Abstract
In the following text, we want to study the behavior of one point compactification operator in the chain := Metrizable, Normal, , KC, SC, US, , , , , Top of subcategories of category of topological spaces, Top (where we denote the subcategory of Top, containing all topological spaces with property , simply by ). Actually we want to know, for and , the one point compactification of topological space belongs to which elements of . Finally we find out that the chain Metrizable, , KC, SC, US, T1, , , , Top is a forwarding chain with respect to one point compactification operator.
1. Introduction
The concept of forwarding and backwarding chains in a category with respect to a given operator has been introduced for the first time in [1] by the first author. The matter has been motivated by the following sentences in [1]: “In many problems, mathematicians search for theorems with weaker conditions or for examples with stronger conditions. In other words they work in a subcategory of a mathematical category, namely, , and they want to change the domain of their activity (theorem, counterexample, etc.) to another subcategory of like such that or according to their need.” Most of us have the memory of a theorem and the following question of our professors: “Is the theorem valid with weaker conditions for hypothesis or stronger conditions for result?” The concept of forwarding, backwarding, or stationary chains of subcategories of a category tries to describe this phenomenon.
In this text, Top denotes the category of topological spaces. Whenever is a topological property, we denote the subcategory of Top containing all the topological spaces with property , simply by . For example, we denote the category of all metrizable spaces by Metrizable.
We want to study the chain {Metrizable, Normal, , KC, SC, US, , TD, TUD, , Top} of subcategories of Top in the point of view of forwarding, backwarding, and stationary chains’ concept with respect to one point compactification or Alexandroff compactification operator.
Remark 1. Suppose is a partial order on . We call (i)a chain, if for all , we have ;(ii)cofinal, if for all , there exists such that .
In the following text, by a chain of subcategories of category , we mean a chain under “” relation (of subclasses of ). We recall that if is a chain of subcategories of category such that is closed under (multivalued) operator , then we call (i)a forwarding chain with respect to ; if for all , we have (i.e., );(ii)a full-forwarding chain with respect to ; if it is a forwarding chain with respect to and for all distinct , we have where, for multivalued function , by , we mean that at least one of the values of belongs to ;(iii)a backwarding chain with respect to ; if for all , we have ;(iv)a full-backwarding chain with respect to ; if it is a backwarding chain with respect to and for any distinct , we have where, for multivalued function , by , we mean that at least one of the values of belongs to ;(v)a stationary chain with respect to if it is both forwarding and backwarding chains with respect to .
Basic properties of forwarding, backwarding, full-forwarding, full-backwarding, and stationary chains with respect to given operators have been studied in [1]. We refer the interested reader to [2] for standard concepts of the Category Theory.
We recall that by we mean the set of all natural numbers ; also is the least infinite ordinal (cardinal) number and is the least infinite uncountable ordinal number. Here ZFC and GCH (generalized continuum hypothesis) are assumed (note: by GCH for infinite cardinal number , there is not any cardinal number with , i.e., ).
We call a collection of subsets of a filter over if ; for all we have ; for all and with we have . If is a maximal filter over (under relation), then we call it an ultrafilter over . If for all , we have ; then we call a uniform ultrafilter over .
We end this section by the following two examples.(I)For , let and with and for . Then is full-forwarding with respect to [1, Example 2.2].(II)Let ON denote the class of all ordinal numbers; CN denotes the class of all cardinal numbers; for every set by we mean cardinal number of , and for each cardinal number ON, ON, ON. Define ONON with for and CN. Then is a full-backwarding chain of subclasses of ON with respect to [1, Example 2.3].
2. Basic Definitions in Separation Axioms
In this section we bring our basic definitions in Top.
Convention 1. Henceforth in the topological space suppose . So (see [3, 4]) is a topological basis on . The space with topological basis is called one point compactification or Alexandroff compactification of .
Let
By the operator , in this text we mean the above mentioned operator.
