Topol. Algebra Appl. 2015; 3:11–25
Research Article
Open Access
A. Dorantes-Aldama, R. Rojas-Hernández, and Á. Tamariz-Mascarúa*
The partially pre-ordered set of
compactifications of C p (X, Y)
DOI 10.1515/taa-2015-0002
Received September 19, 2014; revised January 16, 2015; accepted April 3, 2015
Abstract: In the set of compactifications of X we consider the partial pre-order defined by (W , h) ≤X (Z, g)
if there is a continuous function f : Z → W , such that (f ∘ g)(x) = h(x) for every x ∈ X. Two elements
(W , h) and (Z, g) of K(X) are equivalent, (W , h) ≡X (Z, g), if there is a homeomorphism h : W → Z such
that (f ∘ g)(x) = h(x) for every x ∈ X. We denote by K(X) the upper semilattice of classes of equivalence of
compactifications of X defined by ≤X and ≡X . We analyze in this article K(C p (X, Y)) where C p (X, Y) is the
space of continuous functions from X to Y with the topology inherited from the Tychonoff product space Y X .
We write C p (X) instead of C p (X, R).
We prove that for a first countable space Y, K(C p (X, Y)) is not a lattice if any of the following cases happen:
(a) Y is not locally compact,
(b) X has only one non isolated point and Y is not compact.
Furthermore, K(C p (X)) is not a lattice when X satisfies one of the following properties:
(i) X has a non-isolated point with countable character,
(ii) X is not pseudocompact,
(iii) X is infinite, pseudocompact and C p (X) is normal,
(iv) X is an infinite generalized ordered space.
Moreover, K(C p (X)) is not a lattice when X is an infinite Corson compact space, and for every space X,
K(C p (C p (X))) is not a lattice. Finally, we list some unsolved problems.
Keywords: Compactifications of spaces of continuous functions, lattices, b-lattices, k-embedded subsets, Cembedded subsets, retracts, weakly pseudocompact spaces
MSC: Primary 54C35, 54D35, 03G10, Secondary 54D45, 54C45.
1 Introduction, basic definitions and notations
Throughout this article all topological spaces are considered Tychonoff and with more than one point if the
contrary is not specified. The symbol ω denotes both the first infinite cardinal number and the natural numbers. For an ordinal number α, [0, α) denotes the space of ordinals strictly less than α with its order topology,
and α denotes the discrete space of cardinality |α|; in particular 2 means the discrete space {0, 1}. For a space
X and a subset A of X, clX A, or simply clA, if there is no possibility for confusion, is the closure of A in X. A
subset A of a space X is a G δ subset of X if there is a countable collection {U n : n < ω} of open subsets of X
*Corresponding Author: Á. Tamariz-Mascarúa: Departamento de Matemáticas, Facultad de Ciencias, UNAM, Circuito exterior s/n, Ciudad Universitaria, CP 04510, México D. F., México, E-mail: [email protected]
A. Dorantes-Aldama: Departamento de Matemáticas, Facultad de Ciencias, UNAM, Circuito exterior s/n, Ciudad Universitaria,
CP 04510, México D. F., México, E-mail: [email protected]
R. Rojas-Hernández: Departamento de Matemáticas, Facultad de Ciencias, Universidad Autónoma del Estado de México, Instituto Literario No. 100, Col. Centro, C.P. 50000, Toluca, Estado de México, México, E-mail: [email protected]
© 2015 A. Dorantes-Aldama et al., licensee De Gruyter Open.
This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 License.
12 | A. Dorantes-Aldama, R. Rojas-Hernández, and Á. Tamariz-Mascarúa
⋂︀
such that A = n<ω U n . A subset A of a space X is G δ -dense in X if every non-empty G δ subset of X intersects
A.
Recall that a partial order (Z, ≤) is a lattice if every pair of elements a, b ∈ Z has a least upper bound and
a greatest lower bound. A lattice (Z, ≤) is complete if every non-empty subset of Z has a least upper bound
and a greatest lower bound.
A pair (Z, g) (respectively, (gX, g)) is a compactification of X if Z (resp., gX) is a compact space and g :
X → Z (resp., g : X → gX) is an embedding such that g[X] is dense in Z (resp., in gX). Let (W , h) and (Z, g)
be two compactifications of X. We say that (W , h) ≤X (Z, g) (or just (W , h) ≤ (Z, g)), if there is no confusion
regarding the identity of the space X compactified by (W , h) and (Z, g) if there is a continuous function f :
Z → W such that g = f ∘ h. These compactifications (W , h) and (Z, g) are equivalent if (W , h) ≤X (Z, g) and
(Z, g) ≤X (W , h). We denote this fact by W ≡X Z (or by W ≡ Z if there is no possibility for confusion). Observe
that two compactifications (W , h) and (Z, g) of X are equivalent if there is a homeomorphism f : Z → W
such that (h ∘ f )(x) = g(x) for every x ∈ X. We denote by K(X) the ordered set of ≡X -equivalence classes
of compactifications of X with the partial order defined by ≤X . This partial order in K(X) will be denoted by
the same symbol ≤X . The partially ordered set (K(X), ≤X ) is a complete upper semilattice. The symbol (βX, β)
denotes the Stone-Čech compactification of X, and (αX, α) denotes the one-point compactification of X for a
locally compact space X.
A compactification (bX, b) of a space X is primary if it is obtained by collapsing a compact subspace
of the Stone-Čech remainder βX \ X to a single point. (In some literature this class of compactifications are
called simple. In this paper we follow the terminology used in [12]). A space X is weakly pseudocompact if
X is G δ -dense in one of its compactifications, and is primary pseudocompact if X is G δ -dense in a primary
compactification.
We say that (K(X), ≤X ) is a b-lattice if the primary compactifications of X are dense in (K(X), ≤X ); that is,
if for each (bX, b) ∈ K(X), there is a compact set K ⊆ βX \ X such that βX/K ≤X bX. Observe that if X is locally
compact then K(X) is a b-lattice because the one-point compactification of X is a primary compactification.
We denote by C p (X, Y) the space of continuous functions from X to Y with the topology of the pointwise
convergence which is that inherited from the product topology in Y X . The product topology is generated by
the sets of the form:
[x1 , . . . , x n ; B1 , . . . , B n ] = {f ∈ Y X : f (x i ) ∈ B i , i = 1, . . . , n}
where n < ω, {x1 , . . . , x n } ⊆ X, and B1 , . . . , B n are open subsets of Y, as a base. So, the topology in
C p (X, Y) is generated by the collection {[x1 , . . . , x n ; B1 , . . . , B n ] ∩ C p (X, Y) : n < ω, {x1 , . . . , x n } ⊆
X, and B1 , . . . , B n are open subsets of Y } as a base. When Y is equal to the real line, we just write C p (X)
instead of C p (X, R).
In [5] some new results about weakly pseudocompact spaces were obtained, and the authors improved
several results on this topic obtained by F.W. Eckertson [6] and S. García-Ferreira and A. García-Máynez [8].
Part of the discussions made in [5] are related to the structure of the partially ordered set (K(X), ≤X ). In particular, in [5] the following results were proved:
Proposition 1.1. Let X be a space. If K(X) is a b-lattice, then X is weakly pseudocompact if and only if X is
primary pseudocompact.
Corollary 1.2. If X is not locally pseudocompact and K(X) is a b-lattice, then X is not weakly pseudocompact.
Theorem 1.3. Let X be a first countable space. Then K(X κ ) is a b-lattice (resp., lattice, complete lattice) if and
only if X is locally compact and either κ is finite or X is compact. In particular, K(ω κ ) and K(Rκ ) are b-lattices
(resp., lattice, complete lattice) iff κ < ω, and K(Sκ ) is not a b-lattice for every κ > 0, where S is the Sorgenfrey
line.
Proposition 1.4. Let D be a dense subspace of ω κ . Then, K(D) is b-lattice (resp., lattice, complete lattice) if
and only if κ < ω.
The partially pre-ordered set of compactifications of C p (X, Y)
|
13
The main goal in this article is to analyze (K(C p (X, Y)), ≤C p (X,Y) ) where Y is a first countable space. In particular, our concern will be centered in knowing when (K(C p (X, Y)), ≤C p (X,Y) ) is a b-lattice.
