# Study of a Forwarding Chain in the Category of Topological Spaces betweenT0 and T2 with respect to One Point Compactification Operator

Authors: Fatemah Ayatollah Zadeh Shirazi; Meysam Miralaei; Fariba Zeinal Zadeh Farhadi
Journal: Chinese Journal of Mathematics
 (2014)

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## Abstract

In the following text, we want to study the behavior of one point compactification operator in the chainΞ := {Metrizable, Normal, T2, KC, SC, US, T1, TD, TUD, T0, Top} of subcategories of category of topological spaces, Top (where we denote the subcategory of Top, containing all topological
spaces with property P , simply by P). Actually we want to know, for P∈Ξ and X∈P, the one point compactification of topological space X
belongs to which elements of Ξ. Finally we find out that the chain {Metrizable, T2, KC, SC, US, T1, TD, TUD, T0, Top} is a forwarding chain with respect to one point compactification operator.


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## Body

## 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 subcategory𝒟of a mathematical category, namely,𝒞, and they want to change the domain of their activity (theorem, counterexample, etc.) to another subcategory of𝒞like𝒦such that𝒦⊆𝒟or𝒟⊆𝒦according 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 𝒞 tries to describe this phenomenon.In this text, Top denotes the category of topological spaces. WheneverP is a topological property, we denote the subcategory of Top containing all the topological spaces with property P, simply by P. For example, we denote the category of all metrizable spaces by Metrizable.We want to study the chain {Metrizable, Normal,T2, KC, SC, US, T1, TD, TUD, T0, 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 A. We call B⊆A(i)
achain, if for all x,y∈B, we have x≤y∨y≤x;(ii)
cofinal, if for all x∈A, there exists y∈B such that x≤y.In the following text, by a chain of subcategories of category𝒞, we mean a chain under “⊆” relation (of subclasses of 𝒞). We recall that if ℳ is a chain of subcategories of category 𝒞 such that ⋃ℳ is closed under (multivalued) operator ψ, then we call ℳ(i)
aforwarding chain with respect toψ; if for all C∈ℳ, we have ψ((⋃ℳ)∖C)∩C=⌀ (i.e., ψ((⋃ℳ)∖C)⊆(⋃ℳ)∖C);(ii)
afull-forwarding chain with respect toψ; if it is a forwarding chain with respect to ψ and for all distinct C1,C2,C3∈ℳ, we have
(1)C1⊆C2⊆C3⟹(∃X∈C2∖C1ψ(X)∈C3∖C2),
where, for multivalued function ψ, by ψ(X)∈C3∖C2, we mean that at least one of the values of ψ(X) belongs to C3∖C2;(iii)
abackwarding chain with respect toψ; if for all C∈ℳ, we have ψ(C)⊆C;(iv)
afull-backwarding chain with respect toψ; if it is a backwarding chain with respect to ψ and for any distinct C1,C2,C3∈ℳ, we have
(2)C1⊆C2⊆C3⟹(∃X∈C3∖C2ψ(X)∈C2∖C1),
where, for multivalued function ψ, by ψ(X)∈C2∖C1, we mean that at least one of the values of ψ(X) belongs to C2∖C1;(v)
astationary chain with respect toψ if it is both forwarding and backwarding chains with respect to ψ.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 byℕ we mean the set of all natural numbers {1,2,…}; also ω={0,1,2,…} is the least infinite ordinal (cardinal) number and Ω is the least infinite uncountable ordinal number. Here ZFC and GCH (generalized continuum hypothesis) are assumed (note: by GCH for infinite cardinal number α, there is not any cardinal number β with α<β<2α, i.e., α+=2α).We call a collectionℱ of subsets of X a filter over X if ⌀∉ℱ; for all A,B∈ℱ we have A∩B∈ℱ; for all A∈ℱ and B⊆X with A⊆B we have B∈ℱ. If ℱ is a maximal filter over X (under ⊆ relation), then we call it an ultrafilter over X. If for all A∈ℱ, we have card(A)=card(X); then we call ℱ a uniform ultrafilter over X.We end this section by the following two examples.(I)
Forn∈ℕ, let Cn:=[0,1/n] and ψ:[0,1]→[0,1] with ψ(0)=0 and ψ(x)=n2x+1-n for x∈Cn∖Cn+1=(1/(n+1),1/n]. Then {Cn:n∈ℕ} is full-forwarding with respect to ψ [1, Example 2.2].(II)
Let ON denote the class of all ordinal numbers; CN denotes the class of all cardinal numbers; for every setA by |A| we mean cardinal number of A, and for each cardinal number α∈ ON, Dα={γ∈ ON:|γ|=α}, Cα={γ∈ ON:|γ|<α}. Define ψ: ON→ON with ψ(γ)=γ-α for γ∈Dα and α∈ CN. Then {Cα:α≥ω} is a full-backwarding chain of subclasses of ON with respect to ψ [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 X suppose ∞∉X. So (see [3, 4])
(3)ℬ≔{U⊆X:UisanopensubsetofX}∪{V∪{∞}:V⊆X,X∖Viscompactandclosed}
is a topological basis on X∪{∞}. The space X∪{∞} with topological basis ℬ is called one point compactification or Alexandroff compactification of X.Let(4)A(X)≔{onepointcompactificationofXXisnotcompact,XXiscompact.By the operatorA, in this text we mean the above mentioned operator.Remark 2.
