uniform space is metrizable,
that is, it has a numerable base for the uniformity and the topology generated by the uniformity is T2 , then we can solve the problem directly
for a metric space (X, d): for a metric space (X, d) with a bornology B
we want to nd a uniformly equivalent metric d0 such that the bornology B coincides with the metric bornology for d0 , that is, B = Bd0 (X).
We have two cases:
1. suppose that (X, d) is a metric space such that the bornology
B contains X . Then every subset of X is bounded. If we put
d0 = min{1, d}, d0 is a bounded uniformly equivalent metric, so
B = Bd0 (X);
2. now suppose that X ∈/ B.
Denition 8. Let A ⊂ X and x ∈ X , then d(x, A) is the distance
from x to A and if A = ∅ then we agree that d(x, A) = ∞. We denote
the -enlargement of A by A where
A = {x ∈ X : d(x, A) < } =
[
x∈A
24
B (x)
and B(x)is the open ball of radius around x.
Lemma 8. Let (X, d) be a metric space and B a bornology on it such
that X ∈/ B. If B has a countable base {Bn : n ∈ N} such that ∃δ > 0,
with
Bn δ ⊆ Bn+1 ,
∀n ∈ N, then there exists a uniformly continuous characteristic func-
tion
χ : X → [0, ∞) such that
B = {E ⊂ X : ∃K > 0, χ(E) < K}.
Proof. For n ∈ N dene φn : X → [0, 1] by φn(x) = min{1, 1δ d(x, Bn)}.
Then φn (Bn ) = 0, φn (X − Bn+1 ) = 1, 0 ≤ φn ≤ 1 andPφn is uniformly
continuous. Put χ = φ1 + φ2 + φ3 + .... Then χ | Bn = n−1
i=2 φi , so that:
1. χ | Bn is uniformly continuous ∀n ∈ N;
2. χ(Bn ) ⊆ [0, n − 1] ∀n ∈ N.
Moreover χ is uniformly continuous: ∀ > 0 ∃δ 0 > 0 such that δ 0 < δ
and δ 0 < δ , and ∀x, y ∈ X with d(x, y) < δ 0 , then ∃n ∈ N such that
x ∈ Bn − Bn−1 and y ∈ Bnδ , then
2
|χ(x) − χ(y)| = |φn (x) − φn (y) + φn+1 (x) − φn+1 (y)| < d(x, y) < 2.
δ
Now let
E ∈ B then there exists an n ∈ N such that E ⊂ Bn . So
Pn−1
χ(E) = i=1 φi (E) ⊆ [0, n − 1]. Conversely if χ(E) is bounded then
for some n ∈ N, E ⊆ Bn because otherwise for every n ∈ N there exists
an x ∈ E such that χ(x) ≥ n.
Theorem 9. Let B a bornology on (X, d). The following conditions
are equivalent:
1. B = Bd (X) for some uniformly equivalent metric d0 on (X, d)
0
2. B has a countable base {Bn : n ∈ N} such that ∃δ > 0, with
Bn δ ⊆ Bn+1 ,
∀n ∈ N.
25
Proof. 1. ⇒ 2.) If B = Bd0 (X), xed a certain z ∈ X the basis
{Bn = {x ∈ X : d(x, z) < n}, n ∈ N}
and δ < 12 satises the thesis.
2. ⇒ 1.)The case X ∈ B is of course trivial because B = P(X)
and we have d0 = min{1, d}. So that, suppose X ∈/ B and take the
bounded metric d∗ = min{1, d}, which is uniformly equivalent to d.
Applying the previous lemma we dene a new metric d0 : X 2 → R,
d0 (x, y) = d∗ (x, y) + |χ(x) − χ(y)|
where χ is the characteristic function associated to the base of the
bornology.
To show that d0 and d are uniformly equivalent it is enough to proof
that d0 and d∗ = min{1, d} are uniformly equivalent. On one hand
d∗ ≤ d0 , so if d0 (x, y) < then d∗ (x, y) < and then Ud is coarser than
Ud∗ . On the other hand, using the uniform continuity of χ, ∀ > 0,
∃δ > 0 such that ∀x, y ∈ X with
d∗ (x, y) < δ ⇒ d0 (x, y) = d∗ (x, y) + |χ(x) − χ(y)| < + δ,
and we have that the equivalence of the uniformities follows and also
the uniform equivalence of the metrics.
Now we want to show that B ∈ B i B ∈ Bd0 (X). If B ∈ B then it
is clearly bounded for the metric d∗ and χ(B) is bounded in R by the
lemma, so obviously B ∈ Bd0 (X). If B ∈ Bd0 (X) then there exists a
k ∈ R, k > 0 such that ∀x, y ∈ B then d0 (x, y) < k . As d∗ is bounded
metric then χ(B) is bounded in R so then B ∈ B by the lemma.
26
3.4 The Heine-Borel property for uniformly equivalent metrics.
As an application of the last theorem now we come back to the study
of the compact bornology CB and to the study of proper metrics as in
the second chapter. We have (X, U) a uniform such that its uniformity
generates a T2 topology.
Denition 9. [13] [1] The uniform space (X, U) is uniformly locally
compact if there is an entourage U ∈ U such that U [x] is a compact
subspace of X for every x ∈ X .
This denition implies that X with the topology generated by U is
locally compact because U generates a T2 topology.
