Paracompact like Spaces ……………………………………....… Talal Ali Al- Hawary
Paracompact Like Spaces
Received: 22/2/2004 Accepted: 11/5/2004
Talal Ali Al- Hawary *
* / Department of Mathematics, Mu’tah University1 Introduction
Let be a topological space (or simply, a space). A subset is called semi-open (simply, SO) if there exists an open set such that , where denotes the closure of O in X. Clearly A is a semi-open set if and only if . A complement of a semi-open set is called semi-closed (simply, SC). A is called preopen (simply, PO) if . A is called regular-open (simply, RO) if . Clearly, A is RO if for some closed subset F. Complements of regular-open sets are called regular-closed (simply, RC). Clearly A is regular-closed if and only if or if for some open subset O. A is called if . A is called a regularly semi-open (simply, RSO) if there exists an RO set U such that . Finally, A is called semi-preopen (simply, SPO) if , which is equivalent to say that is an RC set. The collection of all SO (resp., RO, RC, PO and -open) subsets of X is denoted by (resp., , , and ). We remark that is a topology on X and . A space is called extremally disconnected (simply e.d.) if is open for every open subset U, which is equivalent to say that every disjoint open subsets, have disjoint closures.
In [6], a space is called S-closed if every SO cover of X (i.e. every cover of X by SO subsets) contains a finite subfamily whose union is dense in X. In [2], is called I-compact if every RC cover of X (i.e. every cover of X by RC subsets) contains a finite subfamily whose interiors cover X. In [3], is called countably S-closed if every countable RC cover of X contains a finite subcover for X. In [1], is called countably I-compact if every countable RC cover of X contains a finite subfamily whose interiors cover X. Also in [3], is called feebly compact if every countable open cover of X contains a finite subfamily whose union is dense in X.
By , we mean the set of all natural numbers. Recall that a space is compact if and only if every open cover of X has a finite open refinement that covers X, see for example [4]. Thus we have the following result:
Lemma 1 A space is countably S-closed if and only if every RC countable cover of X has a finite RC refinement that covers X.
Proof. Let be an RC cover of X. Then it has a finite RC refinement that covers X. Thus for every i=1,2,...,k, and as covers X, is a finite subcover for X.
The converse is obvious. ■
Another immediate result is next.
Lemma 2 A space is countably I-compact if and only if every countable RC cover of X has a finite RC refinement such that .
Proof. Follows from the fact that every finite collection is locally finite. ■
Analogous to [1,2,3,6], we study several classes of paracompact like spaces, namely: S-paracompact, countably S-paracompact, countably I-paracompact and feebly paracompact spaces and characterize them. Our main goal is to focus on countably I-paracompact spaces and show that a space is countably I-paracompact if and only if it is countably S-paracompact and extremally disconnected. Moreover, we study maps of countably I-paracompact spaces.
2 Countably I-paracompact spaces
We begin this section by characterizing S-paracompact spaces for better understanding of countably I-paracompact spaces.
Definition 1 A space is called S-paracompact if it is and for every semiopen cover of X, there exists a locally finite semiopen refinement of such that .
Theorem 1 A space is S-paracompact if and only if it is and for every RC cover of X, there exists a locally finite RC refinement of that covers X.
Proof. Let be an RC cover of the space X. Then as every RC-set is an SO, is an SO cover of X. Thus as X is S-paracompact, there exists a locally finite SO refinement of such that . Since is locally finite, is locally finite. As for every , is an RC-set (being the closure of an open set) and as for some in , . Thus is a locally finite RC refinement of . Moreover as for every , , we have and hence . That is covers X.
Conversely, let be an SO cover of the space X. Then is an RC cover of X. Thus has an RC locally finite refinement such that . Clearly is a locally finite refinement of that covers X. Hence and therefore X is S-paracompact. ■
Next, countably S-paracompactness, countably I-paracompactness and feebly paracompactness are introduced.
Definition 2 A space is called countably S-paracompact if it is and for every countable RC cover of X, there exists a locally finite RC refinement of that covers X.
Definition 3 A space is called countably I-paracompact if it is and for every countable RC cover of X, there exists a countable locally finite RC refinement of such that .
Definition 4 A space is called feebly paracompact if it is and for every countable open cover of X, there exists a locally finite open refinement of such that .
Next, we show that every countably S-paracompact space is feebly paracompact.
Lemma 3 A countably S-paracompact space is feebly paracompact.
Proof. Let be an open cover of X. Then is an Rc cover of X and hence there exists a countable RC locally finite refinement of such that . Thus is a countable RC locally finite refinement of . Moreover, . Therefore, is feebly paracompact. ■
The following result is a direct consequence of Lemma 1 and the fact that every finite collection is locally finite.
Corollary 1 A countably S-compact space that is is countably S-paracompact.
The following result follows directly from Lemma 2.
Corollary 2 A countably I-compact space that is is countably I-paracompact.
Next, we prove one of our main results that will be a useful tool throughout this paper. In this result, we give a decomposition of countably I-paracompact spaces in terms of countably S-paracompact spaces.
Theorem 2 A space is countably I-paracompact if and only if it is countably S-paracompact and e.d.
Proof. Let be countably I-paracompact space. Obviously, X is countably S-paracompact. Suppose X is not e.d. Then there exists an open subset U of X such that is not open. So and hence is a countable RC cover of X. Then it has a countable RC locally finite refinement such that . Thus . But is open and hence and so . That is and so . Therefore, is open, a contradiction.