We call a topological spaceX (if A⊆X, by A′, we mean the set of all limit points of A in X)(i)
T0; if for all distinct x,y∈X, there exist open neighborhood U of x and open neighborhood V of y such that x∉V or y∉U;(ii)
TUD; if for all x∈X, {x}′ is a union of closed subsets of X;(iii)
TD; if for all x∈X, {x}′ is a closed subset of X;(iv)
T1; if for all x∈X, {x} is a closed subset of X;(v)
US if any convergent sequence has a unique limit;(vi)
SC; if for any convergent sequence(xn:n∈ω) to x∈X, {xn:n∈ω}∪{x} is a closed subset of X;(vii)
KC if any compact subset ofX is closed;(viii)
T2 (or Hausdorff); if for all distinct x,y∈X there exist open neighborhood U of x and open neighborhood V of y with U∩V=⌀;(ix)
normal; if it isT2 and for every disjoint closed subsets A,B of X, there exist disjoint open subsets U,V of X with A⊆U and B⊆V;(x)
k-space; if for all A⊆X, A is closed if and only if for all closed compact subset K of X, A∩K is closed.Regarding [5], we have T2⊆ KC ⊆ SC ⊆ US ⊆T1. Also by [6] we have T1⊆TD⊆TUD⊆T0; therefore.Metrizable⊆ Normal ⊆T2⊆ KC ⊆ SC ⊆ US ⊆T1⊆TD⊆TUD⊆T0⊆ Top.In this section, we want to study the operatorA on the above chain. However, it has been proved in [1, Lemma 3.1 and Corollary 3.2] that the chain T1⊆TD⊆TUD⊆T0 is stationary with respect to the operator A; therefore, the main interest is on Metrizable ⊆ Normal ⊆T2⊆ KC ⊆ SC ⊆ US ⊆T1.Note 1.
A topological spaceX is KC if and only if {U⊆X:U is an open subset of X}∪{V∪{∞}:V⊆X and X∖V is a compact subset of X} is a topological basis on X.Remark 3.
SupposeX is noncompact space and A(X)=X∪{∞} is one point compactification of X. We have the following.(1)
IfX is KC, then A(X) is US (therefore A(X) is T1 too) [7, Theorem 4].(2)
IfX is KC, then A(X) is KC if and only if X is a k-space [7, Theorem 5].(3)
A(X) is T2 if and only if X is T2 and locally compact [4]; thus A(X) is T2 if and only if it is normal.(4)
A(X) is an embedding of X.(5)
IfA(X) is KC, then X is KC too (hint: if K is a compact subset of X, then K is a compact subset of A(X) by (2). If A(X) is KC, then K is a closed subset of A(X), and again by (2), K is a closed subset of X, so X is KC).(6)
AT2 space is a k-space if it is either first countable or locally compact so every metrizable space is k-space [3, 7].(7)
X is T1 if and only if A(X) is T1 [3]. Moreover if X is T0 (and noncompact), then(8)
X is TD if and only if A(X) is TD [1, Lemma 3.1].(9)
X is TUD if and only if A(X) is TUD [1, Lemma 3.1].(10)
A(X) is T0 [1, Lemma 3.1].For topological spacesX,Y, by X⊔Y, we mean topological disjoint union of X and Y.Lemma 4.
LetX1 is a compact topological space, X2 is a noncompact topological space, then A(X1⊔X2)=X1⊔A(X2).Proof.
SupposeX1∩X2=⌀ and U is an open subset of A(X1⊔X2)=X1∪X2∪{∞}. Using the following cases, U is an open subset of X1⊔A(X2)(=X1∪X2∪{∞}) too. (i)
Consider∞∉U. In this case, U is an open subset of X1⊔X2, so U1:=U∩X1 is an open subset of X1 and U2:=U∩X2 is an open subset of not only X2 but also A(X2) using the definition of one point compactification. Then the set U1∪U2 is an open subset of X1⊔A(X2), since U=U1∪U2, U is an open subset of X1⊔A(X2).(ii)
Consider∞∈U. In this case, (X1∪X2)∖U is a closed compact subset of X1⊔X2. Since X1 and X2 are two closed subsets of X1⊔X2, ((X1∪X2)∖U)∩X1=X1∖U is a closed subset of X1 and ((X1∪X2)∖U)∩X2=X2∖U is a closed subset of X2 and (X1∪X2)∖U, so X2∖U is a closed compact subset of X2. Therefore, U1:=X1∖(X1∖U)=X1∩U is an open subset of X1 and U2:=A(X2)∖(X2∖U)=A(X2)∩U(=(X2∩U)∪{∞}) is an open subset of A(X2). Then the set U1∪U2 is an open subset of X1⊔A(X2), since U=U1∪U2, U is an open subset of X1⊔A(X2). Conversely, if V is an open subset of X1⊔A(X2), then, using the following cases, V is an open subset of A(X1⊔X2) too.(iii)
Consider∞∉V. In this case, V1:=V∩X1 is an open subset of X1. Also V2:=V∩A(X2)=V∩X2 is an open subset of A(X2) and X2. Thus, V=V1∪V2 is an open subset of X1⊔X2; hence it is an open subset of A(X1⊔X2).(iv)
Consider∞∈V. In this case, V1:=V∩X1 is an open subset of X1 by Remark 3(4). Using the compactness of X1, X1∖V1 is a closed compact subset of X1. Also V2:=V∩A(X2) is an open subset of A(X2) containing ∞; thus X2∖V2 is a closed compact subset of X2. Since X1∖V1 and X2∖V2 are two closed compact subsets of X1⊔X2, (X1∖V1)∪(X2∖V2)=(X1∪X2)∖V is a closed compact subset of X1⊔X2 too. Hence V is an open subset of A(X1⊔X2).Lemma 5.
IfY is a closed subset of X, then A(X) is an embedding of A(Y).Proof.
IfY is compact, then A(Y)=Y and by Remark 3(4) we are done. If Y is not compact, X∖Y is an open subset of X and A(X); thus Y∪{∞} is a closed compact subset of A(X). Suppose F⊆Y∪{∞}; we prove that F is a closed subset of Y*:=Y∪{∞} as a subspace of A(X) if and only if F is a closed subset of A(Y)=Y∪{∞} as one point compactification of Y. However, we mention that Y∪{∞} in both topologies is an embedding of Y by Remark 3(4).