Remark 2. We call a topological space (if , by , we mean the set of all limit points of in )(i); if for all distinct , there exist open neighborhood of and open neighborhood of such that or ;(ii)TUD; if for all , is a union of closed subsets of ;(iii)TD; if for all , is a closed subset of ;(iv); if for all , is a closed subset of ;(v)US if any convergent sequence has a unique limit;(vi)SC; if for any convergent sequence to , is a closed subset of ;(vii)KC if any compact subset of is closed;(viii) (or Hausdorff); if for all distinct there exist open neighborhood of and open neighborhood of with ;(ix)normal; if it is and for every disjoint closed subsets of , there exist disjoint open subsets of with and ;(x)-space; if for all , is closed if and only if for all closed compact subset of , is closed.
Regarding [5], we have KC SC US . Also by [6] we have ; therefore.
Metrizable Normal KC SC US Top.
In this section, we want to study the operator on the above chain. However, it has been proved in [1, Lemma 3.1 and Corollary 3.2] that the chain is stationary with respect to the operator ; therefore, the main interest is on Metrizable Normal KC SC US .
Note 1. A topological space is KC if and only if is an open subset of and is a compact subset of is a topological basis on .
Remark 3. Suppose is noncompact space and is one point compactification of . We have the following.(1)If is KC, then is US (therefore is too) [7, Theorem 4].(2)If is KC, then is KC if and only if is a -space [7, Theorem 5].(3) is if and only if is and locally compact [4]; thus is if and only if it is normal.(4) is an embedding of .(5)If is KC, then is KC too (hint: if is a compact subset of , then is a compact subset of by (2). If is KC, then is a closed subset of , and again by (2), is a closed subset of , so is KC).(6)A space is a -space if it is either first countable or locally compact so every metrizable space is -space [3, 7].(7) is if and only if is [3]. Moreover if is (and noncompact), then(8) is TD if and only if is TD [1, Lemma 3.1].(9) is TUD if and only if is TUD [1, Lemma 3.1].(10) is [1, Lemma 3.1].
For topological spaces , by , we mean topological disjoint union of and .
Lemma 4. Let is a compact topological space, is a noncompact topological space, then .
Proof. Suppose and is an open subset of . Using the following cases, is an open subset of too. (i)Consider . In this case, is an open subset of , so is an open subset of and is an open subset of not only but also using the definition of one point compactification. Then the set is an open subset of , since , is an open subset of .(ii)Consider . In this case, is a closed compact subset of . Since and are two closed subsets of , is a closed subset of and is a closed subset of and , so is a closed compact subset of . Therefore, is an open subset of and is an open subset of . Then the set is an open subset of , since , is an open subset of . Conversely, if is an open subset of , then, using the following cases, is an open subset of too.(iii)Consider . In this case, is an open subset of . Also is an open subset of and . Thus, is an open subset of ; hence it is an open subset of .(iv)Consider . In this case, is an open subset of by Remark 3(4). Using the compactness of , is a closed compact subset of . Also is an open subset of containing ; thus is a closed compact subset of . Since and are two closed compact subsets of , is a closed compact subset of too. Hence is an open subset of .
Lemma 5. If is a closed subset of , then is an embedding of .
Proof. If is compact, then and by Remark 3(4) we are done. If is not compact, is an open subset of and ; thus is a closed compact subset of . Suppose ; we prove that is a closed subset of as a subspace of if and only if is a closed subset of as one point compactification of . However, we mention that in both topologies is an embedding of by Remark 3(4).
First, suppose is a closed subset of . Using the following two cases, is a closed subset of too. (i)Consider . In this case, is an open subset of ; therefore it is an open subset of , so is a closed subset of .(ii)Consider . In this case, is a closed subset of since it is a closed subset of and is closed in . Therefore, is an open subset of with . So is a closed compact subset of . Therefore, is a closed compact subset of . Since , is a closed compact subset of , so is closed in . Conversely, suppose is a closed subset of . Using the following two cases, is a closed subset of too.(iii)Consider . In this case, is an open subset of ; therefore, there exists an open subset of with . is an open subset of too; thus is an open subset of ; therefore is a closed subset of .(iv)Consider . In this case, is a closed compact subset of with ; thus is a closed compact subset of . Hence, is a closed compact subset of , and is an open subset of . Therefore, is an open subset of , so is a closed subset of .