The partially ordered set (K(X), ≤X ) has been studied in several papers (see for example [9] and [13]). In
this article we closely follow [14] and [15]. It is worth mentioning that two classic texts that develop the theory
of compactifications of Hausdorff spaces are [3] and [11].
In Section 2, we include some basic results about K(C p (X, Y)) and we prove that if Y is first countable
and X is countable, then K(C p (X, Y)) is a lattice if and only if Y is locally compact and either X is finite or Y
is compact and X is discrete. Sections 3 and 4 are devoted to introducing and analyzing k- and k̃-embedded
subspaces (Y ⊆ X is k-embedded in X if for every compactification bY of Y, there is a compactification aX of
X such that cl aX Y ≤ bY. A space Y is ̃︀
k-embedded in X if Y is homeomorphic to a k-embedded subspace of X)
and we prove that all closed subsets of a locally compact space and all retracts are k-embedded subspaces. As
a consequence we obtain that for a first countable non-locally compact space Y, K(C p (X, Y)) is not a lattice,
no matter what the space X is, and that K(C p ([0, α), Y)) and K(C p (Ψ(A), Y)) are not lattices for every ordinal
number α > ω and every noncountable almost disjoint family A, where Y is first countable. In Section 5 we
prove that K(C p (X, Y)) is not a lattice when Y is a non-compact first countable space and X is either a nonpseudocompact zero-dimensional space, or X has only one non isolated point. The main theorems appear in
Section 6: The semilattice K(C p (X)) is not a lattice when X has at least one of the following properties: (i) X
has a non-isolated point with countable character, (ii) X is not pseudocompact, (iii) X is pseudocompact and
C p (X) is normal, or (iv) X is an infinite generalized ordered space. Additionally, K(C p (X)) is not a lattice when
X is an infinite Corson compact space, and for every space X, K(C p (C p (X))) is not a lattice. Finally, in Section
7 we list some unsolved problems.
2 Some first remarks about K(C p (X, Y))
Lemma 2.1. [14] If K(X) is a b-lattice, then K(X) is a lattice.
Theorem 2.2. [15] If X is a first countable space, then X is locally compact if and only if K(X) is a complete
lattice.
Corollary 2.3. If X is a first countable space then the following are equivalent:
1. X is locally compact,
2. K(X) is a complete lattice,
3. K(X) is a lattice, and
4. K(X) is a b-lattice.
Related to Theorem 1.3, we have the following two propositions and one example. We will obtain a more
general result in Corollary 3.15 below.
Proposition 2.4. For a zero-dimensional space X, K(C p (X, Z)κ ) (respectively, K(C*p (X, Z)κ )) is a b-lattice
(resp., lattice, complete lattice) if and only if X and κ are finite.
Proof. Indeed, on the one hand C p (X, Z)κ (resp., C*p (X, Z)κ ) is a dense subset of Z|X|·κ . If either X or κ is
infinite, [0, 1]|X|·κ is a connected compactification of C p (X, Z)κ (resp., C*p (X, Z)κ )). This last space is dense in
Z |X|·κ . So, K(C p (X, Z)κ ) (resp., K(C*p (X, Z)κ )) is not a b-lattice (see Corollary 4.2 and Theorem 4.5 in [5]). On
the other hand, if |X | and κ are finite, then C p (X, Z)κ = Z|X|·κ is locally compact and so, K(C p (X, Z)κ ) is a
b-lattice.
Example 2.5. A σ-compact space Z for which K(Z) is not a b-lattice: If X is an infinite Eberlein compact zerodimensional space, then C p (X, Z) is σ-compact (see Theorem 4.18 in [4]) but K(C p (X, Z)κ ) is not a lattice for
every cardinal κ.
14 | A. Dorantes-Aldama, R. Rojas-Hernández, and Á. Tamariz-Mascarúa
With similar arguments to those given in the proof of Proposition 2.4, we obtain:
Proposition 2.6. For every zero-dimensional space X and every cardinal number κ, K(C p (X, Q)κ ) (resp.,
K(C p (X, Q)κ ), K(C p (X, ω ω ))κ , K(C*p (X, ω ω ))κ ) is not a lattice.
If X is a connected space and Y is a totally disconnected space such that K(Y) is a b-lattice, then, since
C p (X, Y) ∼
= Y, trivially K(C p (X, Y)) is a b-lattice. So, in order to avoid trivial cases, and inspired in Propositions 2.4 and 2.6, we are tempted to request that X and Y be spaces that satisfy that C p (X, Y) be dense in X Y .
Nevertheless, this request will only be explicitly necessary in a few cases (see, for example Lemma 2.7 and
Theorem 2.8, below) as we will confirm throughout this article.
Lemma 2.7. Assume that C p (X, Y) is dense in Y X . Then, C p (X, Y) is locally compact if and only if Y is locally
compact and either X is finite or Y is compact and X is discrete.
Proof. Assume that C p (X, Y) is locally compact. Let y ∈ Y. Let V be an open set with y ∈ V. Since C p (X, Y)
is dense in Y X , we can take g ∈ [x; V] ∩ C p (X, Y). Then, there exist n < ω, x1 , ..., x n ∈ X and open subsets of
Y, A1 , ..., A n such that g ∈ [x1 , ..., x n ; A1 , ..., A n ] ∩ C p (X, Y) ⊆ clC p (X,Y) ([x1 , ..., x n ; A1 , ..., A n ] ∩ C p (X, Y)) ⊆
[x; V] ∩ C p (X, Y) where clC p (X,Y) ([x1 , ..., x n ; A1 , ..., A n ] ∩ C p (X, Y)) is compact. Again, using the fact that
C p (X, Y) is dense in Y X we can prove that
T = clC p (X,Y) ([x0 , ..., x k ; A0 , ..., A k ] ∩ C p (X, Y))
is equal to
{h ∈ C p (X, Y) : h(x i ) ∈ clY A i for every i ∈ {1, ..., k}};
⋂︀
that is, T = [x0 , ..., x k ; clY A0 , ..., clY A k ] ∩ C p (X, Y). So, T is closed in Y X and it is dense in i≤k π −1
x i [clY A i ].
⋂︀
⋂︀
⋂︀
Then, T = i≤k π −1
[cl
A
]
is
compact.
Hence,
each
cl
A
is
compact,
x
∈
A
⊆
cl
A
⊆
V, and
xi
Y i
Y i
i≤n i
i≤n Y i
either X = {x1 , ..., x n } or Y is compact and X is discrete. Moreover, since y and V were taken arbitrarily, Y is
locally compact.
Theorem 2.8. Let Y be a first countable topological space and X be a countable space. Then the following are
equivalent:
1. K(C p (X, Y)) is a complete lattice,
2. K(C p (X, Y)) is a lattice,
3. K(C p (X, Y)) is a b-lattice,
4. C p (X, Y) is locally compact,
5. Y is locally compact and either X is finite or Y is compact and X is discrete.
Proof. Since X is countable, it is zero-dimensional ([7], Corollary 6.2.8), and so C p (X, Y) is dense in Y X . Then,
χ(C p (X, Y)) ≤ |X | · χ(Y) (see [3], Corollary 4.4), and C p (X, Y) is first countable. Thus, by Corollary 2.3 the
statements (1), (2), (3) and (4) are equivalent. The equivalence (4) ⇔ (5) is Lemma 2.7.
3 k- and strongly k-embedded subspaces
The following result is essential for our purposes. It was already proved in [10] for locally compact spaces.
Theorem 3.1. Let X and Y be two homeomorphic spaces. If K(X) is a b-lattice (respectively, lattice, complete
lattice), then K(Y) is a b-lattice (resp., lattice, complete lattice).
Proof. We are only going to prove the theorem for the case when K(X) is a b-lattice. Let h : Y → X be a homeomorphism. Let (bY , b) be a compactification of Y. There exist a compactification (T, l) of X and a homeomorphism H : bY → T such that H b[Y] = l ∘ h. Since K(X) is a b-lattice, there are a compact space
K ⊆ βX \ β X [X] and a continuous function f : T → βX/K such that f (l(x)) = β X (x) for all x ∈ X. Additionally,
The partially pre-ordered set of compactifications of C p (X, Y)
|
15
there is a homeomorphism φ : (βY , β Y ) → (βX, β X ) such that φ β Y [Y] = β X ∘ h. Take M = φ−1 [K]. M is a
compact subspace of βY \ β Y [Y]. We only have to prove that there is a continuous function g : bY → βY /M
such that g(b(y)) = β Y (y) for every y ∈ Y. Define g(z) = (φ−1 ∘ f ∘ H)(z). Of course, g is continuous. It remains to prove that g b[Y] = β Y Y. In fact, if y ∈ Y then g(b(y)) = (φ−1 ∘ f )(H(b(y)))) = φ−1 (f (l(h(y)))) =
φ−1 (β X (h(y))) = β Y (y).
