1. Introduction
In 1943, Hewitt [1] has introduced the notion of resolvable space as follows: A topological space is said to be resolvable if it has two disjoint dense subsets. Hence a topological space X is resolvable if and only if X is written as a union of two disjoint dense subsets. Hewitt in [1] has also called a topological space X maximally irresolvable if each dense subset of X is open. Nowadays, maximally irresolvable spaces are called submaximal spaces.
Recently, Belaid et al. [2], were interested in spaces such that their compactifications are submaximal. They proved that if X is a topological space and
is a compactification of X, then the following statements are equivalent:
1)
is submaximal.
2) For each dense subset D of X, the following properties hold:
a) D is co-finite in K(X);
b) for each
,
is closed.
It is clear that a compactification of resolvable spaces is resolvable. Hence the following question is natural:
“Characterize spaces X such that a compactification
of X is a resolvable space?”
The first section is devoted to a brief study of spaces X such that their compactification is a resolvable space. The particular case of the one-point compactification is given.
The purpose of the second section is to give an intrinsic topological characterization of spaces X such that the Wallman compactification
of X is a resolvable space.
2. Resolvable Space and Compactifications
First, recall that a compactification of a topological space X is a couple
, where
is a compact space and
is a continuous embedding (e is a continuous one-to-one map and induces a homeomorphism from X onto
) such that
is a dense subspace of
. When a compactification
of
is given,
will be identified with
and assumed to be dense in
.
Let us give some basic facts about space such that its compactification is a resolvable space.
Lemma 2.1 Let X be a topological space,
be a compactification of X and
be a subset of
. If X is an open set of
, then the following statements are equivalent:
1) A is a dense subset of
;
2)
is a dense subset of
.
Proof. 1)
2) Let
be an open set of
. Since
is an open set of
,
is an open set of
. Hence
. Thus
; so that
is a dense set of
.
2)
1) Let
be an open set of
. Since
is a non-empty open set of
,
. Then
. Therefore
is a dense set of
.
An immediate consequence of Lemma 1.1 is the following.
Proposition 2.2 Let X be a topological space and
be a compactification of
. If X is an open set of
, then the following statements are equivalent:
1)
is resolvable;
2)
is resolvable.
Let us recall the construction of the one-point compactification: For any non-compact space X the one-point compactification of
is obtained by adding one extra point
(called a point at infinity) and defining the open sets of
to be the open sets of X together with the sets of the form
, where
is an open set of X such that
is a closed compact set of X. The one point compactification
of X is also called the Alexandroff compactification of X [3].
The following result characterizes space such that its one point compactification is a resolvable space. Its proof follows immediately from Proposition 2.2; thus it is omitted.
Proposition 2.3 Let X be a non-compact topological space. Then the following statements are equivalent:
1) The one-point compactification
of
is resolvable;
2)
is resolvable.
3. Resolvable Space and Wallman Compactification
First, recall that the Wallman compactification of
- space was introduced, in 1938, by Wallman [4] as follows:
Let
be a class of subsets of a topological space
which is closed under finite intersections and finite unions.
A
-filter on
is a collection
of nonempty elements of
with the properties:
1)
is closed under finite intersections;
2)
implies
.
A
-ultrafilter is a maximal
-filter. When
is the class of closed sets of X, then the
-filters are called closed filters.
The points of the Wallman compactification
of a space
are the closed ultrafilters on
. For each closed set
, define
to be the set
. Thus
is a base for the closed sets of a topology on
.
Let
be an open set of
, we define
, it is easily seen that the class
is a base for open sets of the topology of
. The following properties of
are frequently useful:
Proposition 3.1 Let
be a
-space and
the Wallman compactification of
. Then the following statements hold:
1)
is a
-space;
2) For
and
. Then
is an embedding of X into
(
will be identified to
).
3) If
is an open set of
, then
.
4) If
and
are two open sets of
, then
and
.
Recall that Kovar in [5] has characterized space with finite Wallman compactification remainder as following:
Proposition 3.2 Let
be a
-space. Then the following statements are equivalent:
1)
;
2) There exists a collection of
pairwise disjoint non-compact closed sets of X and every family of noncompact pairwise disjoint closed sets of X contain at most
elements.
The following proposition follows immediately from Proposition 3.2 and Proposition 3.1-1).
Proposition 3.3 Let X be a
-space and
such that every family of non-compact pairwise disjoint closed sets of X contains at most
elements. Then X is resolvable if and only if
is resolvable.
The following lemma has been given in [2] as Remark 4.5 and Remark 4.9.
Lemma 3.4 Let X be a
-space. Then the following properties hold:
1) If
is a closed non-compact subset of
, then there exists
such that
.
2)
. Then for each
,
is a non-compact closed set of
.
The following result is an immediate consequence of Lemma 3.4.
Corollary 3.5 Let X be a
-space,
be the Wallman compactification of X and
be an open set of X. Then the following statements are equivalent:
1)
;
2) There exists a non compact closed set
of
such that
.
Now, we are in a position to give a characterization of spaces such that their Wallman compactification is resolvable.
Theorem 3.6 Let X be a
-space. Then the following statements are equivalent:
1) The Wallman compactification
of X is resolvable;
2) There exist two disjoint subsets
and
of X such that:
a)
.
b) For
and for each non empty open set
, there exists a non compact closed set
of
such that
.
Proof. 1)
2) Since
is a resolvable space, there exist two disjoint dense sets A1 and A2 of
such that
is the union of A1 and A2. Set
and
.
Let
and
be a non empty open set of
such that
. Set
such that
. Since
is a dense subset of
,
. Now,
implies that
. It follows that there exists
, and thus
. According to Corollary C4 there exists a non compact closed set
of
such that
and
.
2)
1) Let
be two disjoint subsets of
satisfying the condition b) and such that
. Let
in
and we define
![](https://www.scirp.org/html/9-5300428\99c2579a-9721-4f7a-a2d2-80beb0aa3c48.jpg)
It is immediate that
.
Now, let
be a open set of
. We consider two cases:
Case 1:
. Then
. So
.
Case 2:
. Then
. By condition b), there exists a non compact closed F of X such that
. Let
such that
. Hence
. Thus
.
Therefore
is a dense set of
; so that
is a resolvable space.
Example 3.7 Let
be the set of all rational numbers equipped with the natural topology
. Let
equipped with the topology
. It is immediate that the topological space
satisfies the condition 2) of the Theorem 2.6. Then
is a resolvable space.
The previous result incites us to ask the following question.
Question 3.8 Let X be a space. We denote by
(resp.
) the
-compactifcation of X introduced by Herrlich in [6] (resp. the Stone Cech compactification). When is
(resp.
) a resolvable space?
4. Acknowledgements
This paper has been supported by deanship of scientific research of University of Dammam under the reference 2011085.