Now we ask when for a metric space (X, d) there exists an uniform
equivalent proper metric d0 , that is, (X, d0 ) has the Heine-Borel property. As in the second chapter we give a similar answer:
Theorem 10. Let (X, d) a metric space. The following cases are equivalent:
1. there exists a uniform equivalent proper metric d0;
2. there exists a uniform equivalent metric such that CBd(X) =
Bd0 (X);
3. CBd(X) has a countable base {Bn : n ∈ N} such that ∃δ > 0,
∀n ∈ N with
Bnδ ⊆ Bn+1 ;
4. (X, d) is uniformly locally compact and second countable.
Proof. Everything is similar to the proof of Theorem 4. in the second
chapter. We have to control only the fact of the uniformly locally
compactness.
S
3. ⇒ 4.) Recall that B δ = x∈B Bδ (x). As for every x ∈ X there
exists nx ∈ N such that x ∈ Bnδ x then Bδ (x) ⊂ Bnδ x so Bδ (x) ∈ CBd (X)
27
so that Bd (x) is compact. Then as (X, d) is a T2 space then it is
uniformly locally compact.
4. ⇒ 3.) If (X, d) is uniformly locally compact then there exists
a entourage V ∈ U, such that V [x] is compact for every x ∈ X .
By denition of base of a uniformity there exists a δ > 0 such that
U = {(x, y) ∈ X 2 : d(x, y) < δ} ⊂ V , so that U [x] are compact
neighborhoods of every x. Let B1 ∈ CB then its closure S
is compact so
B1 is covered by a nite subcover U [xk ]. If we put B2 = k U [xk ] then
◦
δ
it is an open set in CB, such that B1 ⊂ B 2 . The rest of the proof
follows the homologous proof in the second chapter.
It is a known result that C is a compact subset of (X, U) i it
is totally bounded and complete. Now I want to give a similar
result but for the more general notion of bounded set in the sense of
Bourbaki. This is going to be done comparing the compact bornology
with bornology of the bounded sets in the sense of Bourbaki.
Lemma 11. [1] Let (X, U ) be a uniform space not necessarily T2 then
if (X, U) is uniformly locally compact, then A is bounded in the sense
of Bourbaki X i A is a compact subset of X .
Proof. We know that there exists an entourage V ∈ U such that V [x]
is compact for every x ∈ X and suppose that B ⊂ X is compact. Take
a symmetric entourage U ∈ U such that U 2 ⊆ V then there exists a
nite subset F ⊂ X with B ⊆ U [F ]. So
U [B] ⊆ U 2 [F ] ⊆ V [F ].
As V [F ] is compact then it is also U 2 [F ]. By induction we can conclude
that U n [F ] is compact in X for every n ∈ N and every nite subset F
of X .
So we can conclude with the next theorem
Theorem 12. Let (X, U) a uniformly locally compact space then the
compact bornology CBU(X) coincides with BBU(X), the Bourbaki
bornology.
28
Soon, in the last chapter, we are going to study the problem of
metrization of the bornology of bounded sets in the sense of Bourbaki,
that is, when, for a metric space (X, d), we have a uniformly equivalent
metric d0 such that BBd (X) = Bd0 (X). Obviously by the theorem of
uniform metrization of bornologies, we already have the answer but as
the Bourbaki bounded sets aren't easily visualized, then to work with a
base of this sets is not very practic. So we want to see other equivalents
solutions that maybe are easier to work with. In this last part of this
chapter we have a nice answer:
Theorem 13. Let (X, d) be a uniformly locally compact metric space.
Then there exists a uniformly equivalent metric d0, such that, (X, d0)
has the Heine-Borel property i BBd(X) = Bd (X).
0
Remark 7. If (X, d) is a uniformly locally compact then BBd(X) =
Bd (X) i d is a proper metric.
29
Chapter 4
Realcompactication v.s
compactication.
Now we are going to introduce the notions of compactication and
realcompactication of a topological space to compare them. But rst
we are going to give some basic notions about a Tychono space because as we will see later we can speak only about compactication
or realcompactication of a Tychono space. All this notions can be
studied in more detail in citar [12] and [3].
Denition 10. A Tychono space is topological space which is both
T1 and completely regular, where a topological space X is completely
regular i whenever F is a closed set in X and x ∈/ F there is a realvalued continuous function f : X → [0, 1] such that f (x) = 0 and
f (F ) = 1.
Remark 8. Subspaces of a Tychono space are Tychono and non
empty product of Tychono spaces is Tychono.
Remark 9. All the T4 spaces that is normal spaces that are also T1
are Tychono spaces.
Denition 11. Let {fα : α ∈ A} be a collection of functions on X
to spaces Xα , then {fα : α ∈ A} separates points and closed sets
in X if given a closed subset F of X and x ∈/ F there exists some
f ∈ {fα : α ∈ A} such that f (x) ∈
/ f (F ).
30
Theorem 14. Immersion's lemma. If X is a T1 space and
{fα : α ∈ A} is a collection of continuous functions on X to spaces
Xα , which separates points from closed sets, then the evaluation map
e:X→
Y
Xα
α∈A
dened as
e(x) = (fα (x))α∈A
is a homeomorphic embedding.
Corollary 15. A topological space X is a Tychono space i it is
homeomorphic to some subspace of some cube, that is, of some product
of closed real intervals.
Proof. In fact as X is a Tychono space then the family of real-valued
bounded continuous functions on X , C ∗ (X), separates points from
closed sets. As f (X) is a bounded set in R for every f ∈ C ∗ (X),
take If a closed interval in R such that f (X) ⊆ If .