Example 3.2. 1. The identity function from the Sorgenfrey line S to R is a condensation, K(R) is a b-lattice
but K(S) is not a lattice (see Theorem 1.3).
2. Let E(ω1ω ) be the absolute of ω1ω . Because of Theorem 3.3 in [6], K(E(ω1ω )) is a b-lattice but K(ω1ω ) is not
a lattice, and ω1ω is a perfect and irreducible image of E(ω1ω ).
Recall that a subspace A of a space X is a retract of X if there is a continuous function r : X → A such that
r(a) = a for all a ∈ A. In this case, the function r is called retraction. Observe that if A is a retract of X, then
A is C-embedded in X.
Proposition 3.3. Let S be a set and for each s ∈ S let X s be a topological space. Assume that for each s ∈ S,
∏︀
∏︀
C s is a retract of X s . Then, s∈S C s is a retract in s∈S X s .
∏︀
∏︀
Proof. Let r s : X s → C s be a retraction for each s ∈ S. Then the function r : s∈S X s → s∈S C s defined by
r((x s )s∈S ) = (r s (x s ))s∈S is a retraction.
Let X and Y be two topological spaces and y ∈ Y. We denote by ct y the constant function that sends all x ∈ X
to the point y ∈ Y. The collection of all the constant functions from X to Y is denoted by Cts(X, Y).
Proposition 3.4. For all spaces X and Y, Cts(X, Y) is a retract in Y X . In particular, Cts(X, Y) is a retract of
C p (X, Y).
Proof. Let z0 ∈ X be fixed. The function r : Y X → Cts(X, Y) defined by r(f ) = ct f (z0 ) is a retraction.
As we have already agreed, for each locally compact and non compact space Y, αY is the one point compactification of Y. We denote by p Y the point in αY not belonging to Y which compactifies Y.
Remark 3.5. 1. For every pair of topological spaces X and Y, Cts(X, Y) is homeomorphic to Y. Indeed, the
function ϕ Y : Y → Cts(X, Y) defined by ϕ Y (y) = ct y is a homeomorphism.
2. If Y is locally compact and non-compact, then
αCts(X, Y) = Cts(X, Y) ∪ {ct p Y } = Cts(X, αY).
We can generalize Remark 3.5.(2):
Lemma 3.6. Let Y be a non-compact space. Let (bY , b) be a compactification of Y. Then, Cts(X, bY) =
cl(bY)X Cts(X, b[Y]).
Proof. Since Cts(X, bY) is homeomorphic to bY (Remark 3.5.(1)), Cts(X, bY) is compact. Moreover, Cts(X, b[Y]) ⊆
Cts(X, bY). Thus, cl(bY)X Cts(X, b[Y]) ⊆ Cts(X, bY). Let z ∈ bY \ b[Y]. We are going to prove that ct z ∈
cl(bY)X Cts(X, b[Y]). Let [x1 , ..., x n ; A1 , ..., A n ] be a canonical open subset of (bY)X which contains ct z . Then
⋂︀
⋂︀
⋂︀
z ∈ i≤n A i . So, i≤n A i is a non-empty open subset of bY. Thus, we can take a point y0 ∈ i≤n A i ∩ b[Y]. It
happens then that ct y0 ∈ [x1 , ..., x n ; A1 , ..., A n ] ∩ Cts(X, b[Y]).
The previous lemma motivates the introduction of the following natural concepts.
Definition 3.7. 1. [5] A subspace Y of a space X is k-embedded in X if for each compactification (K, i) of
Y, there exists a compactification (M, j) of X such that (clM j[Y], j Y) ≤ (K, i); that is, there exists a
continuous function h : K → clM Y such that h(i(y)) = j(y) for every y ∈ Y.
2. A subspace Y of a space X is strongly k-embedded in X if for each compactification (K, i) of Y there is a
compactification (M, j) of X such that K ≡Y clM Y; that is, there exists a homeomorphism h : K → clM Y
such that h(i(y)) = j(y) for every y ∈ Y. (Observe that the relations of being k-embedded and strongly
k-embedded are transitive.)
16 | A. Dorantes-Aldama, R. Rojas-Hernández, and Á. Tamariz-Mascarúa
Proposition 3.8. Let X ⊆ Y and define φ : K(Y) → K(X) by φ((Z, a)) = (clZ a[X], a X). Then, φ is a
homomorphism of partially ordered sets with φ[K(Y)] dense in K(X) if and only if X is k-embedded in Y .
Proof. Assume that X is k-embedded in Y . We are going to prove that φ is a homomorphism. Let (W , a), (Z, b)
be two compactifications of Y and suppose that (W , a) ≤ (Z, b). Then, there exists a continuous function
f : Z → W such that f ∘ b = a. Observe that f [clZ b[X]] ⊆ clW f [b[X]] = clW a[X]. Therefore, the function
f clZ b[X] witnesses that φ((W , a)) ≤ φ((Z, b)).
Now we are going to prove that φ[K(X)] is dense in K(Y). Let (W , a) be a compactification of X, since X
is k-embedded in Y , there is a compactification (Z, b) of Y such that (clZ b[X], b X) ≤ (W , a). Therefore,
φ((Z, b)) ≤ (W , a).
Suppose that φ is a dense homomorphism of partially ordered sets, and let (W , a) be a compactification
of X; then there is a compactification (Z, b) of Y such that φ((Z, b)) ≤ (W , a); this means that (clZ b[X], b X) ≤ (W , a), therefore X is k-embedded in Y .
Proposition 3.9. 1. Every closed subset of a locally compact space X is k-embedded in X.
2. Every closed and C* -embedded subspace of X is strongly k-embedded in X.
Proof. (1) Let F be a closed subset of the locally compact space X. Let (K, j) be a compactification of F. We
claim that clαX F ≤ (K, j). There are two cases, if clαX F = F, then F has to be compact because it remains
closed in αX. In this case K = F. The second possibility happens when clαX F = {p X } ∪ F. In this case it is easy
to see that clαX F ≡ αF and clαX F ≤ K.
(2) Let A be a closed and C* -embedded subspace of X. Choose a compactification (K, b) of A. Since A is C* embedded in X, clβX A is equivalent to the Stone-Čech compactification of A. Denote by b̃ the continuous
extension of b to clβX A, b̃ : clβX A → K. Consider the partition D = {{x} : x ∈ A ∪ (βX \ clβX A)} ∪ {b̃−1 (y) :
y ∈ K \ b[A]}. Denote by q and Y the quotient map and the quotient space induced by D on βX, respectively.
Then q identifies two points x and y if and only if x, y ∈ clβX A \ A and b̃(x) = b̃(y).
Let us observe that the space Y is compact. We shall verify that Y is a Hausdorff space. Choose x, y ∈
βX with q(x) ≠ q(y). If x, y ∈ βX \ clβX A choose disjoint open neighborhoods U and V of x and y in βX,
respectively, such that U, V ⊆ βX \ clβX A. Then q(U) and q(V) are disjoint open sets in Y with q(x) ∈ q(U)
and q(y) ∈ q(V). If x ∈ clβX A and y ∈ βX \ clβX A choose disjoint open sets U and V in βX with clβX A ⊆ U
and y ∈ V. It follows that q(U) and q(V) are disjoint open sets in Y with q(x) ∈ q(U) and q(y) ∈ q(V). The
case x ∈ βX \ clβX A and y ∈ clβX A is similar. Finally, assume that x, y are both in clβX A. Let us observe
that q(x) ≠ q(y) implies b̃(x) ≠ b̃(y). Choose open sets U1 and V1 in K such that b̃(x) ∈ U1 , b̃(y) ∈ V1
and clU1 ∩ clV1 = ∅. It follows that clb̃−1 (U1 ) ∩ clb̃−1 (V1 ) = ∅. Choose open sets U2 and V2 in βX with
U2 ∩ clβX A = b̃−1 (U1 ) and V2 ∩ clβX A = b̃−1 (V1 ). Finally, let U = U2 \ (clU2 ∩ clV2 ) and V = V2 \ (clU2 ∩ clV2 ).