Lemma 6. Suppose Metrizable, Normal, , , , , , , , , Top}; also consider topological spaces . We have the following. (1) if and only if .(2)Consider two closed subsets of with and . So if and only if .
Proof. (1) has a formal proof, so we deal with (2). If and is a closed subspace of , then . Suppose ; are closed subspaces of with and . We prove .
First, note the fact that if is an open subset of (resp. ) with , then is an open subset of , since is an open subset of and is an open subset of . Now consider the following cases for . (i)Consider Metrizable. If are metrizable subspaces of , then there exist metrics , respectively, on such that , the metric topology induced from on is subspace topology on induced from , and the metric topology induced from on is subspace topology on induced from . Define with
Then the metric topology induced from on coincides with ’s original topology. (ii)Consider . Suppose are Hausdorff subspaces of and are two distinct points of . Consider the following cases:(1) and ; in this case, and are disjoint open neighborhoods of, respectively, and ;(2); there exist disjoint open subsets of with and . Suppose ; thus is an open subset of . There exists an open subset of with . Hence, are disjoint open subsets of with and .
Using the above cases, is Hausdorff. (iii)Consider Normal. If are normal subspaces of , then are Hausdorff and, using the case “”, is Hausdorff. Now suppose are disjoint closed subsets of ; also we may suppose . Let , , . and . There are disjoint open subsets of containing, respectively, . Also there are disjoint open subsets of containing, respectively, . There are open subsets of with and . Let and ; then are disjoint open subsets of containing, respectively, .(iv)Consider KC. Suppose are KC and is a compact subset of . Since are closed, are compact too. Since is a compact subset of and is KC, is a closed subset of . Since is a closed subset of and is a closed subset of , is closed subset of . Similarly, is a closed subset of . Thus is a closed subset of and is KC.(v)Consider SC. Suppose are SC and is a sequence in converging to . Using the following cases, is a closed subset of .(1)Consider . Suppose . In this case, is an open neighborhood of in , so there exists such that for all . Hence is a converging sequence to in . Since is SC, is a closed subset of . Therefore, is a closed subset of . For each if (resp. ), is a closed subset of (resp. ) since (resp. ) is SC and in particular . Thus for all , is a closed subset of . By closeness of and in , the set is closed in .(2)Consider and there exists such that or . Suppose there exists with . In this case. is a converging sequence to in , and, using the same argument as in the second paragraph of the case “”, is closed in .(3)Consider none of the above two cases. In this case, converges to and it has two subsequences and such that , , and . Using item (2), and are two closed subsets of ; thus is a closed subset of .(vi)Consider US. If are US and is not US, consider converging sequence in to with . Let ; we may suppose . The set is an open neighborhood of . Thus there exists with and . So is a converging sequence to in and ; thus is not US which is a contradiction.(vii)Consider . Suppose and are T1; let . We may suppose . Since is T1, is a closed subset of . Since is a closed subset of and is a closed subset of , is a closed subset of .(viii)Use similar methods for the rest of the cases of .
Lemma 7. Suppose Metrizable, Normal, , , , , , , , , Top}; also consider topological spaces . We have if and only if .
Proof. By Lemma 6 and Lemma 4, it is clear if or is compact. So we may suppose and are two disjoint noncompact topological spaces. Since and are two open subset of , two sets and are two closed subsets of with . By Lemma 6, if and only if . By Lemma 5, is homeomorphic to and is homeomorphic to ; hence if and only if .
Lemma 8. In topological space , if is SC, then is US.