Obviously C ∗ (X) = {f ∈ C ∗ (X) : f : X → If }. So by the Q
Inmersion's
lemma we can conclude that the evaluation map e : X → f ∈C ∗ (X) If
dened as
e(x) = (f (x))f ∈C ∗ (X)
is an homeomorphic embedding of X in the cube f ∈C ∗ (X) If .
The converse is obvious because the interval [0, 1] is a Tychono
space.
Q
The way in which has been proved the previous corollary, that is,
using the Inmersion's lemma, is going to be repeated to construct compactications and realcompactications, as we will see in the following
sections.
31
4.1 Compactication of a topological space.
Denition 12. Let X be a topological space. Then a compactication of it is a compact space cX and a map c : X → cX which is an
homeomorphic embedding of X in cX in such a way that c(X) = cX .
Theorem 16. A topological space X has a compactication i X is a
Tychono space.
Proof. ⇒) If X has a compactication then X is homeomorphic to
a subspace of a compact space cX . As cX is compact then it is a
Tychono space so X is a Tychono space.
⇐) Let X be a Tychono space then it is homeomorphic to a subspace of a cube. Every cube is compact so if we take the closure of X
in the cube then it is a compactication of X .
Denition 13. Given two compactications of X we say that c1X ≤
c2 X i there exists a continuous mapping h : c2 X → c1 X such that
h ◦ c2 X = c1 X . Thus the inequality c1 X ≤ c2 X means that c2 X can
be mapped onto c1X in such a way that every point of the space X ,
considered as a subspace of both c2 and c1, is mapped onto himself.
Two compactications c1X and c2X are equivalent compactications
i h is a homeomorphism. This is equivalent to say that c1X ≤ c2X
and c2X ≤ c1X .
The previous denitions gives us an order ≤ is the familly of all the
compactications of a Tychono space. In particular we have a largest
element in this familly.
Denition 14. Let X be a Tychono. Then its biggest compactication is the Stone-ƒech Compactication, denoted by βX .
Theorem 17. Every continuous function f : X → Z of a Tychono
space X to a compact space Z is extendable to a mapping F : βX → Z .
32
Proof. By the Immersion's lemma c = β × f : X → βX × Z is a
homeomorphic embedding, so that cX = c(X) ⊂ βX × Z is a compactication of X . By the maximality of βX there exists a mapping
h : βX → cX such that h ◦ β = c. Let p : cX → Z be the restriction
of the projection of βX × Z onto Z to cX and let F = p ◦ h : βX → Z .
Since F β = phβ = pc = f , the mapping F is an extension of f .
Theorem 18. If every continuous function of a Tychono space X to
a compact space is continuously extendable over a compactication cX
of X , then cX is equivalent to the Stone-ƒech Compactication of X .
Proof. If a compactication of X has the property of the hypothesis
then there exists an extension B : cX → βX of the embedding
β : X → βX . We have then that B ◦ c = β , that is, βX ≤ cX . As
βX is the biggest compactication of X then cX = βX .
Recall that when we have proved that every Tychono space is
homeomorphic to some subspace of some cube we have construct indeed the Stone-ƒech Compactication of X . We know that e(X) is
a compactication of X as it has been set when it have proved that
a Tychono space has a compactication. There the evaluation map
∗
e : X → RC (X) was given by
e(x) = (f (x))f ∈C ∗ (X) .
Now we are going to prove that every bounded continuous function
f : X → R can be extended to a continuous function f˜ : e(X) → R.
For every f ∈ C ∗ (X) let f˜ : e(X) → R be the restriction of the
continuous projection
pf : R C
∗
(X)
→R
such that
pf ((y)f ∈C ∗ (X) ) = y ∈ R,
pf (e(x)) = f (x).
Then f˜ is also continuous and then f˜ ◦ e = pf ◦ e = f . By the
previous theorem, as every bounded continuous function f ∈ C ∗ (X) is
33
a continuous function from X to the compact interval in R, If ⊃ f (X),
we have that βX = e(X). In particular the continuous extension
f˜ : βX → R is also a bounded function.
Remark 10. Obviously X is a compact i βX = X .
Repeating this argument for any subfamily Γ ⊆ C ∗ (X) of continuous
bounded real-valued functions separating points from closed sets we
have always a compactication of the Tychono space X .
Theorem 19. Let X be a Tychono space and let Γ ⊆ C ∗(X) be
a family of continuous bounded real-valued functions separating points
from closed sets. Then e(X) ⊂ RΓ, where e : X → RΓ is the evaluation
map given by
e(x) = (f (x))f ∈Γ ,
is the smallest compactication such that every f ∈ Γ can be extended
to continuous bounded function f˜ : e(X) → R.
Proof. We have to prove only that in fact it is the smallest. Suppose
that there exists another compactication cX with the same property of
extendability in the hypothesis, then we have to prove that cX ≥ e(X).
In fact there exists a continuous mapping h : cX → e(X) such that
h ◦ c = e, dened in the next way:
c
h(y) = (f (y))f ∈Γ ∈
Y
Rf = RΓ
f ∈Γ
where f c : cX → R is the continuous extension of f ∈ Γ. Then for
every f ∈ Γ let pf : RΓ → R be the continuous projections dened by
pf ((y)f ∈Γ ) = y ∈ R,
pf (e(x)) = f (x).
In particular pf ◦ h = f c so
h(y) ∈ h(c(X)) ⊂ h ◦ c(X) = e(X).