Then U ∩ clβX A = b̃−1 (U1 ) and V ∩ clβX A = b̃−1 (V1 ) and U ∩ V = ∅. Therefore, q(U) and q(V) are disjoint open
neighborhoods of q(x) and q(y) in Y, respectively.
Notice that A closed in X implies X ⊆ A ∪ (βX \ clβX A). Then {{x} : x ∈ X } ⊆ D and so q X is an
embedding. It follows that (Y , q X) is a compactification of X. We claim that (clY q(A), q A) is equivalent
to (A, b). The map q clβX A : clβX A → clY q(A) is a quotient map and the function b̃ : clβX A → K is constant
on the fibers of q clβX A, then there exists a continuous map d : clY q(A) → K such that b̃ = d ∘ q clβX A. Let
us observe that d is one-to-one and hence a homeomorpism. In addition, d(q(x)) = b̃(x) = b(x) for any x ∈ A.
Hence (clY (q(A)), q A) is equivalent to (A, b). So we have proved that A is strongly k-embedded in X.
Observe that the statement in (1) of the previous proposition cannot be strengthened. An example: Let A be a
mad family such that βΨ(A) = αΨ(A) where Ψ(A) is the Mrówka-Isbell space defined by A. Then A is closed
in Ψ(A) but it is not a strongly k-embedded subset of Ψ(A).
Corollary 3.10. Every retract of a space X is C- and strongly k-embedded in X.
Corollary 3.11. If X is normal and A is closed in X then A is strongly k-embedded in X.
Proposition 3.12. Suppose that X ⊆ Y ⊆ Z. If X is k-embedded in Z, then X is k-embedded in Y .
The partially pre-ordered set of compactifications of C p (X, Y)
|
17
Proof. Let (K, a) be a compactification of X. There is a compactification (Z ′ , c) of Z such that (clZ ′ c[X], c X) ≤ (K, a). Observe that (clZ ′ c[Y], c Y) is a compactification of Y . Let Y ′ = clZ ′ c[Y]. Hence clY ′ c[X] =
clZ ′ c[X] ∩ Y ′ = clZ ′ c[X]. Therefore (clY ′ c[X], c X) ≤ (K, a).
In [5] the following result was proved.
Theorem 3.13. If Y is k- and C* -embedded in X and K(X) is a b-lattice, then K(Y) is a b-lattice.
The converse of Theorem 3.13 is not true: K(Rω ) is not a lattice but [0, 1]ω is a compact retract of Rω .
We can also obtain:
Theorem 3.14. If Y is homeomorphic to a k-embedded subspace of X and K(X) is a lattice, then K(Y) is a
lattice.
Proof. Let (M, b) and (N, c) be two compactifications of Y. By hypothesis, there is an embedding i : Y → X,
such that i[Y] is k-embedded in X. Hence there are compactifications (K, a), (K ′ , a′ ) of X, and continuous
functions f : M → clK a[i[Y]], h : N → clK ′ a′ [i[Y]] such that f ∘ b = a ∘ i and h ∘ c = a′ ∘ i.
Since K(X) is a lattice, there exists a compactification (Z, z) of X and continuous functions h : K → Z,
h′ : K ′ → Z such that h ∘ a = z and h′ ∘ a′ = z.
Now observe that h ∘ f : M → clZ z[i[Y]] and h ∘ g : N → clZ z[i[Y]] are continuous functions such that
h ∘ f ∘ b = z ∘ i and h ∘ g ∘ c = z ∘ i. Therefore (clZ z[i[Y]], z ∘ i) ≤ (M, b) and (clZ z[i[Y]], z ∘ i) ≤ (N, c).
We conclude that K(Y) is a lattice.
We have already seen that Cts(X, Y) is a retract of C p (X, Y), so Cts(X, Y) is a strongly k-embedded subspace of C p (X, Y). Hence, we obtain a generalization of Propositions 2.4 and 2.6 above.
Corollary 3.15. Let Y be a first countable space which is not locally compact. Then, for every topological
space X, K(C p (X, Y)) is not a lattice.
Proof. If K(C p (X, Y)) were a lattice, then K(Y) is a lattice (Theorem 3.1). But Y is first countable, so by Corollary 2.3, Y must be locally compact.
Because of Theorem 3.14, we obtain:
Theorem 3.16. Let X be a topological space and Y be a non-compact first countable space. Then K(C p (X, Y ω ))
(and K(C p (X × ω, Y))) is not a lattice.
Proof. Assume that K(C p (X, Y ω )) is a lattice. Then, it happens that K(Cts(X, Y ω )) is a lattice (Theorem 3.14
and Corollary 3.10). By Remark 3.5.(1) and Theorem 3.14, K(Y ω ) is a lattice. But this last assertion contradicts
Theorem 2.8.
The partially ordered set K(C p (X × ω, Y)) is not a lattice either because
ω
C p (X × ω, Y) ∼
= C p (X, Y ).
What can we say when X has a dense k-embedded subspace? The following results give us some answers.
Proposition 3.17. Let X ⊆ Y . If X is locally compact and dense in Y , then X is k-embedded in Y if and only if
either Y = X or Y = αX.
Proof. Assume that X is k-embedded in Y . There is a compactification bY ∈ K(Y), such that clbY X ≤X αX.
Since X is dense, clbY X = bY and since bY is a compactification of X, then αX ≤X bY . Therefore bY ≡X αX.
Now suppose that Y = αX. Let bX be a compactification of X, then αX ≤ bX.
Corollary 3.18. Let X ⊆ Y . If X is locally compact and dense in Y , and if Y is compact, then X is k-embedded
in Y if and only if Y = αX.
Proposition 3.19. Let X ⊆ Y . If X is dense in Y and |Y \ X | ≥ 2, then X is not k-embedded in Y .
18 | A. Dorantes-Aldama, R. Rojas-Hernández, and Á. Tamariz-Mascarúa
Proof. Let x, y be two distinct points in Y \ X and K = {x, y}. Let (aY , a) be a compactification of Y . Then
(aY /a[K], π ∘ a X) is a compactification of X where π : aY → aY /a[K] is the natural projection. If X is kembedded in Y , there is a compactification (cY , c) of Y and there is a continuous function f : aY /a[K] → cY
such that f ∘ π ∘ a X = c X. Observe that f ∘ π ∘ a = c, because X is dense in Y . Since K ⊆ Y , c(x) ≠ c(y).
But we also have π(a(x)) = π(a(y)), therefore f (π(a(x))) = f (π(a(y))), hence c(x) = c(y), a contradiction.
The converse of the previous result is not true. Indeed, let p ∈ βω \ ω. Observe that ω is dense and C* embedded in X = ω ∪ {p}, but ω is not k−embedded in X because ω is locally compact and X is not the one
point compactification of ω.
Corollary 3.20. Let A ⊆ X. If A is k-embedded in X, then | clX A \ A| ≤ 1.
Proof. Since A is k-embedded in X, A is k-embedded in clX A (Proposition 3.12), then | clX A \ A| ≤ 1.
Proposition 3.21. Let X ⊆ Y . If X is dense and k-embedded in Y and if K(X) is a lattice, then K(Y) is a lattice.
Proof. Let (aY , a), (bY , b) be compactifications of Y . Then (aY , a X), (bY , b X) are compactifications of
X. Since K(X) is a lattice, there exists a compactification (cX, c) of X, and continuous functions f : aY → cX
and g : bY → cX, such that
f ∘ a X = c and g ∘ b X = c.
Since X is k-embedded in Y , there is a compactification (dY , d) of Y and there is a continuous function h :
cX → dY = cldY X such that h ∘ c = d X. Therefore h ∘ f ∘ a X = d X and h ∘ g ∘ b X = d X.
Now take y ∈ Y . There is a net (x λ )λ∈Λ in X such that limλ∈Λ (x λ ) = y.
Hence
h ∘ f ∘ a(y) = lim h ∘ f ∘ a(x λ ) = lim d(x λ ) = d(lim(x λ )) = d(y).
λ∈Λ
λ∈Λ
λ∈Λ
Therefore (dY , d) ≤ (aY , a). Analogously (dY , d) ≤ (bY , b). Then K(Y) is a lattice.
