TWO-DIMENSION-LIKE FUNCTIONS DEFINED ON THE CLASS OF ALL TYCHONOFF SPACES
I. TSERETELI
Abstract. Two-dimension-like functions are constructed on the class of all Tychonoff spaces. Several of their properties, analogous to those of the classical dimension functions, are established.
1. Introduction. All topological spaces discussed in this paper are assumed to be Tychonoff spaces.
As usual, Ind, ind, and dim denote the classical dimension functions (the large inductive, small inductive, and covering dimensions, respectively).
LetN′be the union of all integers≥ −1 and of one element set consisting of a single formal symbol “+∞” provided with an essential order relation.
The set of all natural numbers is denoted byN.
Throughout the paper for anyn∈N, the symbolIn denotes a standard
n-cube In ≡[0; 1]n, I0≡ {0}, andI−1 ≡∅where∅stands for the empty set.
The family of all Tychonoff spaces is denoted byT.
The class K of topological spaces is said to be permissible if K satisfies the following conditions:
1) for any integern≥ −1 In∈ K;
2) ifX ∈ KandA⊆X, thenA∈ K;
3) ifX1, X2∈ K, thenX1×X2∈ KwhereX1×X2is the usual product of spaces.
The function d defined on a permissible class K of topological spaces with values inN′ is called the generalized dimension-like function (GDF) if a)d∅=−1 and b)dX=dY wheneverX is homeomorphic toY.
The GDFddefined on a permissible classK of topological spaces is said to be of the Tumarkin type if the following conditionsTK
1 −T8Kare satisfied:
TK
1 ) for any integern≥ −1dIn=n;
1991Mathematics Subject Classification. 54F45.
Key words and phrases. Dimension, dimension-like function.
201
TK
2 ) ifX ∈ KandAis a locally closed subspace ofX (i.e., ifA=F∪G, whereF is closed andGis open inX), thendA≤dX;
TK
3 ) ifX ∈ Kand X =
∞
∪
i=1Ai where for anyi∈NAi is a closed subset of the spaceX, thendX ≤ sup
1≤i<+∞
{dAi};
TK
4 ) for any space X ∈ K there exists a Hausdorff compactification bX of the spaceX such thatdbX≤dX;
TK
5 ) for every pair of spacesX1, X2∈ Kat least one of which is nonempty we haved(X1×X2)≤dX1+dX2;
TK
6 ) ifX ∈ K andX=A∪B, thendX ≤dA+dB+ 1;
TK
7 ) ifX ∈ Kand there exists a nonnegative integernsuch thatdX≤n, then the spaceX can be represented as the union ofn+ 1 pairwise disjoint subsetsX1, X2, . . . , Xn+1withdXi≤0 for anyi= 1,2, . . . , n+ 1;
TK
8 ) for any X ∈ K and an arbitrary subspace Aof the space X there exists aGδ-setH inX such thatA⊆H ⊆X anddH ≤dA.
It is well-known fact that on the class of all separable metrizable spaces the classical dimension dim is a GDF of the Tumarkin type. On the other hand, as proved by L. Zambakhidze [1], there exists no GDF of the Tumarkin type on the classT. Moreover, there exists no GDF on T even satisfying the conditionsTT
1 , T2T, T3T, T4T, T5T simultaneously [1]. Also, there is no GDF onT satisfying the conditionsTT
1 ,T2T,T8T simultaneously. We say that a subcollection {TT
i1, . . . ,TiTk} (1 ≤ i1 < · · · < ik ≤ 8,
k = 1, . . . ,8) of the collection {TT
1 , . . . ,T8T} is realized if there exists a GDF onT which satisfies all conditions{TT
i1, . . . ,TiTk}simultaneously.
Clearly, if a subcollection {TT
i1, . . . ,TiTk} (1 ≤ i1 < · · · < ik ≤ 8, k =
1, . . . ,8) of the collection{TT
1 , . . . ,T8T} is realized, then any subcollection of{TT
i1, . . . ,TiTk}is realized, too. Also, if{T
T
i1, . . . ,TiTk}is not realized, then
no subcollection of the collection{TT
1 , . . . ,T8T} containing the given one is realized.