And we have it.
34
Denition 15. Let Γ = U C ∗(X) the family of all the uniformly continuous bounded real-valued functions from a metric space (X, d). Then
as in the previous example sXe(X) ⊂ RU C (X) is a compactication of
(X, d) called Samuel Compactication. Then sX can be characterized as the smallest compactication of (X, d) such that every uniformly
continuous bounded real-valued function from f : X → R is extended
to a continuous (bounded) function f˜ : sX → R.
∗
4.2 Realcompactication of a topological space.
Denition 16. A topological space X is called a realcompact space
if X is a Tychono space and there is no Tychono space X̃ which
satises the following two conditions:
1. there exists a homeomorphic embedding α : X → X̃ such that
α(X) 6= α(X) = X̃ ;
2. for every continuous real-valued function f : X → R there is a
continuous function f˜ : X̃ → R such that f˜ ◦ α = f .
Denition 17. A topological space X is called pseudocompact if X
is a Tychono space and every continuous real-valued function dened
on X is bounded.
Theorem 20. A topological space is compact i it is a pseudocompact
realcompact space.
Proof. ⇒) This is not dicult to prove from the denitions.
⇐) Suppose that X is not compact but realcompact. Then, since
X 6= βX , there exists a function f : X → R which cannot be continuously extended over βX by condition 2. of the denition of realcompact
space, as βX is Tychono. Clearly the function is not bounded, because
if it were bounded then f : X → f (X) ⊂ If ⊂ R where If is a closed
and bounded interval in R. So If is compact and then f : X → If ⊂ R
35
should be extendable to a continuous function f˜ : βX → If ⊂ R. We
can concluded then that X is not pseudocompact.
Theorem 21. A topological space is realcompact i it is homeomorphic
to a closed subspace of a power RΛ of the real line.
Proof. Let X be a realcompact space. As X is a Tychono space then
we can apply the last Inmersion's lemma to the set of all the continuous
real-valued functions on X , C(X). So the evaluation map
Y
e:X→
Rf
f ∈C(X)
is a homeomorphic embedding of X in
Q
f ∈C(X) Rf
:= RC(X) . Let
X̃ = e(X) ⊂ RC(X) .
For every continuous real-valued function f : X → R there exists a
function f˜ : X̃ → R such that f˜Q
e = f . Infact, it is the restriction
f˜ = pf | X̃ of the projection pf : f ∈C(X) Rf → Rf . Obviously f˜ is
continuous functions by the continuity of the projections maps pf Now,
by denition of realcompact space, we have that X̃ = e(X) and then,
the space X is homeomorphic to the closed subspace e(X) of RC(X) .
Q
Conversely, let X be a closed subspace of RΛ := λ∈Λ Rλ . Obviously
it is a Tychono space and suppose that it is no realcompact. Then
it has an extension X̃ and α a homeomorphic embedding α : X → X̃
with α satisfying the condition 2. of the denition of realcompact space,
that is, α(X) 6= α(X) = X̃ . For every λ ∈ Λ there exists a function
p̃λ : X̃ → Rλ = R such that p̃λ α = pλ , where
pλ (x) = pλ ((xλ )λ∈Λ ) = xλ ∈ Rλ = R. Let
F : X̃ →
Y
Rλ
λ∈Λ
dened as
F (y) = (p̃λ (y))λ∈Λ .
36
Since F α(x) = x for every x ∈ X , we have that
F (X̃) = F (α(X)) ⊂ F α(X) = X = X,
then F maps X̃ onto X . For every x ∈ X we have that αF (α(x)) =
α(x), so αF : X̃ → X̃ when restricted to α(X) coincides with idα(X) .
Since α(X) is dense in X̃ then αF = idX̃ . As αF (X̃) ⊂ α(X), it
follows that α(X) = X̃ . Hence there is no Tychono space X̃ satisfying
1. and 2., i.e., X is a realcompact space.
Remark 11. As an immediate corollary R with the usual topology is
realcompact and so any power of it, RΛ.
Corollary 22. Every closed subspace of a realcompact space is realcompact
Q
Corollary 23. The non empty product s∈S Xs is realcompact i Xs
is realcompact for every s ∈ S .
Denition 18. In the same way that for compactication, by realcompactication of a topological space X we intend a realcompact space
αX and a map α : X → αX which is an homeomorphic embedding of
X in αX in such a a way that α(X) = αX .
Remark 12. As every compact space is realcompact then every compactication of a space is a realcompactication. But there exist also
realcompactications which are not compactications.
Theorem 24. A topological space has a realcompactication i it is a
Tychono space.
Proof. Suppose that X has a realcompactication X̃ . Then it is home-
omorphic to a subspace of X̃ which is a Tychono space. Then X is
also a Tychono space.
Conversely suppose that X is a Tychono space. Then we can apply
the Immersion's lemma to C(X) and then X ∼
= e(X) ⊂ e(X) ⊂ RC(X) ,
where the evaluation map e : X → RC(X) , given by
e(x) = (f (x))f ∈C(X)
37
is an homeomorphic embedding of X in RC(X) . Obviously e(X) is a
realcompactication of X because is a closed subspace of RC(X) .
Denition 19. The realcompactication constructed in the previous
theorem is the Hewitt Realcompactication of a Tychono space
X . It is denoted by υX . Obviously if X is realcompact i X = υX .