Proof. Let be a noncompact SC space. Suppose is a sequence in converging to . We have the following cases. (i)Consider . In this case, is an open neighborhood of in ; hence there exists such that for all . Therefore, is a converging sequence in to . Since is SC, is US and .(ii)Consider . In this case, there exists such that for all . Therefore, is a converging sequence in to . Thus is a closed subset of . So is a compact closed subset of and is an open neighborhood of which is a contradiction by for all and by converging to . So this case does not occur.
Using the above cases, we have , and is US.
3. The Main Table
See Figure 1; then we have Table 1 which we prove in this Section and where:
The mark “” indicates that in the corresponding case, there exists such that , and the mark “—” indicates that in the corresponding case for all we have .
Let By Remark 3(7) in Table 1, the mark “—” for cases in which “, ” or “ , ” is evident. However, it has been proved in [1, Lemma 3.1 and Corollary 3.2] that the chain is stationary with respect to the operator , so corresponding marks of the cases in which are obtained. Thus it remains to discuss cases in which .
Since the subspace of a metrizable (resp. , SC, and US) space is metrizable (resp. , SC, and US) using Remark 3(4) and (5), if is, respectively, metrizable , KC, SC, or US, then is too. Hence we obtain “—” for the following cases too (choose and from the same rows of Table 2).
Proof (proof of the rest of the cells of Figure 1).
First Row. Here we have and the following cases for .
(i) Consider . Consider two spaces (with induced metric from Euclidean space ) and (with induced metric from Euclidean space ); then is homeomorphic to ; moreover ; therefore .
(ii) Consider . Consider with discrete topology. is compact Hausdorff, so it is normal. If is a topological basis for , then for all we have . Therefore is uncountable and compact space is not metrizable. Thus and .
(iii) Consider . Use Remark 3(3).
(iv) Consider . Consider as the set of all rational numbers as a subspace of Euclidean space . Since is not locally compact, by Remark 3(3), is not Hausdorff. Suppose is a compact subset of ; in order to show that is KC, we show is a closed subset of . We have the following two cases.
Case 1. If , then is a compact subset of ; since is a metric space, is a closed subset of too. Therefore, is an open subset of . Hence, is a closed subset of .
Case 2. If , we claim that is an open subset of and so an open subset of ; otherwise (since is metrizable) there exists a one-to-one sequence in () converging to a point (in metric space ). For all , is a compact closed subset of , and is an open subset of . Since , . Using the compactness of , there exists such that . Since , we have which is a contradiction. Thus is an open subset of , and is a closed subset of .
Finally we have and .
(v) Consider . If is a metric space, then it is a -space and, by Remark 3(2), is KC; hence .
(vi) Consider or . We claim that if is a metric space, then is SC. First, note the fact that, by Remark 3(1), is US and hence . Suppose is a sequence in converging to ; we show that is a closed subset of . Consider the following cases.
Case 1. ; in this case, is an open neighborhood of , thus there exists such that for all and converges to in metric space (by Remark 3(4), as a subspace of has its original topology). Thus is a closed compact subset of ; therefore is an open subset of . Finally is a closed subset of and since is , is a closed subset of too.
Case 2. and is finite. In this case, is a finite subset of ( space) and it is closed.
Case 3. and is infinite. In this case, we may assume for all . If is not a closed subset of , then there exists a subsequence of converging to . Thus converges to in too (use Remark 3(4)). Since converges to and is US, we have which is a contradiction with . Therefore is a closed subset of , so
is an open subset of and . Finally is a closed subset of .
Using the above three cases, is a closed subset of and we are done.
Second Row. Here we have and the following cases for .
(i) Consider . Suppose is the least uncountable ordinal number. Consider (with order topology). Since is well ordered, it is normal. However, is not metrizable and .
(ii) Consider . If and are , then, by Remark 3(3), is normal.
(iii) Consider . Consider as disjoint union of with order topology and as the set of all rational numbers with induced metric from Euclidean space . The topological space is normal since are normal. Moreover, is disjoint union of and , so has nonmetrizable subspace , thus is nonmetrizable and . Since is compact, by Lemma 4 we have . Considering case “, ”, we have . Using Lemma 6(1), we have ; thus .