First, supposeF is a closed subset of Y*. Using the following two cases, F is a closed subset of A(Y) too. (i)
Consider∞∈F. In this case, U:=Y*∖F=Y∖F is an open subset of Y; therefore it is an open subset of A(Y), so F=A(Y)∖U is a closed subset of A(Y).(ii)
Consider∞∉F. In this case, F is a closed subset of A(X) since it is a closed subset of Y* and Y* is closed in A(X). Therefore, U:=A(X)∖F is an open subset of A(X) with ∞∈U. So A(X)∖U is a closed compact subset of X. Therefore, (A(X)∖U)∩Y*=F is a closed compact subset of Y*. Since (A(X)∖U)∩Y*=(A(X)∖U)∩Y, F is a closed compact subset of Y, so F is closed in A(Y). Conversely, suppose F is a closed subset of A(Y). Using the following two cases, F is a closed subset of Y* too.(iii)
Consider∞∈F. In this case, U:=A(Y)∖F=Y∖F is an open subset of Y; therefore, there exists an open subset V of X with V∩Y=U. V is an open subset of A(X) too; thus V∩Y* is an open subset of Y*; therefore Y*∖(V∩Y*)=Y*∖(V∩Y)=Y*∖U=F is a closed subset of Y*.(iv)
Consider∞∉F. In this case, F is a closed compact subset of A(Y) with ∞∉F; thus F is a closed compact subset of Y. Hence, F is a closed compact subset of X, and U=A(X)∖F is an open subset of A(X). Therefore, U∩Y*=Y*∖F is an open subset of Y*, so F is a closed subset of Y*.Lemma 6.
Suppose𝒞∈{Metrizable, Normal, T2, KC, SC, US, T1,
TD,
TUD, T0, Top}; also consider topological spaces X,Y. We have the following. (1)
X⊔Y∈𝒞 if and only if X,Y∈𝒞.(2)
Consider two closed subsetsA,B of X with A∪B=X and A∩B={t}. So A,B∈𝒞 if and only if X∈𝒞.Proof.
(1) has a formal proof, so we deal with (2). IfX∈𝒞 and E is a closed subspace of X, then E∈𝒞. Suppose A,B∈𝒞; A,B are closed subspaces of X with A∩B={t} and A∪B=X. We prove X∈𝒞.
First, note the fact that ifV is an open subset of A (resp. B) with t∉V, then V is an open subset of X, since V is an open subset of A∖{t} and A∖{t}(=X∖B) is an open subset of X. Now consider the following cases for 𝒞. (i)
Consider𝒞= Metrizable. If A,B are metrizable subspaces of X, then there exist metrics d1,d2, respectively, on A,B such that d1,d2≤1, the metric topology induced from d1 on A is subspace topology on A induced from X, and the metric topology induced from d2 on B is subspace topology on B induced from X. Define d:X×X→[0,+∞) with
(5)d(x,y)={d1(x,y)x,y∈A,d2(x,y)x,y∈B,2otherwise.Then the metric topology induced from d on X coincides with X’s original topology. (ii)
Consider𝒞=T2. Suppose A,B are Hausdorff subspaces of X and x,y∈X are two distinct points of X. Consider the following cases:
(1)
x∈X∖A=B∖{t} and y∈X∖B=A∖{t}; in this case, B∖{t} and A∖{t} are disjoint open neighborhoods of, respectively, x and y;(2)
x,y∈A; there exist disjoint open subsets U1,U2 of A with x∈U1 and y∈U2. Suppose t∉U1; thus U1 is an open subset of X. There exists an open subset U of X with U∩A=U2. Hence, U1,U are disjoint open subsets of X with x∈U1 and y∈U.
Using the above cases,X is Hausdorff. (iii)
Consider𝒞= Normal. If A,B are normal subspaces of X, then A,B are Hausdorff and, using the case “𝒞=T2”, X is Hausdorff. Now suppose E,F are disjoint closed subsets of X; also we may suppose t∉E. Let EA:=E∩A, EB:=E∩B, FA:=F∩A. and FB:=F∩B. There are disjoint open subsets UE,UF of A containing, respectively, EA,FA. Also there are disjoint open subsets VE,VF of B containing, respectively, EB,FB. There are open subsets U,V of X with UF=A∩U and VF=V∩B. Let WE:=(UE∖{t})∪(VE∖{t}) and WF:=U∪V; then WE,WF are disjoint open subsets of X containing, respectively, E,F.(iv)
Consider𝒞= KC. Suppose A,B are KC and K is a compact subset of X. Since A,B are closed, A∩K,B∩K are compact too. Since A∩K is a compact subset of A and A is KC, A∩K is a closed subset of A. Since A∩K is a closed subset of A and A is a closed subset of X, A∩K is closed subset of X. Similarly, B∩K is a closed subset of X. Thus K=(A∩K)∪(B∩K) is a closed subset of X and X is KC.(v)
Consider𝒞= SC. Suppose A,B are SC and (xn:n∈ω) is a sequence in X converging to x. Using the following cases, {xn:n∈ω}∪{x} is a closed subset of X.