Applying the Inmersion's Lemma for a subcollection of continuous
functions Γ ⊂ C(X) separating points form closed sets in a Tychono
space X ,we can construct other realcompactications of X .
Theorem 25. If X is a Tychono space and Γ is a collection of realvalued functions from X which separates points from closed sets then
X̃ = e(X) ⊂ RΓ is a realcompactication of X where the evaluation
map e : X → RΓ dened as e(x) = (f (x))f ∈Γ is the homeomorphic
embedding. In particular every function f ∈ Γ can be extended to a
continuous function f˜ : X̃ → R such that f˜ ◦ e = f .
Proof. As X̃ is a closed subspace of RΓ it is obviously a realcompact-
ication of X . The continuousQextension of f ∈ Γ is f˜ = pf | X̃ where
pf is the projection pf : RΓ = f ∈Γ Rf → Rf = R
Remark 13. The following cases are equivalent:
1. υX = βX ;
2. υX is pseudocompact;
3. X is pseudocompact;
4. C(X) = C ∗(X).
As for the compactications, in the set of all the realcompactications of a given Tychono space X there is an order for which we can
say, given two compactications, if one of them is smaller than the
other or if they are equivalent.
Denition 20. Given two realcompactications of X we say that
α1 X ≤ α2 X i there exists a continuous mapping h : α2 X → α1 X
38
such that h ◦ α2 = α1, leaving X xed . Moreover α1X and α2X
are equivalent realcompactications i h is a homeomorphism. This is
equivalent to say that α1X ≤ α2X and α2X ≤ α1X .
Theorem 26. Let Γ ⊂ C(X) a collection of continuous real-valued
functions separating points from closed sets of X . Then
X̃ = e(X) ⊂ RΓ is the smallest realcompactication of X such that
every function f ∈ Γ can be extended to a a continuous function f˜ :
X̃ → R.
Proof. The proof is analogous that for the compactications.
4.3 Realcompactications as sets of lattice homomorphisms.
Now, we are going to give an equivalent way to construct realcompactications. All the theory needed can be found in [4]. Let X again
a Tychono space and L ⊂ C(X) be a unital vector lattice, that is:
Denition 21. A unital vector lattice is a partially ordered unital
vector space together with the two operators supremum and inmum,
∨ and ∧.
Denition 22. A lattice homomorphism is a map ϕ : L → R which
satises:
1. ϕ(λf + µg) = λϕ(f ) + µϕ(g), for all f, g ∈ L and λ, µ ∈ R;
2. ϕ(|f |) = |ϕ(f )| for all f ∈ Γ
3. ϕ(1) = 1
Remark 14. Please note that in this case ϕ is positive, that is,
ϕ(f ) ≥ 0 when f ≥ 0. ϕ preserves the supremum and the inmun of
functions, that is, ϕ(f ∨ g) = ϕ(f ) ∨ ϕ(g) and ϕ(f ∧ g) = ϕ(f ) ∧ ϕ(g).
39
Denition 23. The structure space H(L) is the set of all the lattice
homomorphisms ϕ : L → R, considered as a topological subspace of the
product RL.
Theorem 27. If L ⊂ C(X) separates points from closed sets in X
then H(L) is the smallest realcompactication of X such that for every
f ∈ L can be extended to a continuous function fˆ on H(L), given by
fˆ(ϕ) = ϕ(f ) for each ϕ ∈ H(L).
Proof. In fact H(L) is homeomorphic to X̃ = e(X) where e : X → RL
was the evaluation map given by e(x) = (f (x))f ∈Γ . The homomorphism is given by the next commutative diagrams
e
X → e(X) ,→ X̃ = e(X)
δ
X → δ(X) ,→ H(Γ) = δ(X)
where δ is dened as δ(x) = δx and δx is the point evaluation map
δx (f ) = f (x) for every f ∈ L. We shall prove that δ(X) is in fact dense
in H(L). Given ϕ ∈ H(L), f1 , ..., fn ∈ L and > 0 there exists an
x ∈ X such that
Pn |δx (fi ) − ϕ(fi )| < for all i = 1, ..., n. Otherwise the
function g = i=1 |fi − ϕ(fi )| ∈ L would satises g ≥ and ϕ(g) = 0.
And this is imposible since ϕ is monotone. As L separates points from
closed sets in X and δ is continuous than by the Immersion's Lemma
δ : X → H(L) is an homeomorphic embedding of X in H(L). In
particular if fˆ : H(L) → R is an extension of f ∈ L ⊂ C(X) such that
fˆ(ϕ) = ϕ(f ), then if x ∈ X fˆ(x) = fˆ(δx ) = δx (f ) = f (x) = pf (x)
whereQpf : X̃ → Rf is the continuous projection over restricted to
X̃ ⊂ f ∈L Rf .
If L ⊂ C(X) then L∗ ⊂ L is the unital vector lattice of all the
bounded functions of L, and so L ⊂ C ∗ (X).
Example 9. By the last theorem it is obvious that H(C(X)) = υX
and that H(C ∗(X)) = βX .
Corollary 28. If L∗ ⊂ C ∗(X) is a unital vector lattice of continuous
real-valued bounded functions separating points from closed sets then
H(L∗ ) is a compactication of X .
40
Proof. By the last theorem H(L∗) ≈ X̃ = e(X) with the evaluation
map e : X → RL . As The f ∈ L∗ are bounded then f (X) Q
⊂ If where
If is a closed interval of R. So really we have that e : X → f ∈L∗ If ⊂
Q
Q
L∗
R
=
R
.