(iv) Consider . Let be an uncountable set and . Consider under Fortissimo topology with particular point , that is, under the topology is countable (see [4, counterexample 25]). For distinct , suppose . Then are disjoint open subsets of and is Hausdorff. On the other hand, if are disjoint closed subsets of , suppose , then and are disjoint open subsets of containing and . Thus is normal.
Moreover, if is an open neighborhood of , then is countable; therefore is uncountable and . Therefore, . Let be a sequence of elements of . The sequence does not converge to , since is an open neighborhood of . The space is not metrizable since and there is not any sequence in converging to . So .
On the other hand, using the definition of one point compactification, any subset of containing is a compact subset of . Therefore, is a compact subset of , but it is not a closed subset of ; thus is not KC. We claim that is SC. Suppose is a sequence in converging to . We have the following cases.
Case 1. Consider . In this case, is an open neighborhood of and there exists such that for all we have . Using Remark 3(1), is ; thus is a closed subset of .
Case 2. Consider . The set is an open neighborhood of ; thus there exists such that for all we have (thus for all we have ). Using Remark 3(1), is ; thus is a closed subset of .
Case 3. Consider . In this case, is an open subset of ; therefore it is an open subset of . Thus is a closed subset of .
Using the above three cases, is a closed subset of , and is SC.
Since is SC and it is not KC, .
(v) Consider . Suppose is a uniform ultrafilter over . Consider under topology . If are distinct with , then are disjoint open subsets of containing and is Hausdorff. If are disjoint closed subsets of , suppose . Therefore, are disjoint open subsets of and is normal. Since is a uniform ultrafilter over , it does not contain any finite subset of . Since all of the elements of are infinite, is a limit point of and . Consider a sequence in . We have the following cases.
Case 1. has a constant subsequence like . Since is Hausdorff and every sequence converges to at most one point, converges to its constant value and does not converge to . Thus does not converge to .
Case 2. does not have any constant subsequence. Suppose is a one-to-one subsequence of . Since is an ultrafilter over , and , we have or . Suppose . Since is an ultrafilter over , . Therefore is an open neighborhood of and does not converge to .
Since and by the above two cases, there is not any sequence in converging to ; is not metrizable. Thus .
Now pay attention to the following claims.
Claim 1. The sequence converges to in . Suppose is an open neighborhood of in . Since is , is too. Therefore, is an open neighborhood of in ; thus is a compact (and closed) subset of . Also is finite, since is discrete and is a compact subset of (use Remark 3(4)). Suppose . For all , we have . Hence converges to .
Claim 2. is not a closed subset of . Using the fact that , we have ; hence is not a closed subset of .
Regarding Claims 1 and 2, is not SC. Since is normal, it is KC; so using Remark 3(1), is US. Therefore, .
(vi) Consider . If is Normal, then it is KC and by Remark 3(1), is US. Thus .
Third Row. Here we have and the following cases for . (i)Consider . Suppose is a Hausdorff locally compact nonnormal topological space. Since is not normal, it is not metrizable and . By Remark 3(3), is normal. By Remark 3(4), is not metrizable. Hence . Moreover , where and have their order topology and equipped with product topology (deleted Tykhonoff plank [4, counterexample 87]) is Hausdorff locally compact nonnormal topological space and is an example for this case.(ii)Consider . Use a similar method described for and .(iii)Consider . Consider under topology is an open subset of in its Euclidean topology and , then (Smirnov’s deleted sequence topology [4, counterexample 64]). Also is first countable; therefore it is a -space by Remark 3(6). By Remark 3(2), is KC. Moreover, is not Hausdorff, since is not locally compact in (use Remark 3(3)). Hence .(iv)Consider . Consider as disjoint union of and , where(1) is an uncountable set under Fortissimo topology with particular point , that is, under topology is countable (see [4, counterexample 25] and proof of Table 1 regarding case “, ”);(2) under the topology is an open subset of in its Euclidean topology and (see Smirnov’s deleted sequence topology [4, counterexample 64] and proof of Table 1 regarding case “, ”).