(1)
Considerx≠t. Suppose x∈A∖{t}. In this case, A∖{t} is an open neighborhood of x in X, so there exists N∈ω such that xn∈A∖{t} for all n≥N. Hence (xn:n≥N) is a converging sequence to x in A. Since A is SC, {xn:n≥N}∪{x} is a closed subset of A. Therefore, {xn:n≥N}∪{x} is a closed subset of X. For each n∈ω if xn∈B (resp. xn∈A), {xn} is a closed subset of B (resp. A) since B (resp. A) is SC and in particular T1. Thus for all n∈ω, {xn} is a closed subset of X. By closeness of {xn:n≤N} and {xn:n≥N}∪{x} in X, the set {xn:n∈ω}∪{x} is closed in X.(2)
Considerx=t and there exists N∈ω such that {xn:n≥N}⊆A or {xn:n≥N}⊆B. Suppose there exists N∈ω with {xn:n≥N}⊆A. In this case. (xn:n≥N) is a converging sequence to x in A, and, using the same argument as in the second paragraph of the case “x≠t”, {xn:n∈ω} is closed in X.(3)
Consider none of the above two cases. In this case,(xn:n∈ω) converges to t and it has two subsequences (xnk:k∈ω) and (xmk:k∈ω) such that {xnk:k∈ω}⊆A, {xmk:k∈ω}⊆B, and {nk:k∈ω}∪{mk:k∈ω}=ω. Using item (2), {xnk:k∈ω}∪{x} and {xmk:k∈ω}∪{x} are two closed subsets of X; thus {xn:k∈ω}∪{x}={xnk:k∈ω}∪{x}∪{xmk:k∈ω}∪{x} is a closed subset of X.(vi)
Consider𝒞= US. If A,B are US and X is not US, consider converging sequence (xn:n∈ω) in X to x,y with x≠y. Let x≠t; we may suppose x∈A. The set A∖{t}(=X∖B) is an open neighborhood of x. Thus there exists N∈ω with {xn:n≥N}⊆A∖{t} and y∈{xn:n≥N}¯⊆A. So (xn:n≥N) is a converging sequence to x,y in A and x≠y; thus A is not US which is a contradiction.(vii)
Consider𝒞=T1. Suppose A and B are T1; let x∈X. We may suppose x∈A. Since A is T1, {x} is a closed subset of A. Since A is a closed subset of X and {x} is a closed subset of A, {x} is a closed subset of X.(viii)
Use similar methods for the rest of the cases of𝒞.Lemma 7.
Suppose𝒞∈{Metrizable, Normal, T2, KC, SC, US, T1,
TD,
TUD, T0, Top}; also consider topological spaces X,Y. We have A(X⊔Y)∈𝒞 if and only if A(X),A(Y)∈𝒞.Proof.
By Lemma6 and Lemma 4, it is clear if X or Y is compact. So we may suppose X and Y are two disjoint noncompact topological spaces. Since X and Y are two open subset of X⊔Y, two sets X*:=A(X⊔Y)∖Y(=X∪{∞}) and Y*:=A(X⊔Y)∖X(=Y∪{∞}) are two closed subsets of A(X⊔Y) with X*∪Y*=A(X⊔Y). By Lemma 6, A(X⊔Y)∈𝒞 if and only if X*,Y*∈𝒞. By Lemma 5, X* is homeomorphic to A(X) and Y* is homeomorphic to A(Y); hence A(X⊔Y)∈𝒞 if and only if A(X),A(Y)∈𝒞.Lemma 8.
In topological spaceX, if X is SC, then A(X) is US.Proof.
LetX be a noncompact SC space. Suppose (xn:n∈ω) is a sequence in A(X)=X∪{∞} converging to x,y∈A(X). We have the following cases. (i)
Considerx,y∈X. In this case, X is an open neighborhood of x,y in A(X); hence there exists N∈ω such that xn∈X for all n≥N. Therefore, (xn:n≥N) is a converging sequence in X to x,y. Since X is SC, X is US and x=y.(ii)
Considerx∈X,y=∞. In this case, there exists N∈ω such that xn∈X for all n≥N. Therefore, (xn:n≥N) is a converging sequence in X to x. Thus {xn:n≥N}∪{x} is a closed subset of X. So {xn:n≥N}∪{x} is a compact closed subset of X and V:=A(X)∖({xn:n≥N}∪{x}) is an open neighborhood of ∞(=y) which is a contradiction by xn∉V for all n≥N and by converging (xn:n∈ω) to y. So this case does not occur.
Using the above cases, we havex=y, and A(X) is US.


## 3. The Main Table

See Figure1; then we have Table 1 which we prove in this Section and where:Table 1
P
Q
C
1
C
2
∖
C
1
C
3
∖
C
2
C
4
∖
C
3
C
5
∖
C
4
C
6
∖
C
5
C
7
∖
C
6
C
8
∖
C
7
C
9
∖
C
8
C
10
∖
C
9
C
11
∖
C
10
C
1
√
√
—
√
—
—
—
—
—
—
—
C
2
∖
C
1
—
√
—
√
√
√
—
—
—
—
—
C
3
∖
C
2
—
√
—
√
√
√
—
—
—
—
—
C
4
∖
C
3
—
—
—
√
√
√
—
—
—
—
—
C
5
∖
C
4
—
—
—
—
√
√
—
—
—
—
—
C
6
∖
C
5
—
—
—
—
—
√
√
—
—
—
—
C
7
∖
C
6
—
—
—
—
—
—
√
—
—
—
—
C
8
∖
C
7
—
—
—
—
—
—
—
√
—
—
—
C
9
∖
C
8
—
—
—
—
—
—
—
—
√
—
—
C
10
∖
C
9
—
—
—
—
—
—
—
—
—
√
—
C
11
∖
C
10
—
—
—
—
—
—
—
—
—
—
√Figure 1
Consider the previous classes of topological spaces (in order to be more convenient, note the right diagram).The mark “√” indicates that in the corresponding case, there exists X∈P such that A(X)∈Q, and the mark “—” indicates that in the corresponding case for all X∈P we have A(X)∉Q.Let(6)E:={C1,C2∖C1,C3∖C2,C4∖C3,C5∖C4,C6∖C5,C7∖C6},F:={C8∖C7,C9∖C8,C10∖C9,C11∖C10}.