As
X̃
is
closed
in
∗
f
f ∈L
f ∈L∗ If then it is compact.
∗
Theorem 29. H(L) is a topological subspace of H(L∗).
Proof. Consider the restriction map r : H(L) → H(L∗) dened by
r(ϕ) = ϕ | L∗ . We are going to see that r is a topological embedding,
by means of which we shall consider H(L) as topological subspace of
H(L∗ ).
First note that if ϕ, ψ ∈ H(L) coincide on L∗ , then ϕ = ψ . Indeed,
let f ∈ L and α, β ∈ R such that α < ϕ(f ) < β and α < ψ(f ) < β .
Then for g = (f ∨ α) ∧ β ∈ L∗ , we have that ϕ(f ) = ϕ(g) = ψ(g) =
ψ(f ), as required, This shows that the map r is one-to one. It is clear
that r is continuous since it is the restriction of the natural projection
∗
from RL onto RL . In order to check that r : H(L) → H(L∗ ) is an
open map, it is enough to observe that given f1 , ..., fn ∈ L and real
numbers α1 < β1 , ..., αn < βn , we have
{ϕ ∈ H(L) : αi < ϕ(fi ) < βi , i = 1, ..., n} =
= {ϕ ∈ H(L) : αi < ϕ(gi ) < βi , i = 1, ..., n}
where gI = (fi ∨ αi ) ∧ βi ∈ L∗ for i = 1, ..., n.
Theorem 30. Let X be a Tychono space. Then any realcompactication is of the form H(L) with L ⊂ C(X) a unital vector sublattice
separating points from closed sets of X .
Proof. Let αX a realcompactication of X where α : X → αX is the
embedding. Then if we put L = C(αX) | X Then H(L) is homeomorphic ,that is equivalent to αX . In fact L separates points from closed
sets in X because αX is a Tychono space. Again the homomorphism
is given by the next squares:
α
X → α(X) ,→ αX = α(X)
41
δ
X → δX ,→ H(L) = δ(X)
Corollary 31. Let X be a Tychono space. Then any realcompactication is of the form X̃ = e(X) ⊂ RL where L ⊂ C(X) separates
points from closed sets in X and e : X → RL is the evaluation map
given by e(x) = (f (x))f ∈L.
Theorem 32. Taimanov extension theorem. Lat A a dense subspace of a topological space and f a continuous mapping of A to a
compact space Y . The mapping f has a continuous extension over
X i for every pair F1 , F2 of disjoint closed subsets of Y the inverse
images f −1(F1) and f −1(F2) have disjoint closures in the space X .
Remark 15. Please recall for the next theorem that if f ∈ L, we can
regard it as a continuous function from X into R ∪ {∞} (the one-point
compactication of R).
Theorem 33. Every f ∈ L can be extended to a (unique) continuous
function f ∗ : H(L∗) → R ∪ {∞}. Moreover,
H(L) = {ϕ ∈ H(L∗ ) : f ∗ (ϕ) 6= ∞, ∀f ∈ L}.
Proof. Let f : X → R ∪ {∞} be a function in L. By the Taimanov
extension theorem, f can be extended to a (unique) f ∗ : H(L∗ ) →
R ∪ {∞} i for every a < b < c < d ∈ R, the sets f −1 ([b, c]) and
f −1 (R − (a, d)) have disjoint closures in H(L∗ ). Let h : R → R be the
polygonal function dened by h(t) = 0 when t ∈ R−(a, d), h(t) = 1 for
t ∈ [c, d], h(t) = (t − a)/(b − a) for t ∈ (a, b), and h(t) = (t − d)/(c − d)
for t ∈ (c, d). Since L is a unital vector lattice , it is not dicult to check
that L is closed under composition with a real polygonal functions, so
h ◦ f ∈ L. In addition h ◦ f is bounded so by characterization of H(L∗ )
can be extended to a continuous real function on H(L∗ ). Note that
this extension takes the value 1 on f −1 ([b, c]) and 0 on f −1 (R − (a, b),
and therefore these sets have disjoint closure in H(L∗ ), as we required.
Now, let ϕ ∈ H(L∗ ) be such that f ∗ (ϕ) 6= ∞ for all f ∈ L. Then
ϕ can be extended to a homomorphism on L by dening f → f ∗ (ϕ).
42
This shows that {ϕ ∈ H(L∗ ) : f ∗ (ϕ) 6= ∞, ∀f ∈ L} ⊂ H(L). The
converse inclusion is clear since for every f ∈ L, the functions f ∗ and
fˆ are continuous extensions of f to H(L).
Remark 16. By the two last theorems we can conclude that for every
continuous function f : X → R admits a unique continuous extension
f β : βX → R ∪ {∞}.
Let L ⊂ C(X) be a unital vector lattice separating points from
closed sets . Now, let δ : X → H(L∗ dened as previously. Then it
has a continuous extension δ̃ : βX → H(L∗ ) that is a quotient map
by compactness. For every f ∈ L, since f ∗ ◦ δ̃ and f β are continuous
extensions of f to βX , by uniqueness, it is clear that f ∗ ◦ δ̃ = f β ,
Hence we can describe H(L∗ ) as the quotient space of βX given by the
following equivalent relations: given ξ, η ∈ βX ,
δ̃(ξ) = δ̃(η) ⇔ f ∗ (δ̃)(ξ) = f ∗ (δ̃(η))∀f ∈ L∗
⇔ f β (ξ) = f β (η))∀f ∈ L∗
⇔ f β (ξ) = f β (η))∀f ∈ L
where the last equivalence follows from the fact that, for every continuous real function f and every ξ ∈ βX , we have
f β (ξ) = limn→∞ [f ∨ (−n) ∧ n]β (ξ).