4 k̃- and k* -embedded subspaces
Theorem 3.14 suggests that we introduce the following definitions.
Definition 4.1. 1. A space Y is k̃-embedded in a space X if there is an embedding j : Y → X such that j[Y]
is k-embedded in X.
2. A space Y is k* -embedded in a space X if there is an embedding j : Y → X such that j[Y] is C* - and
k-embedded in X.
Example 4.2. There are spaces X containing a subspace Y such that Y is k̃-embedded in X but Y is not kembedded in X.
Proof. The interval (0, 1) is not k-embedded in R (Propositions 3.9.(1) and 3.17) but it is k̃-embedded in R.
Observe, however, that (0, 1] is k-embbedded in R.
Remark 4.3. 1. The relation of being k̃-embedded is not reflexive.
2. The relation of being k̃-embedded is transitive.
Proof. Indeed, let X = [0, 1) and Y = [0, 1]. Then X is k̃-embedded in Y and Y is k̃-embedded in X. But X is
not homeomorphic to Y.
Also, the statement (2) is easy to prove.
As a consequence of Lemma 4.12 in [5] we obtain:
Theorem 4.4. Let D, X and Y be spaces and y0 be an element of Y . If X × {y0 } ⊆ D ⊆ X × Y, and D is dense
in X × Y, then X × {y0 } is C- and k-embedded in D. In particular, X × {y0 } is C- and k-embedded in X × Y.
Corollary 4.5. Let X and Y be topological spaces. Then X is k* -embedded in X × Y .
The partially pre-ordered set of compactifications of C p (X, Y)
|
19
Theorem 4.6. If X is k̃-embedded in Y and Z is homeomorphic to X, then Z is k̃-embedded in Y .
Proof. Let i : X → Y be an embedding witnessing that X is k̃-embedded in Y (that is, i[Y] is k-embedded in
X). Take an homeomorphism h : Z → X. Then i ∘ h : Z → Y is an embedding and (i ∘ h)[Z] = i[h[Z]] = i[Y]. So,
(i ∘ h)[Z] is k-embedded in X. This means that Z is k̃-embedded in X.
Corollary 4.7. If f : X → Y is a homeomorphism and A ⊆ X, then A is C- and k-embedded in X if and only if
f [A] is C- and k-embedded in Y.
Related to Theorem 3.13 we have:
Theorem 4.8. If Y is k* -embedded in X and K(X) is a b-lattice, then K(Y) is a b-lattice.
Proof. There is an embedding j : Y → X such that j[Y] is k- and C* -embedded in X. By Theorem 3.13, K(j[Y])
is a b-lattice, therefore K(Y) is a b-lattice.
The following is a consequence of Theorem 3.14 and Corollary 4.5.
Corollary 4.9. Let Y be a first countable space. Let X be the free topological sum of two spaces A and B.
Assume that |A| = ℵ0 , and either A is not discrete or Y is not compact. Then, K(C p (X, Y)) is not a lattice.
Proof. We have C p (X, Y) ∼
= C p (A, Y) × C p (B, Y). Then C p (A, Y) is k̃-embedded in C p (X, Y), and so if
K(C p (X, Y)) is a lattice, then K(C p (A, Y)) is a lattice. But this is not possible because of Theorem 2.8.
Example 4.10. 1. For every first countable space Y and every ordinal number α > ω, K(C p ([0, α), Y)) is not
a lattice (compare with Proposition 6.8).
2. For every first countable space Y and every almost disjoint family A, K(C p (Ψ(A), Y)) is not a lattice.
Proof.
1. Indeed, we can apply Corollary 4.9 because
[0, α) = [0, ω] ⊕ [ω + 1, α).
2.
Let A0 be an element of A. It happens that A′ = A \ (A0 ∪ {A0 }) is an independent family in ω \ A0 . Moreover,
Ψ(A) = (A0 ∪ {A0 }) ⊕ Ψ(A′ ). Now, we apply Corollary 4.9 and we finish the proof.
Corollary 4.11. Let Y be a non-compact first countable space. If X contains an infinite discrete clopen subset,
then K(C p (X, Y)) is not a lattice.
5 Spaces which are almost the free topological sum of a family of
spaces
Let X be the free topological sum of the family {X s : s ∈ S}. Let Y be a topological space. A function f ∈ Y X
such that f is constant in each X s will be called ladder in Y X . The set of ladders in Y X will be denoted by
Lad(X, Y).
Claim (2) in the following lemma is a consequence of Proposition 3.3 and Corollary 3.10.
Lemma 5.1. 1. Let X be the free topological sum of the family {X s : s ∈ S}. Let Y be a topological space.
Then Lad(X, Y) ⊆ C p (X, Y) and the function ϕ : Lad(X, Y) → Y S defined by ϕ((cts r s )s∈S ) = (r s )s∈S is a
homeomorphism.
2. Let Y be a topological space. Let X be the free topological sum of the family {X s : s ∈ S}. Then Lad(X, Y)
is C- and k-embedded in C p (X, Y).
Because of Theorem 3.14 and Corollary 2.3 we have:
Corollary 5.2. Let Y be a non-compact first countable space. Let X =
K(C p (X, Y)) is not a lattice.
⨁︀
{X s : s ∈ S} with |S| ≥ ℵ0 . Then
20 | A. Dorantes-Aldama, R. Rojas-Hernández, and Á. Tamariz-Mascarúa
Corollary 5.3. Let Y be a non-compact first countable space and X a non pseudocompact zero-dimensional
space. Then, K(C p (X, Y)) is not a lattice. (Compare to Theorem 6.9 below.)
Example 5.4. For every cardinal number κ, Sκ and Nκ are examples of zero-dimensional non-pseudocompact
spaces. Then, if Y is a first countable non-compact space, K(C p (Sκ , Y)) and K(C p (Nκ , Y)) are not lattices.
Let κ be an infinite cardinal number. A filter F on κ is uniform if |F | = κ for every F ∈ F. Let F be a uniform free
filter on κ. We denote by AF
κ the topological space κ ∪ { p } where p ∉ κ, κ is the discrete space of cardinality
κ, and the neighborhoods of p are the subsets V of AF
κ containing p and such that V ∩ κ ∈ F.
⨁︀
Proposition 5.5. Let Y be a topological space. Let X = X0 ∪ {p} where X0 =
{X s : s ∈ S}, p ∉ X0 and
⋃︀
|S| ≥ ω. Suppose that there is a uniform free filter F on S such that the family {{p} ∪ {X s : s ∈ F } : F ∈ F}
is a local base at the point p. Then C p (X, Y) contains a C- and strongly k-embedded copy of C p (AF
κ , Y), where
κ = | S |.
Proof. Pick a point y s ∈ X s for each s ∈ S and let Z = {y s : s ∈ S} ∪ {p}. Consider the function r : X → Z
given by r(x) = y s if x ∈ X s and s ∈ S and r(p) = p. Notice that r is a retraction. Let us observe that Z
F
is homeomorphic to AF
κ , where κ = | S |. It follows that C p (Z, Y) is homeomorphic to C p (A κ , Y). Therefore,
C p (X, Y) contains a C- and strongly k-embedded copy of C p (AF
κ , Y) (Corollary 3.10).
Example 5.6. Let Y be a non-compact first countable space, let κ be an infinite cardinal and let F be a uniform
free filter on κ. Then, K(C p (AF
κ , Y)) is not a lattice.
Proof. If Y is not locally compact, we use Corollary 3.15. Assume that Y is locally compact. For κ = ω and F
equal to the Fréchet filter, this proposition has already been proved in Theorem 2.8.
For κ > ω and if F is not the Fréchet filter on κ, then the conclusion was proved in Corollary 4.11.
Now suppose that κ > ω and F is the Fréchet filter on κ. In this case, we can consider AF
κ as { p } ∪ ⊕i<ω X i
⋃︀
where X i = {a s : s ∈ S i } ⊆ κ and i<ω X i = κ, X i is not empty for every i < ω and X i ∩ X j = ∅ if i ≠ j. Observe
that for every neighborhood V of p, there is n0 < ω such that X i ⊆ V for all i ≥ n0 . Because of Proposition 5.5,
G
C p (AF
κ , Y) contains a C- and k-embedded copy of C p (A ω , Y) where G is the Fréchet filter on ω. This means
that K(C p (A κ , Y)) is not a lattice.