By Remark 3(7) in Table 1, the mark “—” for cases in which “P∈E, Q∈F” or “P∈F, Q∈E” is evident. However, it has been proved in [1, Lemma 3.1 and Corollary 3.2] that the chain T1⊆TD⊆TUD⊆T0 is stationary with respect to the operator A, so corresponding marks of the cases in which P,Q∈F are obtained. Thus it remains to discuss cases in which P,Q∈E.Since the subspace of a metrizable (resp.T2, SC, and US) space is metrizable (resp. T2, SC, and US) using Remark 3(4) and (5), if A(X) is, respectively, metrizable T2, KC, SC, or US, then X is too. Hence we obtain “—” for the following cases too (choose P and Q from the same rows of Table 2).Table 2
P
Q
Reason of omitting this case
C
2
∖
C
1
,
C
3
∖
C
2
,
C
4
∖
C
3
,
C
5
∖
C
4
,
C
6
∖
C
5
,
C
7
∖
C
6
C
1
IfA(X) is metrizable, then X is metrizable too
C
4
∖
C
3
,
C
5
∖
C
4
,
C
6
∖
C
5
,
C
7
∖
C
6
C
2
∖
C
1
,
C
3
∖
C
2
IfA(X) is T2, then X is T2 too
C
5
∖
C
4
,
C
6
∖
C
5
,
C
7
∖
C
6
C
4
∖
C
3
IfA(X) is KC, then X is KC too
C
6
∖
C
5
,
C
7
∖
C
6
C
5
∖
C
4
IfA(X) is SC, then X is SC too
C
7
∖
C
6
C
6
∖
C
5
IfA(X) is US, then X is US tooProof (proof of  the  rest  of the cells  of  Figure1).
First Row. Here we have P=C1 and the following cases for Q.
(i) ConsiderQ=C1. Consider two spaces X:=(0,1) (with induced metric from Euclidean space ℝ) and 𝕊1={(x,y)∈ℝ2:x2+y2=1} (with induced metric from Euclidean space ℝ2); then A(X) is homeomorphic to 𝕊1; moreover X,𝕊1∈C1; therefore X,A(X)∈C1.
(ii) ConsiderQ=C2∖C1. Consider X:=(0,1) with discrete topology. A(X) is compact Hausdorff, so it is normal. If 𝒟 is a topological basis for A(X), then for all t∈(0,1) we have {t}∈𝒟. Therefore 𝒟 is uncountable and compact space A(X) is not metrizable. Thus X∈C1 and A(X)∈C2∖C1.
(iii) ConsiderQ=C3∖C2. Use Remark 3(3).
(iv) ConsiderQ=C4∖C3. Consider X as the set of all rational numbers as a subspace of Euclidean space ℝ. Since X is not locally compact, by Remark 3(3), A(X) is not Hausdorff. Suppose M is a compact subset of A(X)); in order to show that A(X) is KC, we show M is a closed subset of A(X). We have the following two cases.
Case  1. If ∞∉M, then M is a compact subset of X; since X is a metric space, M is a closed subset of X too. Therefore, A(X)∖M is an open subset of A(X). Hence, M is a closed subset of A(X).
Case  2. If ∞∈M, we claim that X∖M is an open subset of X and so an open subset of A(X); otherwise (since X is metrizable) there exists a one-to-one sequence (xn:n∈ω) in X∖(X∖M) (=X∩M) converging to a point x∈X∖M (in metric space X). For all m∈ω, {xn:n≥m}∪{x} is a compact closed subset of X, and Um:=A(X)∖({xn:n≥m}∪{x}) is an open subset of A(X). Since x∉M, M⊆⋃{Um:m≥0}. Using the compactness of M, there exists m≥1 such that M⊆U0∪U1∪⋯∪Um. Since xm∈M∖(U0∪U1∪⋯∪Um)=M∖Um, we have M⊈U0∪U1∪⋯∪Um which is a contradiction. Thus X∖M is an open subset of X, and M=A(X)∖(X∖M) is a closed subset of A(X).
Finally we haveX∈C1 and A(X)∈C4∖C3.
(v) ConsiderQ=C5∖C4. If X is a metric space, then it is a k-space and, by Remark 3(2), A(X) is KC; hence A(X)∉C5∖C4.
(vi) ConsiderQ=C6∖C5 or C7∖C6. We claim that if X is a metric space, then A(X) is SC. First, note the fact that, by Remark 3(1), A(X) is US and hence T1. Suppose (xn:n∈ω) is a sequence in A(X) converging to x∈A(X); we show that {xn:n∈ω}∪{x} is a closed subset of A(X). Consider the following cases.
Case  1.
x
≠
∞; in this case, X is an open neighborhood of x, thus there exists m∈ω such that xn∈X for all n≥m and (xn:n≥m) converges to x in metric space X (by Remark 3(4), X as a subspace of A(X) has its original topology). Thus {xn:n≥m}∪{x} is a closed compact subset of X; therefore A(X)∖({xn:n≥m}∪{x}) is an open subset of A(X). Finally {xn:n≥m}∪{x} is a closed subset of A(X) and since A(X) is T1, {xn:n≥m}∪{x}∪{x0,…,xm}={xn:n∈ω}∪{x} is a closed subset of A(X) too.
Case  2.
x
=
∞ and {xn:n∈ω} is finite. In this case, {xn:n∈ω}∪{x} is a finite subset of (T1 space) A(X) and it is closed.