L. Zambakhidze has shown [1] that the collections{TT
2 ,T3T,T4T,T5T,T6T,
TT
7 ,T8T}, {T1T,T2T,T4T,T5T,T7T}, {T1T,T2T,T3T,T5T,T6T}, {T1T,T2T,T3T,
TT
6 ,T7T} and{T1T,T2T,T5T,T6T,T7T}are realized. In this paper we prove that collections {TT
1 ,T3T,T4T,T5T,T6T,T7T,T8T} and{TT
1 ,T2T,T3T,T5T,T7T}are realized. To this end we construct two GDFs d1andd2onTsuch thatd1satisfies the conditionsT1T,T3T,T4T,T5T,T6T,T7T,
TT
8 andd2satisfies the conditionsT1T,T2T,T3T,T5T,T7T. Moreover, the func-tionsd1 and d2 are the extensions of the classical dimension function dim from the class of all separable metrizable spaces over the classT.
2. GDF d1. Let X ∈T. It is assumed that d1X = dimX ifX has a countable base andd1X = 0 otherwise.
Theorem 1. The GDFd1 satisfies the conditions T1T,T3T,T4T,T5T,T6T,
TT
7 ,T8T, that is to say, the subcollection {T1T,T3T,T4T,T5T,T6T,T7T,T8T} of
the collection{TT
1 , . . . ,T8T} is realized.
Proof. The function d1 obviously satisfies the conditionsT1T,T3T,T4T,T5T,
TT
6 , and T7T. Therefore it remains for us to prove that d1 satisfies the conditionTT
8 .
LetX ∈T andA⊆X. Assume in the first place thatωX ≤ ℵ0(here and belowωX denotes the weight of the spaceX andℵ0stands for a countable cardinal number). Then by Tumarkin’s theorem [2, Ch. 6,§3, Theorem 14] there exists a Gδ-set H in X such thatA ⊆H ⊆X and dimH ≤dimA.
Hence, keeping in mind thatωA≤ ℵ0andωH ≤ ℵ0, we haved1H≤d1A. Now letωX >ℵ0. IfA=∅, it can be assumed thatH =∅. IfA6=∅, then assume thatH =X.
3. GDFd2. We begin by defining the functionHconstructed by Hayashi [3].
Definition 1 ([3]). A subsetX′of the spaceX ∈Tis called quasiclosed inX if there exists a finite family{F1, . . . , Fk}of closed subsets of the space
X such that X′ =F1±F2± · · · ±Fk, where + and−denote respectively
the union and the difference of sets, and whenever±is written one should take either + or−.
Clearly, every closed subset as well as every open subset of the spaceX is quasiclosed inX.
The functionH is defined on the classT as follows:
LetX ∈T. H(X) =−1 iffX =∅;H(X) = 0 iffX 6=∅andX = ∞∪
i=1Xi where for any i ∈ N Xi is quasiclosed in X and indXi ≤ 0; H(X) ≤ n
(n∈N) iffX=X1∪X2, whereH(X1)≤n−1 andH(X2)≤0;H(X) =n (n= 0,1,2, . . .) iffH(X)≤nandH(X)6≤n−1.
Finally, H(X) =∞ iff the inequalityH(X)≤n does not hold for any n=−1,0,1,2, . . ..
Now we shall define the functiond2.
Let X ∈ T. d2(X) = −1 iff X = ∅; d2(X) ≤ n (n = 0,1,2, . . .) if
X = ∞∪
t=1Xt where H(Xt) ≤ 0 for any t ∈ N and n+1
∪
k=1Xtk = X for any n+ 1 pairwise disjoint natural numbers t1, t2, . . . , tn+1; d2(X) = n (n = 0,1,2, . . .) iff d2(X)≤n andd2(X)6≤n−1;d2(X) =∞ifd2(X)nfor anyn=−1,0,1,2, . . ..
Clearly,d2is a GDF on the classT.
Lemma 1. We have d2(X) = dimX for any X ∈ T with a countable
Proof. Let X ∈ T and ωX ≤ ℵ0. Suppose that d2(X) ≤ n. Then X =
∞
∪
t=1Xt, where H(Xt)≤ 0 for any t ≥1 and n+1
∪
i=1Xti =X for any pairwise disjointt1, t2, . . . , tn+1∈N. Sinceω(Xt)≤ ℵ0for anyt∈N, it follows from [3, Theorem 4.3, Corollary 2] that dimXt= indXt=H(Xt)≤0. Therefore
we have X = ∞∪
t=1Xtwhere dimXt ≤0 for any t ∈Nand n+1
∪
i=1Xti =X for any pairwise disjointt1, . . . , tn+1∈N. Hence [4, Theorem 1.5.8] dimX ≤n. Conversely, let ωX ≤ ℵ0 and dimX ≤n, (n ≥0). Then by Ostrand’s theorem [5]X = ∞∪
t=1Xt, where dimXt≤0 for anyt≥1 and n+1
∪
i=1Xti =Xfor anyn+ 1 pairwise disjoint natural numberst1, t2, . . . , tn+1. Applying again [4, Theorem 4.3, Corollary 2], we obtain H(Xt) = dimXt≤0. Hence, by
the definition of the function d2, we haved2(X)≤n.