As a consequence of Proposition 5.5 and Example 5.6, we obtain:
⨁︀
Corollary 5.7. Let Y be a non-compact first countable space. Let X = X0 ∪ {p} where X0 = {X s : s ∈ S} and
⋃︀
p ∉ X0 with |S| ≥ ω. Let F be a uniform free filter on |S|. Suppose that the family {{p}∪ {X s : s ∈ F } : F ∈ F}
is a local base at the point p. Then K(C p (X, Y)) is not a lattice.
Let X, Y and Z be three spaces. If r : X → Z is a continuous function, we denote by r* the continuous function
from C p (Z, Y) to C p (X, Y) defined by r* (g) = r ∘ g.
Theorem 5.8. Let X and Y be spaces. Suppose that there exist a set Z ⊆ X and a retraction r : X → Z, then
C p (X, Y) contains a C- and strongly k-embedded copy of C p (Z, Y).
Proof. Let A = r* [C p (Z, Y)]. Since r* : C p (Z, Y) → C p (X, Y) is an embedding, the map r* : C p (Z, Y) → A is
a homeomorphism. We claim that r* ∘ π Z : C p (X, Y) → A is a retraction. Since r : X → Z is a retraction, Z
is C-embedded in X. Then π Z : C p (X, Y) → C p (Z, Y) is surjective. It follows that r* ∘ π Z is continuous and
onto. Let g ∈ A = r* ∘ π Z [C p (X, Y)], then g is equal to (f Z) ∘ r for some f ∈ C p (X, Y). Let us observe that
f Z = ((f Z) ∘ r) Z. Hence r* ∘ π Z (g) = (g Z) ∘ r = (((f Z) ∘ r) Z) ∘ r = (f Z) ∘ r = g. This shows that
r* ∘ π Z is a continuous retraction onto A. It follows from Corollary 3.10 that A is C- and strongly k-embedded
in C p (X, Y).
Proposition 5.9. Let X = Z × W. Then, for every space Y, C p (X, Y) contains a C- and k-embedded copy of
C p (Z, Y).
The partially pre-ordered set of compactifications of C p (X, Y)
|
21
Proof. Pick w ∈ W and let Z * = Z × {w}. The function r : X → Z * given by r((z, w)) = (z, w) is a retraction. By
Theorem 5.8, C p (X, Y) contains a C- and k-embedded copy of C p (Z * , Y). The result follows from the fact that
Z and Z * are homeomorphic.
Example 5.10. We can find two spaces X and Y which contain k-embedded subsets A and B, respectively,
and however A × B is not k-embedded in X × Y. Indeed, let A be a mad family on ω, then ω × Ψ(A) is not
k-embedded in ω × αΨ(A). The reason for this is that every dense and k-embedded weakly pseudocompact
subspace Z in a space W reflects the weak pseudocompactness to W (Theorem 6.4 in [5]), ω × Ψ(A) is weakly
pseudocompact but ω × αΨ(A) is Lindelöf and locally compact, so it is not weakly pseudocompact. Then,
ω × Ψ(A) is not k-embedded in ω × αΨ(A).
The following result is easy to prove.
Proposition 5.11. Assume that r : Y → Z is a retraction. Then, r* : C p (X, Y) → C p (X, Z) is a retraction, where
r* (f ) = r ∘ f .
Corollary 5.12. If K(C p (X)) is a lattice (resp., b-lattice), then K(C p (X, [0, 1])) is a lattice (resp., b-lattice). And
if K(C p (X, Z)) is a lattice (resp., b-lattice), then K(C p (X, {0, 1})) is a lattice (resp., b-lattice).
Proof. The space [0, 1] (resp., {0, 1}) is a retract of R (resp., Z).
The converse of Corollary 5.12 is not true. Indeed, [0, 1]ω (resp., {0, 1}ω ) is compact but K(Rω ) (resp., K(Zω ))
is not a lattice.
6 The upper semilattice K(C p (X))
Given a space X and a closed subset F of X we denote by Z F (X) the subspace {f ∈ C p (X) : f (F) ⊂ {0}} of
C p (X). If F = {x} ⊂ X we denote Z F (X) simply as Z x (X).
Proposition 6.1. Let F be a uniform free filter on κ. Then K(Z p (AF
κ )) is not a lattice (see the notation before
Proposition 5.5).
F
Proof. Indeed, if κ = ω, K(Z p (AF
ω )) is not a lattice since Z p (A ω ) is first countable and non-locally compact.
F ∼ ω
If κ > ω and F is not the Fréchet filter, then Z p (A κ ) = R × Z p (AFκ\ω
). By Theorem 4.4, Z p (AF
κ ) contains a
κ\ω
k-embedded copy of Rω . So, in this case, K(Z p (AF
κ )) is not a lattice either.
Now assume that κ > ω and F is the Fréchet filter. Consider a partition {K n : n < ω} of κ where each
K n is not empty. Choose a point x n ∈ K n for each n < ω. Now, we can follow a similar argumentation to that
G
given in the proof of Proposition 5.5, and conclude that Z p (AF
κ ) contains a C- and k-embedded copy of Z p (A ω )
where G is the Fréchet filter on ω. Therefore, also in this last possible case K(Z p (AF
κ )) is not a lattice.
⋃︀
Theorem 6.2. Let F be a uniform free filter on κ. If X contains a subspace X0 = {s} ∪ ξ <ω U ξ such that each
U ξ is open in X, the family {clX U ξ : ξ < κ} is a pairwise disjoint family, {ξ < κ : U ξ ⊆ V } ∈ F for every
⋃︀
⋃︀
neighborhood V of s in X, and {s} = clX ( ξ <κ U ξ ) \ ξ <κ clX U ξ , then
1.
2.
the space C p (X) contains a retract homeomorphic to Z p (AF
κ ),
the space C p (X) is homeomorphic to Z p (AF
κ ) × Y for some space Y.
Proof. (1) Pick s ξ ∈ U ξ . Then, S = {s} ∪ {s ξ : ξ ∈ κ} ⊆ X is homeomorphic to AF
κ . It follows that Z s (S)
F
is homeomorphic to Z p (A κ ). For each ξ ∈ κ, we can pick a continuous function f ξ : X → [0, 1] such that
f ξ (s ξ ) = 1 and f ξ (X \ U ξ ) = {0}. Define a function e : Z → C p (X) by
{︃
e(g)(x) =
g(s ξ )f ξ (x)
if x ∈ cl X U ξ for some ξ ∈ κ;
0
otherwise.
The following claim shows that e is well defined.
22 | A. Dorantes-Aldama, R. Rojas-Hernández, and Á. Tamariz-Mascarúa
Claim 1. The function e(g) is continuous for any g ∈ Z s (S).
Pick g ∈ Z s ω (S). Choose x ∈ X and an open set V in R containing e(g)(x). We shall prove that there exists
an open set U in X with x ∈ U and e(g)(U) ⊂ V. Consider three cases.
⋃︀
∑︀
Case 1. If x ∈ ξ <κ clX U ξ \ {s}, let U η be such that x ∈ clX U η . We have that e(g) clX U η = ( n∈ω g(s ξ )f ξ ) clX U η = g(s η )f η clX U η . Let A be an open set of R containing e(g)(x) = g(s η )f η (x). Since g(s η )f η : X → R
is a continuous function in x, there is an open subset B′ of X such that x ∈ B′ and g(s η )f η [B′ ] ⊆ A. Let
⋃︀
⋃︀
B = B′ ∩ X \ ({s} ∪ ξ ≠η clX U ξ ). Since {s} ∪ ξ ≠η clX U ξ is closed, B is open, contains x and is contained in
B′ . Observe that e(g) B = g(s η )f η B; so, e(g)[B] = g(s η )f η [B] ⊆ g(s η )f η [B′ ] ⊆ A. We conclude that e(g) is
continuous in x.
Case 2. If x = s. Let A be an open set in R which contains [e(g)](s) = 0. Choose ϵ > 0 with (−ϵ, ϵ) ⊆ A.
Since g(s) = 0 and g is continuous in s, we can find F ∈ F such that g(s ξ ) ∈ (−ϵ, ϵ) if and only if ξ ∈ F.
⋃︀
Let V s = X \ clX ( ξ ∉F U ξ ). Clearly V s is open and contains s. We will prove that [e(g)](V s ) ⊆ A. Pick y ∈ V s .