Case  3.
x
=
∞ and {xn:n∈ω} is infinite. In this case, we may assume xn∈X for all n∈ω. If {xn:n∈ω} is not a closed subset of X, then there exists a subsequence (xnk:k∈ω) of (xn:n∈ω) converging to y∈X∖{xn:n∈ω}. Thus (xnk:k∈ω) converges to y in A(X) too (use Remark 3(4)). Since (xnk:k∈ω) converges to y,x(=∞) and A(X) is US, we have y=x which is a contradiction with y∈X=A(X)∖{∞}=A(X)∖{x}. Therefore {xn:n∈ω} is a closed subset of X, so
(7)X∖{xn:n∈ω}=X∖({xn:n∈ω}∪{x})=A(X)∖({xn:n∈ω}∪{x})
is an open subset of X and A(X). Finally {xn:n∈ω}∪{x} is a closed subset of A(X).
Using the above three cases,{xn:n∈ω}∪{x} is a closed subset of A(X) and we are done.
Second Row. Here we have P=C2∖C1 and the following cases for Q.
(i) ConsiderQ=C2∖C1. Suppose Ω is the least uncountable ordinal number. Consider X=Ω+1 (with order topology). Since X is well ordered, it is normal. However, X is not metrizable and A(X)=X.
(ii) ConsiderQ=C3∖C2. If X and A(X) are T2, then, by Remark 3(3), A(X) is normal.
(iii) ConsiderQ=C4∖C3. Consider X as disjoint union of X1=Ω+1 with order topology and X2=ℚ as the set of all rational numbers with induced metric from Euclidean space ℝ. The topological space X is normal since X1,X2 are normal. Moreover, X is disjoint union of X1 and X2, so X has nonmetrizable subspace X1, thus X is nonmetrizable and X∈C2∖C1. Since X1 is compact, by Lemma 4 we have A(X)=X1⊔A(X2). Considering case “P=C1, Q=C4∖C3”, we have A(X2)∈C4∖C3. Using Lemma 6(1), we have X1⊔A(X2)∈C4∖C3; thus A(X)∈C4∖C3.
(iv) ConsiderQ=C5∖C4. Let X be an uncountable set and b∈X. Consider X under Fortissimo topology with particular point b, that is, under the topology {U⊆X:b∉U∨(X∖U is countable)} (see [4, counterexample 25]). For distinct x,y∈X, suppose x≠b. Then U={x},V=X∖{x} are disjoint open subsets of X and X is Hausdorff. On the other hand, if E,F are disjoint closed subsets of X, suppose b∉E, then E and X∖E(⊇F) are disjoint open subsets of X containing E and F. Thus X is normal.
Moreover, ifW is an open neighborhood of b, then X∖W is countable; therefore W is uncountable and W∩(X∖{b})≠⌀. Therefore, b∈X∖{b}¯. Let (xn:n∈ω) be a sequence of elements of X∖{b}. The sequence (xn:n∈ω) does not converge to b, since X∖{xn:n∈ω} is an open neighborhood of b. The space X is not metrizable since b∈X∖{b}¯ and there is not any sequence in X∖{b} converging to b. So X∈C2∖C1.
On the other hand, using the definition of one point compactification, any subsetM of A(X) containing ∞ is a compact subset of A(X). Therefore, A(X)∖{b} is a compact subset of A(X), but it is not a closed subset of A(X); thus A(X) is not KC. We claim that A(X) is SC. Suppose (xn:n∈ω) is a sequence in A(X) converging to w. We have the following cases.
Case  1.Consider w∈X∖{b}. In this case, {w} is an open neighborhood of w and there exists N≥1 such that for all n≥N we have xn=w. Using Remark 3(1), A(X) is T1; thus {xn:n∈ω}∪{w}={x0,…,xN,w} is a closed subset of A(X).
Case  2.Consider w=b. The set X∖{xn:n∈ω,xn≠b} is an open neighborhood of w; thus there exists N≥1 such that for all n≥N we have xn∈(X∖{xn:n∈ω,xn≠b}) (thus for all n≥N we have xn=b). Using Remark 3(1), A(X) is T1; thus {xn:n∈ω}∪{w}={x0,…,xN,w} is a closed subset of A(X).
Case  3. Consider w=∞. In this case, A(X)∖({xn:n∈ω}∪{w})=X∖{xn:n∈ω} is an open subset of X; therefore it is an open subset of A(X). Thus {xn:n∈ω}∪{w} is a closed subset of A(X).
Using the above three cases,{xn:n∈ω}∪{w} is a closed subset of A(X), and A(X) is SC.
SinceA(X) is SC and it is not KC, A(X)∈C5∖C4.
(v) ConsiderQ=C6∖C5. Suppose ℱ is a uniform ultrafilter over ℕ. Consider X=ℕ∪{0}(=ω) under topology {A⊆X:0∉A∨A∖{0}∈ℱ}. If x,y∈X are distinct with x≠0, then {x},X∖{x} are disjoint open subsets of X containing x,y and X is Hausdorff. If E,F are disjoint closed subsets of X, suppose 0∉E. Therefore, E,X∖E(⊇F) are disjoint open subsets of X and X is normal. Since ℱ is a uniform ultrafilter over ℕ, it does not contain any finite subset of X. Since all of the elements of ℱ are infinite, 0 is a limit point of X and 0∈X∖{0}¯. Consider a sequence (xn:n∈ω) in X∖{0}. We have the following cases.
Case  1.
(
x
n
:
n
∈
ω
) has a constant subsequence like (xnk:k∈ω). Since X is Hausdorff and every sequence converges to at most one point, (xnk:k∈ω) converges to its constant value and does not converge to 0. Thus (xn:n∈ω) does not converge to 0.
Case  2.