Theorem 34. For each ϕ ∈ H(L) there exists a ξ ∈ βX such that
f β (ξ) 6= ∞ and ϕ(f ) = f β (ξ) for every f ∈ L.
Proof. Given ϕ ∈ H(L) ⊂ H(L∗). choose ξ ∈ βX with δ̃(ξ) = ϕ.
Then, for every f ∈ Γ since the functions f ∗ and fˆ conincide on H(L),
it follows that ϕ(f ) = fˆ(ϕ) = f ∗ (ϕ) = f β (ξ). This means that
f β (ξ) 6= ∞, and that the homomorphism ϕ is given by evaluation at
the point ξ ∈ βX .
43
Now we are going to give a nal result for the characterization of
realcompactications. If we collect the ponits in βX taht by evaluation,
give rise to ahomomorphism on L, we obtain he space
υL X = δ̃ −1 (H(L)) ⊂ βX . Then we have that
υL X = {ξ ∈ βX : f β (ξ) 6= ∞∀f ∈ L}.
Therefore, υL X is arealcompact space and we have
X ⊂ υX ⊂ υL X ⊂ βX.
Moreover, it is not dicult to check that the restricted map
δ̃ | υL X → H(L) is closed by deniion of υL X , and therefore it is again
a quotient map. So we teh following conmutative diagram:
υL X ,→ βX
δ̃
υL (X) → H(L)
H(L) ,→ H(L∗ )
δ̃
βX → H(L∗ ).
Finally we have our last characterization of the realcompactications
of a Tychono space X .
Theorem 35. Every realcompactication of a Tychono space is the
quotient, for a certain L ⊂ C(X), of υL(X) by the next equivalent
realtion: given f ∈ L and
f β : βX → R ∪ {∞}
its unique continuous extension, then ξ ∼ η i f β (ξ) = f β (η) ∀f ∈ L.
44
Chapter 5
Samuel Realcompactication of a
metric space.
5.1 Samuel Realcompactication and Compactication.
Until now we have done a study of the realcompactications only
taking into the count the topology of X and now we want to study realcompactications for unifrom spaces. In particular for metric spaces.
So considere the following results that can be found in [12]:
Denition 24. A collection {fα : α ∈ A} of functions on X will be
said to separates points in X i whenever x 6= y in X , then for
some α ∈ A, fα (x) 6= fα (y).
Theorem 36. Uniform Immersion's lemma . Let Xα be uniform
spaces and fα : X → Xα , for each α ∈ A, a map dened
on the
Q
uniform space X . Then the evaluation map e : X → α∈A Xα is a
uniform embedding i the maps fα separates points and the uniformity
on X is the weak uniformity generated by fα
Recall that the weak uniformity generated by the maps fα : X → Xα
is the weakest uniformity making each fα uniformly continuous.
Let (X, d) be a metric space, and consider the uniformity generated
by its metric, Ud .
45
Denition 25. Let (X, d) be a metric space and U C(X) the set of
the uniformly continuous real-valued functions on (X, Ud). Then the
Samuel Realcompactication of (X, d), srX , is the smallest realcompactication of it such that every uniformly continuous function
f ∈ U C(X) can be extended to a continuous function f˜ : srX → R.
By the last section we know that in fact srX is a realcompactication of (X, d) and srX = e(X) where e, the evaluation map, is a
topological embedding of X in RU C(X) where e(x) = (f (x))f ∈U C(X) .
We want to know if this topological embedding is really a uniform
embedding of (X, Ud ) in RU C(X) with the usual uniformity. By the
Uniform immersion's lemma we have to look for the weak uniformity
generated by the set of functions U C(X)([5]). It is the uniformity U0
which has the subbase
{(x, y) ∈ X × X : |f (x) − f (y)| < }
where f ∈ U C(X) and > 0, and we want to know if it is equivalent
to Ud .
Let
U0 = {(x, y) ∈ X × X : |f (x) − f (y)| < }
for a certain f ∈ U C(X) and > 0. Then as f ∈ U C(X) then there
exists a δ > 0 such that for every x, y ∈ X satisfying d(x, y) < δ then
|f (x) − f (y)| < . So
U = {(x, y) ∈ X × X : d(x, y) < δ} ⊂ U0
and U0 ∈ Ud . That is U0 is coarser that Ud .
The converse is not satised in general. In fact let d be the discrete
metric on a not countable space X , that is, d(x, x) = 0 and d(x, y) = 1
∀x 6= y . Then every function f : (X, Ud ) → R is uniformly continuous.
If U0 is ner than Ud then we will have the equivalent and then U0
will be metrizable. If U0 is metrizable then it has a numerable base for
it. But this is impossible because it can be proved that it can't be nd
a numerable sequence of f ∈ U C(X) that generates U0 .
46
So we conclude that e : (X, U0 ) → RU C(X) is a uniformly continuous
embedding of X ∼
= e(X) in RU C(X) but that e : (X, Ud ) → RU C(X) is
only a topological embedding with e a uniformly continuous function,
because the inverse function e−1 : (e(X), URU C(X) | e(X) ) don't need to
be a uniformly continuous function.