⋃︀
Either y ∉ {cl(U ξ ) : ξ ∈ κ} or there exists a unique η ∈ F for which y ∈ cl(U η ). Hence either e(g)(y) = 0 or
∑︀
[e(g)](y) = ξ ∈κ g(s ξ )f ξ (y) = g(s η )f η (y). In both cases we conclude that [e(g)](y) ∈ (−ϵ, ϵ). Hence [e(g)](V x ) ⊆
(−ϵ, ϵ) ⊆ A. This shows that e(g) is continuous in s.
⋃︀
⋃︀
Case 3. Now, let x be an element in X\clX ξ <κ U ξ . Take U x = X\clX ξ <ω U ξ . It happens that U x is open, x ∈ U x
and e(g)(U x ) = {0} = {e(g)(x)}. Thus, for every open subset A of R which contains 0, e(g)[U x ] = {0} ⊆ A.
Claim 2. The map e : Z s (S) → C p (X) is an embedding.
Let P = e(Z). It is easy to verify that e−1 = π S P and hence e−1 is continuous. So, in order to prove that
e is an embedding we shall only prove that e is continuous. Pick g ∈ Z s (S) and an open set V in C p (X) which
contains e(g). We can suppose that V is a subbasic open set, that is V = [x, B] ∩ C p (X) where x ∈ X and B is
⋃︀
an open subset of R. If x ∈ X \ {f ξ−1 ((0, 1]) : ξ ∈ κ} choose W = Z s (S). It is clear, in this case, that g ∈ W
⋃︀
and e(W) ⊆ V. Now suppose, otherwise, that x ∈ {f ξ−1 ((0, 1]) : ξ ∈ κ}. Then there exists a unique ordinal
∑︀
number η with x ∈ f η−1 ((0, 1]). It follows from e(g) ∈ V that g(s η )f η (x) = n∈ω g(s ξ )f ξ (x) = [e(g)](x) ∈ B.
Since the function m : R → R given by m(r) = rf k (x) is continuous, we can find an open subset B′ of R such
that g(s η ) ∈ B′ and m(B′ ) ⊆ B. Let W = [s η , B′ ] ∩ C p (S) ∩ Z s (S). Clearly g ∈ W. Moreover, for h ∈ W, it follows
∑︀
from h(s η ) ∈ B′ that [e(h)(x)] = ξ ∈κ h(s ξ )f ξ (x) = h(s η )f η (x) ∈ m(B′ ) ⊆ B. Hence in this case we also have
e(W) ⊆ V. Therefore, e is continuous and so we have proved Claim 2.
Let us observe that the map e ∘ π S : C p (X) → P is a continuous retraction. Since the space P is homeomorphic to Z s (S), we have proved (1).
(2) The space C p (X) is homeomorphic to the product Z s (X) × R. Indeed, the map φ : C p (X) → Z s (X) × R given
by φ(g) = (g − g(s), g(s)) is clearly a homeomorphism. Notice that Z s (X) is a topological subgroup of C p (X)
and e[Z] = P ⊆ Z s (X). Consider the function l = e ∘ π S Y : Y → P. Observe that l is a continuous and
linear retraction onto P. Let q : Y → Y be the function given by q(f ) = f − l(f ) for any f ∈ Z s (X). Then q is a
continuous function. Let Q = q(Z s (X)). Notice that l(q(f )) = l(f − l(f )) = l(f ) − l(l(f )) = l(f ) − l(f ) = 0 for any
f ∈ Z s (X). We claim that Z s (X) is homeomorphic to P × Q. Consider the function ϕ = p∆q : Y → P × Q. We
shall prove that ϕ is a homeomorphism. Assume that (l(f ), q(g)) in P × Q. Let h = l(f ) + q(g) ∈ Z s (X). Then
l(h) = l(l(f ) + q(g)) = l(l(f )) − l(q(g)) = l(f ) and q(h) = h − l(h) = (l(f ) + q(g)) − l(f ) = q(g). So ϕ(h) = (l(f ), q(g)).
It follows that ϕ is an onto function. It is clear that ϕ is continuous. Finally let us observe that ϕ−1 = a P × Q,
where a : C p (X) × C p (X) → C p (X) is the sum function. Indeed, a(ϕ(f )) = a((l(f ), q(f ))) = l(f ) + q(f ) = l(f ) + (f −
l(f )) = f . Thus ϕ−1 is continuous and hence ϕ is a homeomorphism. Then Z s (X) is homeomorphic to P × Q.
F
Since P is homeomorphic to Z p (AF
κ ), the space Z s (X) is homeomorphic to Z p (A κ ) × Q. It follows that C p (X) is
homeomorphic to Z p (AF
κ ) × Q × R, and so we have proved (2).
⋃︀
Corollary 6.3. K(C p (X)) is not a lattice when X contains a subspace X0 = {s} ∪ ξ <κ U ξ such that the family
{U ξ : ξ < κ} is a pairwise disjoint family of open subsets of X, and there is a uniform free filter F on κ such
⋃︀
⋃︀
that for every neighborhood V of s in X, {ξ < κ : U ξ ⊆ V } ∈ F, and {s} = (clX ξ <κ U ξ ) \ ξ <κ clX U ξ .
The partially pre-ordered set of compactifications of C p (X, Y)
|
23
Proof. If K(C p (X)) is a lattice, so is K(Z p (AF
κ )). This contradicts Proposition 6.1.
Corollary 6.4. If X contains a non-isolated point of countable character, then K(C p (X)) is not a lattice.
Proof. Let s be a non-isolated point of X with countable character. Then there exists a countable family {B n :
n ∈ ω} of open sets in X which is a local base at the point s. We can assume that clB n+1 is a proper subset of
B n for every n ∈ ω. For each n ∈ ω pick a non-empty open set U n with U n ⊆ clU n ⊆ B n \ clB n+1 . Then X0 =
⋃︀
{s} ∪ i<ω U i ⊆ X and the Fréchet filter F on ω satisfy the conditions requested in Corollary 6.3. Therefore,
K(C p (X)) is not a lattice.
Proposition 6.5. If r : X → X is a retraction and r(X) is infinite and first countable at some non-isolated point,
then K(C p (X)) is not a lattice.
Proof. Assume that r[X] is first countable at some point. Then, by Corollary 6.4, K(π r(X) (C p (X))) is not a lattice.
Choose r̃ = r* ∘ π r(X) : C p (X) → C p (X). As in the proof of Theorem 5.8 we can show that r̃ is a retraction.
Moreover, since r* is an embedding, r̃(C p (X)) is homeomorphic to π r(X) (C p (X)). Hence K(r̃(C p (X))) is not a
lattice. By Corollary 3.10 the space r̃(C p (X)) is k-embedded in C p (X). Therefore, K(C p (X)) is not a lattice.
Corollary 6.6. K(C p (R)) is not a lattice.
Corollary 6.7. K(C p (C p (X))) is never a lattice.
Proof. Let X be an arbitrary space. The space C p (X) is homeomorphic to the product Y × R, where Y = {f ∈
C p (X) : f (x0 ) = 0} is a subspace of C p (X) and the point x0 ∈ X is chosen arbitrarily. Then C p (C p (X)) is
homeomorphic to C p (Y × R). By Proposition 5.9 the space C p (Y × R) contains a C- and k-embedded copy of
C p (R). So we can apply Corollary 4.7 to see that C p (C p (X)) also contains a C- and k-embedded copy of C p (R).
Finally apply Theorem 3.14 and Corollary 6.6 to see that K(C p (C p (X))) is not a lattice.
If X is a generalized ordered topological space, there exist a complete linearly ordered topological space
̃︀ containing X densely and such that the topology inherited by X from X
̃︀ coincides with the topology of X. For
X
̃︀
a cardinal number κ, we say that an increasing sequence (x ξ )ξ <κ in X is precise if x ξ < x η when ξ < η < κ, and
for each ordinal limit η < κ, x η = sup{x ξ : ξ < η}. In a similar way we define a precise decreasing sequence
̃︀ For each point x ∈ X we define lt(x) = min{κ : there is a precise increasing sequence (x ξ )ξ <κ in X
̃︀
(x ξ )ξ <κ in X.
such that x = sup{x ξ : ξ < κ}}; and we define rt(x) = min{κ : there is a precise decreasing sequence (x ξ )ξ <κ
̃︀ such that x = inf {x ξ : ξ < κ}.
in X
Proposition 6.8. If X is an infinite generalized ordered topological space, then K(C p (X)) is not a lattice.