(
x
n
:
n
∈
ω
) does not have any constant subsequence. Suppose (xnk:k∈ω) is a one-to-one subsequence of (xn:n∈ω). Since ℱ is an ultrafilter over X∖{0}(=ℕ), and {xn2k:k∈ω}∩{xn2k+1:k∈ω}=⌀, we have {xn2k:k∈ω}∉ℱ or {xn2k+1:k∈ω}∉ℱ. Suppose {xn2k:k∈ω}∉ℱ. Since ℱ is an ultrafilter over ℕ, ℕ∖{xn2k:k∈ω}∈ℱ. Therefore (ℕ∖{xn2k:k∈ω})∪{0} is an open neighborhood of 0 and (xn:n∈ω) does not converge to 0.
Since0∈X∖{0}¯ and by the above two cases, there is not any sequence in X∖{0} converging to 0; X is not metrizable. Thus X∈C2∖C1.
Now pay attention to the following claims.
Claim  1. The sequence (n:n≥1) converges to ∞ in A(X). Suppose U is an open neighborhood of ∞ in A(X). Since X is T1, A(X) is T1 too. Therefore, V:=U∖{0} is an open neighborhood of ∞ in A(X); thus A(X)∖V=X∖V(⊆X∖{0}) is a compact (and closed) subset of X. Also A(X)∖V is finite, since X∖{0} is discrete and A(X)∖V is a compact subset of X∖{0} (use Remark 3(4)). Suppose N=max⁡(A(X)∖V). For all n>N, we have n∈V⊆U. Hence (n:n≥1) converges to ∞.
Claim  2.
{
n
:
n
≥
1
}
∪
{
∞
} is not a closed subset of A(X). Using the fact that 0∈X∖{0}¯, we have {n:n≥1}∪{∞}¯=A(X); hence {n:n≥1}∪{∞} is not a closed subset of A(X).
Regarding Claims  1 and  2,A(X) is not SC. Since X is normal, it is KC; so using Remark 3(1), A(X) is US. Therefore, A(X)∈C6∖C5.
(vi) ConsiderQ=C7∖C6. If X is Normal, then it is KC and by Remark 3(1), A(X) is US. Thus A(X)∉C7∖C6.
Third Row. Here we have P=C3∖C2 and the following cases for Q. (i)
ConsiderQ=C2∖C1. Suppose X is a Hausdorff locally compact nonnormal topological space. Since X is not normal, it is not metrizable and X∈C3∖C2. By Remark 3(3), A(X) is normal. By Remark 3(4), A(X) is not metrizable. Hence A(X)∈C2∖C1. Moreover X=((ω+1)×(Ω+1))∖{(ω,Ω)}, where ω+1 and Ω+1 have their order topology and (ω+1)×(Ω+1) 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)
ConsiderQ=C3∖C2. Use a similar method described for P=C2∖C1 and Q=C3∖C2.(iii)
ConsiderQ=C4∖C3. Consider X=ℝ under topology {O∖B:O is an open subset of ℝ in its Euclidean topology and B⊆{1/n:n∈ℕ}}, then X∈C3∖C2 (Smirnov’s deleted sequence topology [4, counterexample 64]). Also X is first countable; therefore it is a k-space by Remark 3(6). By Remark 3(2), A(X) is KC. Moreover, A(X) is not Hausdorff, since X is not locally compact in 0 (use Remark 3(3)). Hence A(X)∈C4∖C3.(iv)
ConsiderQ=C5∖C4. Consider X as disjoint union of X1 and X2, where
(1)
X1 is an uncountable set under Fortissimo topology with particular point b∈X1, that is, under topology {U⊆X1:b∉U∨(X1∖U is countable)} (see [4, counterexample 25] and proof of Table 1 regarding case “P=C2∖C1, Q=C5∖C4”);(2)
X2=ℝ under the topology {O∖B:O is an open subset of ℝ in its Euclidean topology and B⊆{1/n:n∈ℕ}} (see Smirnov’s deleted sequence topology [4, counterexample 64] and proof of Table 1 regarding case “P=C3∖C2, Q=C4∖C3”).
SinceX1∈C2∖C1 and X2∈C3∖C2, we have X=X1⊔X2∈C3∖C2 by Lemma 6(1). Moreover A(X1)∈C5∖C4 and A(X2)∈C4∖C3 lead us to A(X)=A(X1⊔X2)∈C5∖C4 by Lemma 7. (v)
ConsiderQ=C6∖C5. Consider X as disjoint union of X1 and X2, where we have the following.
(1)
Supposeℱ is a uniform ultrafilter over ℕ. Consider X1=ℕ∪{0}(=ω) under topology {A⊆X:0∉A∨A∖{0}∈ℱ} (see proof of Table 1 regarding case “P=C2∖C1, Q=C6∖C5”).(2)
X2 is Smirnov’s deleted sequence topological space (see proof of Table 1 regarding case “P=C3∖C2, Q=C4∖C3”).
ThenX∈C3∖C2 by X1∈C2∖C1 and X2∈C3∖C2 and Lemma 6(1). Also A(X)∈C6∖C5 by A(X1)∈C6∖C5, A(X2)∈C4∖C3, and Lemma 7. (vi)
ConsiderQ=C7∖C6. If X is T2, then it is KC and by Remark 3(1), A(X) is US. Thus A(X)∉C7∖C6.
Fourth Row. Here we have P=C4∖C3 and the following cases for Q.
(i) ConsiderQ=C4∖C3. Consider W as the set of all rational numbers as a subspace of Euclidean space ℝ. Using the case “P=C1, Q=C4∖C3” for X:=A(W), we have X∈C4∖C3. Since X is compact, we have A(X)=X∈C4∖C3.
(ii) ConsiderQ=C5∖C4. Consider uncountable set X with countable complement topology {U⊆X:U=⌀∨(X∖U is countable)} [4, counterexamples 20 and 21]. Since every two nonempty open subsets of X have nonempty intersection, X is not Hausdorff. It is clear that X is T1. Moreover, M is a compact subset of X if and only if M is finite. Therefore, every compact subset of X is closed and X is KC. So X∈C4∖C3.