Since and , we have by Lemma 6(1). Moreover and lead us to by Lemma 7. (v)Consider . Consider as disjoint union of and , where we have the following.(1)Suppose is a uniform ultrafilter over . Consider under topology (see proof of Table 1 regarding case “, ”).(2) is Smirnov’s deleted sequence topological space (see proof of Table 1 regarding case “, ”).
Then by and and Lemma 6(1). Also by , , and Lemma 7. (vi)Consider . If is , then it is KC and by Remark 3(1), is US. Thus .
Fourth Row. Here we have and the following cases for .
(i) Consider . Consider as the set of all rational numbers as a subspace of Euclidean space . Using the case “, ” for , we have . Since is compact, we have .
(ii) Consider . Consider uncountable set with countable complement topology is countable [4, counterexamples 20 and 21]. Since every two nonempty open subsets of have nonempty intersection, is not Hausdorff. It is clear that is . Moreover, is a compact subset of if and only if is finite. Therefore, every compact subset of is closed and is KC. So .
Now suppose is an uncountable subset of with uncountable complement. So is not closed. For all compact subset of , the set is finite and closed. Therefore, is not a -space. Using Remark 3(2), is not KC. Using Remark 3(1), is US; we claim that is SC. Suppose is a sequence in , converging to . We have the following cases.
Case 1. Consider . In this case, is an open neighborhood of in . So there exists such that for all we have and . Therefore, is a (finite and) closed subset of .
Case 2. Consider . Since is open in , it is open in too. Thus is closed in .
By the above cases, is closed in and is SC. Hence .
(iii) Consider . Consider as disjoint union of and , where we have the following. (1)Suppose is a uniform ultrafilter over . Consider under topology (see proof of Table 1 regarding case “, ”).(2) is an uncountable set with countable complement topology is countable (see [4, counterexamples 20 and 21] and proof of Table 1 regarding case “, ”).
Then by , , and Lemma 6(1). Also by , , and Lemma 7.
(iv) Consider . If is KC and by Remark 3(1), is US. Thus .
Fifth Row. Here we have and the following cases for . (i)Consider . Consider as with doubling [8]; that is, if , let under topological basis and are ordinal numbers with is an ordinal number with is an ordinal number with is an ordinal number with . Then and is compact which leads to too.(ii)Consider . Consider as disjoint union of and , where we have the following.(1) is with doubling [8] as in case “, ”. Then .(2)For uniform ultrafilter over , consider is equipped with topology . Using case “ , ”, we have and . By , , and Lemma 6(1), we have . By , , and Lemma 7, we have .(iii)Consider . Use Lemma 8.
Sixth Row. Here we have and the following cases for . (i)Consider . Consider as , such that has its usual order topology, is a uniform ultrafilter over , and is an open neighborhood basis for [8, example 1.2]; then is compact and .(ii)Consider . According to [7, example 5], there exists a US topological space such that is not US. By Remark 3(7), is . Hence . Moreover, by Lemma 8, is not SC; thus .
Seventh Row. Here we have and . Suppose as an infinite set with finite complement topology ( is finite)} [4, counterexamples 18 and 19]. Then is compact and .
4. Some Observations in Figure 1
Using Figure 1, we have the following results. The collection , KC, SC, is a full-forwarding chain with respect to . In other words, Table 3 is valid.
In Table 3, the mark “” indicates that in the corresponding case there exists such that , and the mark “—” indicates that
in the corresponding case for all we have . The collection {Metrizable, , KC, SC, US, , TD, TUD, , Top} is a forwarding chain with respect to . The collection , Top} is a stationary chain with respect to .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors would like to take this opportunity to express their special thanks to S. Hassani, H. R. Daneshpajooh, and M. Nayeri for their help and cooperation.