Proof. If X is discrete, the conclusion follows from Theorem 1.3. Assume that X is not discrete. There is a
̃︀ which is complete and such that X is a dense subspace of X.
̃︀ Define
linearly ordered topological space X
A = {x ∈ X : lt(x) is infinite} and B = {x ∈ X : rt(x) is infinite}. Since X is not discrete at least one of the
sets A and B is non-empty. Say A is not empty. Let z ∈ A such that lt(z) = inf {lt(x) : x ∈ A}. We write simply
κ instead of lt(z). Observe that κ is a regular cardinal. For each limit ordinal η < κ consider the following
̃︀
sequence of non-empty open subsets of X (recall that X is a dense subspace of X):
A η = (x η , x η+1 ) ∩ X, ..., A η+n = (x η+2n , x η+2n+1 ) ∩ X, ....
Claim: The family {clX A ξ : ξ < κ} is a pairwise disjoint family, {ξ < κ : A ξ ⊆ V } ∈ F for every neighborhood
⋃︀
V of z in X where F is the free filter on κ generated by all the final segments of κ, and {z} = clX ( ξ <κ A ξ ) \
⋃︀
ξ <κ clX A ξ ,
If ξ0 = η0 + n0 < ξ1 = η1 + n1 where η0 and η1 are limit ordinals, then cl X A ξ0 ⊆ [η0 + 2n0 , η0 + 2n0 + 1]
and cl X A ξ1 ⊆ [η1 + 2n1 , η1 + 2n1 + 1] and
[η0 + 2n0 , η0 + 2n0 + 1] ∩ [η1 + 2n1 , η1 + 2n1 + 1] = ∅.
24 | A. Dorantes-Aldama, R. Rojas-Hernández, and Á. Tamariz-Mascarúa
̃︀ such that
So the collection {clX A ξ : ξ < κ} is a pairwise disjoint family. On the other hand, let a, b ∈ X
a < z < b. By hypothesis, there is ξ < κ such that a < x𝛾 < z for every 𝛾 ∈ κ such that ξ < 𝛾 . Hence
A𝛾 ⊆ (a, b) for every 𝛾 > ξ . Then, for every neighborhood V of z in X we have: {ξ < κ : A ξ ⊆ V } ∈ F. Finally,
⋃︀
⋃︀
clX ( ξ <κ A ξ ) \ ξ <κ clX A ξ ⊆ {z} ∪ {x η : η < κ and η is a limit ordinal}. If x η belongs to X, then lt(x η ) < lt(z)
⋃︀
⋃︀
and this contradicts the election of z. Therefore, {z} = clX ( ξ <κ A ξ ) \ ξ <κ clX A ξ .
Because of Claim and Corollary 6.3, we conclude that K(C p (X)) is not a lattice.
Theorem 6.9. Let X be a non-pseudocompact space. Then K(C p (X)) is not a lattice.
Proof. Indeed, we are going to prove that in this case C p (X) contains a k* -embedded copy of Rω . We use a
similar proof to that of Theorem 6.2.
Since X is not pseudocompact, there is a collection {U n : n ∈ ω} of open subsets of X such that clX U n ∩
⋃︀
⋃︀
clX U m = ∅ for every n, m < ω with m ≠ n, and n<ω clX U n = clX n<ω U n . Pick s n ∈ U n . Then, S = {s n : n ∈
ω} ⊆ X is homeomorphic to ω. Let Z = C p (S). For each n ∈ ω, we can pick a continuous function f n : X →
∑︀
[0, 1] such that f n (s n ) = 1 and f n (X \ U n ) = {0}. Define a function e : Z → C p (X) by e(g) = n∈ω g(s n )f n .
Following similar arguments as those given in the proof of Theorem 6.2, we obtain that e(g) is continuous for
any g ∈ Z, and e is an embedding.
⋃︀
If X = n<ω clX U n , then we have the conclusion of our theorem as a consequence of Corollary 5.2. Now,
⋃︀
⋃︀
assume that X ≠ n<ω clX U n and let s ∈ X \ n<ω clX U n . Let P = e[Z]. We continue the proof as expected:
the space C p (X) is homeomorphic to the product Y × R, where Y = {f ∈ C p (X) : f (s) = 0} is a subspace of
C p (X). Notice that Y is a topological subgroup of C p (X) and P ⊆ Y. The function p = e ∘ π S Y : Y → P is a
retraction onto P. The function q : Y → Y given by q(f ) = f − p(f ) for any f ∈ Y is continuous. Let Q = q(Y).
Y is homeomorphic to P × Q and hence C p (X) is homeomorphic to P × Q × R.
Finally, following similar arguments to those given at the end of the proof of Theorem 6.2, we conclude
that C p (X) contains a C- and k-embedded copy of Rω . Therefore, K(C p (X)) is not a lattice.
For pseudocompact spaces, we have:
Theorem 6.10. If X is an infinite pseudocompact space and C p (X) is normal, then K(C p (X)) is not a lattice.
Proof. Let S be a countable infinite subset of X and let T be the family of all one-to-one maps from 2 to S. For
each t ∈ T choose a continuous map f t : X → R such that f t (t(0)) ≠ f t (t(1)). Let f : X → RT be the diagonal of
the family of maps {f t : t ∈ T }. Let us observe that f S is injective. Then Y = f (X) ⊂ RT is an infinite second
countable compact space. It was proved in [1] that every continuous surjective map from a pseudocompact
space to a second countable space is an R-quotient. Then f is an R-quotient, that is f * (C p (Y)) is a closed
subspace of C p (X). Since C p (X) is normal, we can apply Corollary 3.11 to see that f * (C p (Y)) is strongly kembedded in C p (X). Since Y is an infinite second countable compact space, it follows from Corollary 6.4 that
K(C p (Y)) is not a lattice. Finally we can apply Theorem 3.14 to see that K(C p (X)) is not a lattice.
Corollary 6.11. If F is a closed infinite subspace of a Σ-product of second countable spaces, then K(C p (F)) is
not a lattice.
Proof. If F is not pseudocompact the result follows from Theorem 6.9. If X is pseudocompact, it is known that
C p (X) is normal, so the result follows from Theorem 6.10.
In particular:
Corollary 6.12. If X is an infinite Corson compact space, then K(C p (X)) is not a lattice.
7 Open problems
A zero-dimensional space X is b2 -discrete if every countable subspace N of X is closed, discrete and every
function f : N → {0, 1} can be extended to a continuous function to all X.
The partially pre-ordered set of compactifications of C p (X, Y)
|
25
Problem 7.1. Are there spaces X and Y for which C p (X, Y) is dense in Y X and such that K(C p (X, Y)) is a
b-lattice? In particular, is there a b2 -discrete space X such that K(C p (X, 2)) is a b-lattice?
Problem 7.2. With respect to Example 4.10, is K(C p ([0, ω1 ), Y)) or K(C p (Ψ(A), Y)) a lattice, when A is a mad
family on ω and Y ∈ {2, 2ω , [0, 1]}?
Problem 7.3. Regarding Example 5.4, is K(C p (Sκ , Y)) or K(C p (Nκ , Y)) a lattice, when Y ∈ {2, 2ω , [0, 1]}?
Problem 7.4. Is K(C p (AF
κ , Y)) a lattice for an infinite cardinal κ and a uniform free filter F on κ, when Y ∈
{2, 2ω , [0, 1]}? See Example 5.6.
Problem 7.5. Is it true that if A is k-embedded in C p (X) and B is k-embedded in C p (Y), then A × B is kembedded in C p (X) × C p (Y)?
Problem 7.6. State if the following proposition is true: K(C p (X)) is a lattice iff K(C p (X)) is a b-lattice.
The two last problems must be emphasized on this list because of their importance.
Problem 7.7. Is it true that K(C p (X)) is not a lattice when X contains a convergent sequence?
Problem 7.8. Is K(C p (X)) not a lattice when X is not a countably compact space?
Acknowledgement: The authors would like to thank the referee for her/his guiding and stimulating comments. The first author was supported by CONACyT grant No. 178425/245149. The research of the third author
was supported by PAPIIT IN115312.
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