Equivalently to the Samuel Realcompactication we have already
dened in a previous chapter the Samuel Compactication of a metric
space in the next way:
Denition 26. Let (X, d) be a metric space and U C ∗(X) be the set of
all the bounded uniformly continuous real-valued functions on (X, Ud).
Then the Samuel Compactication of (X, d), sX , is the smallest
compactication of (X, d) such that every f ∈ U C(X)∗ can be extended
to a (bounded)continuous function f˜ : sX → R.
Recall that the weak uniformity generated dy U C ∗ (X) is uniformly
equivalent to the weak uniformity generated by U C(x), U0 . So have
∗
again the uniform embedding e : (X, U0 ) → RU C (X)
We can construct the Samuel Compactication in other way. Let
(X̃, Ũ0 ) be the uniform completion of (X, U0 ). Then it is uniformly
isomorphic to the Samuel Compactication, because as it is compact
then by uniqueness of the completion they have to coincide.
Theorem 37. The next cases are equivalent:
1. srX = sx;
2. U C ∗(X) = U C(X)
47
5.2 Conclusion.
Remember from the last chapter that if (X, d) is a metric space
then its Samuel Realcompactication can be expressed as a quotient of
a subspace of the Stone-ƒech compactication of (X, d): given
f ∈ U C(X) and let
f β : βX → R ∪ {∞}
be its unique continuous extension, then the Samuel Realcompactication of the metric space (X, d) is srX = υU C(X) X/ ∼ where
υU C(X) X = {ξ ∈ βX : f β (ξ) 6= ∞, ∀f ∈ U C(X)} ⊂ βX
and ξ ∼ η i f β (ξ) = f β (η) ∀f ∈ U C(X).
Now suppose that our metric space (X, d) has the property that
every uniformly continuous function f ∈ U C(X) is bounded for the
closed metric balls of the metric space then we have the nest result:
Theorem 38. Let (X, d) be a metric space such that for every
f ∈ U C(X) the image of the closed metric balls are bounded in R and
let x0 ∈ X . Then
υU C(X) X =
∞
[
clβX Bn (x0 ).
n=1
Proof. ⇒Let ξ ∈ clβX Bn(x0) for some n ∈ N. For every f ∈ U C(X),
f is bounded in Bn (x0 ) so f β (ξ) 6= ∞ and by denition of υU C(X) X ,
ξ ∈ υU C(X) X .
⇐ Let ξ ∈ υU C(X) X and consider f (y) = d(y, x0 ) ∈ U C(X). Since
f β (ξ) 6= ∞ we can choose n > f β (ξ). Then, ξ ∈ clβX Bn (x0 ). Otherwise, there exists an open neighborhood V of ξ in βX such that
V ∩ X ⊂ {x ∈ X : d(x, x0 ) > n}. Since ξ ∈ clβX V = clβX (V ∩ X), we
have that f β (ξ) > n, which is a contradiction.
Example 10. Now we give some examples of spaces that have the
porperty that for every f ∈ U C(X) the image of the closed balls in
(X, d) are bounded:
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1. Proper metric spaces, that is, metric spaces for which their metric
has the Heine-Borel Property.
2. Normed spaces.
3. More generally, length spaces.
4. For a normed space (X, k.k) then we can construct a metric space
(X, d) with d(x, y) = kx − yk + |f (x) − f (y)| with f ∈ U C(X)
and such that (X, d) is not necessarily a normed or a length space.
So we are looking for metric space such that for every f ∈ U C(X)
the image of the closed balls in (X, d) are bounded. Remember from
the second section that in a metric space (X, d) we can characterize
the bornology of the bounded sets in the sense of Bourbaki of (X, d),
BBd(X) in the next way:
Let (X, U) be a uniform space and B ⊂ X . Then each real valued
uniformly continuos function f ∈ U C(X) is bounded on B i B is
bounded in the sense of Bourbaki.
So if we ask to a metric space (X, d) that the images of its closed
metric balls by all the uniformly-continuos real-valued functions
f ∈ U C(X) have to be bounded then, as the closed metric balls form
a base for the metric bornology of (X, d), Bd X , this is equivalent to
ask that
Bd(X) = BBd(X).
By the theorem of unifrom metrization in the second section we have
the next corollary:
49
Corollary 39. Let BBd(X) the bornology of the bounded sets in the
sense of Bourbaki of (X, d). The following conditions are equivalent:
1. BBd(X) = Bd0 (X) for some uniformly equivalent metric d0 on
(X, d)
2. BBd(X) has a countable base {Bn : n ∈ N} such that ∃δ > 0
∀n ∈ N, with
Bn δ ⊆ Bn+1
where
Bnδ
= {x ∈ X : d(x, Bn ) < δn } =
[
Bδ (x).
x∈Bn
Obviously BBd (X) = BBd0 (X) for every uniformly equivalent metric d0 in (X, d) so we have our problem result sumarized in the next
theorem:
Theorem 40. Let (X, d) be a metric space, then the following two
results are equivalent:
1. the bornology of the Bourbaki bounded sets of (X, d) is equal to
the metric bornology of (X, d), that is, BBd(X) = Bd(X);
2. the Samuel Realcompactication of (X, d) is the next quotient
space
srX =
∞
[
clβX Bn (x0 )/ ∼
n=1
where ξ ∼ η i f β (ξ) = f β (η) ∀f ∈ U C(X) for the unique
extension f β : βX → R ∪ {∞} of f ∈ U C(X).
50
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