Corollary 1. The GDFd2is the extension of the functiondimfrom the
class of all separable metrizable spaces over the class T.
Corollary 2. The equalities d2(In) = dimIn =n hold for any integer n≥ −1.
Lemma 2. We haved2(X′)≤d2(X)for each X ∈T and an arbitrary
subspaceX′ of the spaceX.
Proof. Assume that X ∈ T and X′ is an arbitrary subspace of X. Let d2(X)≤n (n≥ −1). It will be shown thatd2(X′)≤nholds too. Indeed, sinced2(X)≤n, we have X =
∞
∪
t=1Xtwhere H(Xt)≤0 for each t≥1 and X =n∪+1
i=1Xti for any pairwise disjoint numbers t1, . . . , tn+1. Introduce the notationX′
t≡Xt∩X′. Obviously,X′=
∞
∪
t=1X
′
t.
Further, sinceH(Xt)≤0 for any t≥1, by the definition of the function
H we have Xt =
∞
∪
i=1Xti ≤ 0, where each Xti is quasiclosed in Xt and indXti≤0 (i= 1,2, . . .). Observe thatXt′ =Xt∩X′=
∞
∪
i=1Xti
∩X′= ∞
∪
i=1(Xti∩X
′). Since eachX
tiis quasiclosed inXt,Xti∩(Xt∩X′) =Xti∩X′
will be quasiclosed inXt∩X′=Xt′[3, Theorem 1.4]. Introduce the notation
X′
ti = Xti∩X′. Then Xt′ =
∞
∪
i=1X
′
ti, where each Xti′ is quasiclosed in Xt′.
Moreover, sinceX′
ti=Xti∩X′ ⊆Xti, we have indXti′ ≤indXti≤0.
Now we shall show that X′
t = n+1
∪
i=1X
′
ti for any pairwise disjoint natural
numberst1, . . . , tn+1. Indeed, n+1
∪
i=1X
′
ti =
n+1
∪
i=1(Xti∩X
′) =n∪+1
i=1Xti
∩X′=
Lemma 3. Let X ∈T and X = ∞∪
i=1Xi where each Xi is quasiclosed in X. Also assume there exists a natural number n such thatd2(Xi)≤nfor
any i≥1. Then d2(X)≤n.
Proof. It is obvious that ifn=−1, the assertion is true.
Let us consider the casen= 0. By definition, d2(X) = 0 iffH(X) = 0. Applying Theorem 3.2 from [3], we conclude that the assertion of the lemma is true in this case too.
Now consider the casen≥1. It can be assumed without loss of generality that Xi∩Xj=∅wheneveri6=j. (Indeed, otherwise we have to consider
a new covering{Xi′}∞
i=1 of the spaceX, whereX1′ =X1,Xk′ =Xk\ k−1
∪
i=1X
′
i
fork >1. Then [3, Theorems 1.1 and 1.3] eachX′
i is quasiclosed inX and
X′
i∩Xj′ =∅wheneveri6=j. SinceXk′ ⊆Xk for anyk≥1, by Lemma 2
we have d2(Xk′) ≤d2(Xk)≤n). By the definition of the function d2 and sinced2Xi≤n, we haveXi=
∞
∪
t=1Xit, whereH(Xit)≤0 for eacht≥1 and n+1
∪
j=1Xitj =Xi for any pairwise disjoint natural numbers t1, . . . , tn+1.
We introduce the notationX(t)≡
∞
∪
i=1Xit. It is obvious that
∞
∪
t=1X(t)=X. We shall prove that H(X(t))≤0 for anyt≥1.