Now supposeE is an uncountable subset of X with uncountable complement. So E is not closed. For all compact subset M of X, the set E∩M is finite and closed. Therefore, X is not a k-space. Using Remark 3(2), A(X) is not KC. Using Remark 3(1), A(X) is US; we claim that A(X) is SC. Suppose (xn:n∈ω) is a sequence in A(X), converging to x∈A(X). We have the following cases.
Case  1. Consider x∈X. In this case, U:=(X∖{xn:n∈ω})∪{x} is an open neighborhood of x in A(X). So there exists N≥1 such that for all n≥N we have xn∈U and xn=x. Therefore, {xn:n∈ω}∪{x}={x0,x1,…,xN,x} is a (finite and) closed subset of X.
Case  2. Consider x=∞. Since A(X)∖({xn:n∈ω}∪{x})=X∖{xn:n∈ω} is open in X, it is open in A(X) too. Thus {xn:n∈ω}∪{x} is closed in A(X).
By the above cases,{xn:n∈ω}∪{x} is closed in A(X) and A(X) is SC. Hence A(X)∈C5∖C4.
(iii) ConsiderQ=C6∖C5. Consider X as disjoint union of X1 and X2, where we have the following. (1)
Supposeℱ is a uniform ultrafilter over ℕ. Consider X1=ℕ∪{0}(=ω) under topology {A⊆X:0∉A∨A∖{0}∈ℱ} (see proof of Table 1 regarding case “P=C2∖C1, Q=C6∖C5”).(2)
X2 is an uncountable set with countable complement topology {U⊆X2:U=⌀∨(X2∖U is countable)} (see [4, counterexamples 20 and 21] and proof of Table 1 regarding case “P=C4∖C3, Q=C5∖C4”).
ThenX∈C4∖C3 by X1∈C2∖C1, X2∈C4∖C3, and Lemma 6(1). Also A(X)∈C6∖C5 by A(X1)∈C6∖C5, A(X2)∈C5∖C4, and Lemma 7.
(iv) ConsiderQ=C7∖C6. If X is KC and by Remark 3(1), A(X) is US. Thus A(X)∉C7∖C6.
Fifth Row. Here we have P=C5∖C4 and the following cases for Q. (i)
ConsiderQ=C5∖C4. Consider X as Ω+1 with doubling Ω [8]; that is, if p∉Ω+1, let X=(Ω+1)∪{p} under topological basis {(α,β):α and β are ordinal numbers with α,β<Ω}∪{[0,α):α is an ordinal number with α<ω}∪{(α,Ω]:α is an ordinal number with α<ω}∪{(α,Ω)∪{p}:α is an ordinal number with α<ω}. Then X∈C5∖C4 and X is compact which leads to A(X)∈C5∖C4 too.(ii)
ConsiderQ=C6∖C5. Consider X as disjoint union of X1 and X2, where we have the following.
(1)
X1 is Ω+1 with doubling Ω [8] as in case “P=C5∖C4, Q=C5∖C4”. Then X1=A(X1)∈C5∖C4.(2)
For uniform ultrafilterℱ over ℕ, consider X2=ℕ∪{0}(=ω) is equipped with topology {A⊆Y:0∉A∨A∖{0}∈ℱ}. Using case “P=C2∖C1, Q=C6∖C5”, we have X2∈C2∖C1 and A(X2)∈C6∖C5. By X1∈C5∖C4, X2∈C2∖C1, and Lemma 6(1), we have X=X1⊔X2∈C5∖C4. By A(X1)∈C5∖C4, A(X2)∈C6∖C5, and Lemma 7, we have A(X)=A(X1⊔X2)∈C6∖C5.(iii)
ConsiderQ=C7∖C6. Use Lemma 8.
Sixth Row. Here we have P=C6∖C5 and the following cases for Q. (i)
ConsiderQ=C6∖C5. Consider X as (ω+1)∪{ℱ}, such that ω+1 has its usual order topology, ℱ is a uniform ultrafilter over ω, and {{ℱ}∪U:U∈ℱ} is an open neighborhood basis for ℱ [8, example  1.2]; then X is compact and A(X)=X∈C6∖C5.(ii)
ConsiderQ=C7∖C6. According to [7, example  5], there exists a US topological space X such that A(X) is not US. By Remark 3(7), A(X) is T1. Hence A(X)∈C7∖C6. Moreover, by Lemma 8, X is not SC; thus X∈C6∖C5.
Seventh Row. Here we have P=C7∖C6 and Q=C7∖C6. Suppose X as an infinite set with finite complement topology {U⊆X:U=⌀∨(X∖U is finite)} [4, counterexamples 18 and 19]. Then X is compact and A(X)=X∈C7∖C6.


## 4. Some Observations in Figure1

Using Figure1, we have the following results. The collection {T2, KC, SC, T1} is a full-forwarding chain with respect to A. In other words, Table 3 is valid.Table 3
P
Q
T
2
KC
∖
T
2
SC
∖
KC
T
1
∖
SC
T
2
√
√
√
√
KC
∖
T
2
—
√
√
√
SC
∖
KC
—
—
√
√
T
1
∖
SC
—
—
—
√In Table3, the mark “√” indicates that in the corresponding case there exists X∈P such that A(X)∈Q, and the mark “—” indicates thatin the corresponding case for allX∈P we have A(X)∉Q. The collection {Metrizable, T2, KC, SC, US, T1, TD, TUD, T0, Top} is a forwarding chain with respect to A. The collection {T1,TD,TUD,T0, Top} is a stationary chain with respect to A.



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*Source: 541538-2014-04-30.xml*