Conversely, let be a countably S-paracompact and e.d. space and let be an RC cover of X. Then there exists an RC locally finite refinement of that covers X. Since for each , is an RC-set, we have for some open . As X is e.d., is open which is also an RC-set. Now . Therefore is a locally finite RC refinement of that covers X. ■
For the proof of the following Lemma, see for example [4,pp.246].
Lemma 4 Let be a locally finite collection of subsets of a space X. Then
Using Lemma 4, we can show that in an e.d. space, feebly paracompactness is stronger than countably S-paracompactness.
Lemma 5 Every feebly paracompact e.d. space is countably S-paracompact.
Proof. Let be an RC cover of X. As X is e.d. and as is open, then is open for each and hence is a countable open cover of X. Then there exists a locally finite open refinement of such that . Thus is a locally finite RC refinement of such that by Lemma 4. Therefore, covers X. ■
A stronger result than that in Theorem 2 is given next.
Theorem 3 In an e.d. space , the following are equivalent:
(1) is countably I-paracompact.
(2) is countably S-paracompact.
(3) is feebly paracompact.
Proof. is the content of Theorem 2.
: Let be a countably S-paracompact space and let be an open cover of X. Then is a countable RC cover of X. Thus it has a locally finite RC refinement that covers X. For every , B is an RC-set and so which is open as X is e.d. Then is a locally finite open refinement of such that
Therefore, X is feebly paracompact.
: As X is feebly paracompact and e.d., by Lemma 5, X is countably S-paracompact and by Theorem 2, it is countably I-paracompact ■
The following Lemma follows easily by definitions.
Lemma 6 (1) A subset A of a space X is semipreopen if and only if is RC.
(2) .
Theorem 4 For a space , the following are equivalent:
(1) is countably I-paracompact.
(2) For every SPO countable cover of X, there exists a countable locally finite SPO refinement of such that
(3) For every SO countable cover of X, there exists a countable locally finite SO refinement of such that
(4) For every RSO countable cover of X, there exists a countable locally finite RSO refinement of such that
Proof. follow easily from Lemma 6. ■
3 On constructions of countably I-paracompact spaces
A property P of a space is a semiregular property if: has property P if and only if has property P, where is the topology generated by the set of all RO-subsets in X.
Lemma 7 [3] e.d. is a semiregular property.
Theorem 5 Countably S-Paracompact is a semiregular property.
Proof. If is a space and x and y are two distinct elements in X, then there exist disjoint RO-sets A and B such that and . Since A and B are both open, is .
Conversely, if is and x and y are two distinct elements in X, then there exist disjoint open sets U and V such that and . As and , and . As and are two disjoint RO-sets (otherwise, there exists which implies the existence of an open set W such that and as so . Thus there exists , that is there exists an open set T such that . Hence as , , which is a contradiction). Therefore, is . The rest of the proof is similar to that of Proposition 2.6 in [3]. ■
Combining Lemma 7 and Theorem 5 together with Theorem 3, we obtain the following result:
Corollary 3 Countably I-paracompactness is a semiregular property.
Lemma 8 [3] e.d. is a semiopen hereditary property.
The proof of the following result is similar to that of Proposition 2.9 in [3] and thus omitted.
Lemma 9 Let be a space. If is countably S-paracompact and Y is a regular semiopen subset of X such that , then is countably S-paracompact.
Since every RO-set and every RC-set is an RSO-set, we have the following result:
Corollary 4 Let be a countably I-paracompact space. Then
(1) countably I-paracompact for every RO-subset Y of X.
(2) countably I-paracompact for every RC-subset Y of X.
Lemma 10 [3] Let be a space and A is a PO subset of X. Then .
Theorem 6 If is an RO-set and is a countably I-paracompact subspace of X such that , then X is countably I-paracompact.
Proof. Let be an RC cover of X. Then by Lemma 10, is an RC cover of . Thus for each i=1,2,...,m, there exists a locally finite RC refinement of such that where means the interior of in the subspace topology . Then is an RC locally finite refinement of such that Therefore, X is countably I-paracompact. ■
4 Maps of countably I-paracompact spaces
A map is called irresolute (resp., semicontinuous) if the inverse image of every SO (resp., open) subset of Y is an SO-set in X. f is called almost open if for all .
Lemma 11 Let be an irresolute map from the countably S-paracompact space onto. Then is countably S-paracompact.
Proof. Let be an RC cover of Y. By Lemma 6, each is SO and as f is irresolute, is SO. Again by Lemma 6, is SPO and thus is an RC-set. Then is a countable RC cover of X (as ) and so it has an RC locally finite refinement that covers X. Hence is an RC locally finite refinement of that covers Y (as ) and therefore is countably S-paracompact. ■
Combining Lemma 11 together with Theorem 3, we obtain the following result:
Theorem 7 Let be an irresolute map from the countably I-paracompact space onto the e.d. space . Then is countably I-paracompact.
By , we mean the semi-closure of A. That is the smallest SC-set containing A. We next recall the following Lemma to prove our final result.
Lemma 12 (1) A map is semicontinuous if and only if , see [7].
(2) For an SO-set A in an e.d. space, , see [5].
We end this section by showing that the semicontinuous almost open image of a countably I-paracompact space is countably I-paracompact.