Since Xi∩Xj =∅fori6=j, it is obvious thatXit=Xi∩X(t). Hence due to the quasiclosedness of Xi in X this implies [3, Theorem 1.4] that
Xit is quasiclosed in X(t). On the other hand, since H(Xit)≤0, we have
Xit =
∞
∪
k=1Xitk, where each Xitk is quasiclosed in Xit (and, accordingly, in Xi and X(t) as well [3, Theorem 1.5]) and for any i, t, k ≥ 1 we have indXitk≤0. But it is clear thatX(t)=
∞
∪
i,k=1Xitk for any t≥1. Hence, by the definition of the function H,H(X(t))≤0 for anyt≥1.
Now let us consider natural numberst1, . . . , tn+1such thatti6=tj
when-everi6=j (i, j= 1, . . . , n+ 1). We have
n+1
∪
m=1X(tm)=
n+1
∪
m=1
∞
∪
i=1Xitm
=n∪+1
i=1
n+1
∪
m=1Xitm
= ∞∪
i=1Xi=X.
Hence d2(X)≤n.
Corollary 3. letX ∈T andX = ∞∪
i=1Xi, where eachXi is closed inX.
Also assume that there exists a natural numbern such thatd2(Xi)≤nfor
any i≥1. Then d2(X)≤n.
Lemma 4. IfX1is a quasiclosed subset ofX ∈T andY1is a quasiclosed
Proof. By the assumption
X1=F0±F1± · · · ±Fk−1±Fk
and
Y1= Φ0±Φ1± · · · ±Φs−1±Φs,
where eachFi(o≤i≤k) is a closed subset ofX and each Φj (0≤j≤s) is
a closed subset ofY. The sign “+” denotes the usual union of sets, the sign “−” the usual difference of sets, and so whenever ±is written one should take either + or−.
The lemma will be proved by double induction (with respect tokands). Ifk=s= 0, then X1=F0 andY1= Φ0, whereF0 is a closed subset of the space X and Φ0 is a closed subset of the space Y. ThereforeX1×Y1 will be a closed subset and thus it will also be a quasiclosed subset ofX×Y. Assume that Lemma 4 has already been proved in two cases: 1) 0≤k≤
m−1 and 0 ≤ s ≤ n; 2) 0 ≤ k ≤ m and 0 ≤ s ≤ n−1, and prove it for k = m and s = n. For this note that the following (easily verifiable) point-set equations hold for any setsA, B, C, D:
(a) (A∪B)×(C∪D) = (A×C)∪(A×D)∪(B×C)∪(B×D); (b) (A∪B)×(C\D) ={[(A×C)∪(B×C)]\(A×D)}\(B×D); (c) (A\B)×(C∪D) ={[(A×C)∪(A×D)]\(B×C)}\(B×D); (d) (A\B)×(C\D) = [(A×C)\(A×D)]\(B×C).
Four cases are possible:
(++)
(
X1=F0±F1± · · · ±Fm−1+Fm
Y1= Φ0±Φ1± · · · ±Φn−1+ Φn
,
(+−)
(
X1=F0±F1± · · · ±Fm−1+Fm
Y1= Φ0±Φ1± · · · ±Φn−1−Φn
,
(−+)
(
X1=F0±F1± · · · ±Fm−1−Fm
Y1= Φ0±Φ1± · · · ±Φn−1+ Φn
,
(−−)
(
X1=F0±F1± · · · ±Fm−1−Fm
Y1= Φ0±Φ1± · · · ±Φn−1−Φn
.
Let us consider each of these cases separately. Introduce the notation
F0±F1± · · · ±Fm−1≡Fem−1;
Case (++). Due to (a) we have
X1×Y1= (Fem−1∪Fm)×(Φen−1∪Φn) = (Fem−1×Φen−1)∪
∪(Fem−1×Φen)∪(Fm×Φen−1)∪(Fm)×Φn).
By the assumption of inductionFem−1×Φen−1,Fem−1×ΦnandFm×Φen−1 are quasiclosed subsets of X×Y. Since the sets Fmand Φn are closed in
X andY, respectively, Fm×Φn is closed (and thus is also quasiclosed) in
X×Y. This means that the union of these sets will be quasiclosed inX×Y as well [3, Theorem 1.1].
Case (+−). Applying (b), we have
X1×Y1={[(Fem−1×Φn)∪(Fm×Φen−1)]\
\(Fem−1×Φen−1)}\(Fm×Φn).
By the assumption the sets Fem−1×Φn, Fm×Φen−1 and Fem−1×Φen−1 are quasiclosed inX×Y. The setFm×Φn is obviously closed inX ×Y.
HenceX1×Y1 is quasiclosed inX×Y [3, Theorems 1.1 and 1.3].
Case (−+). By (c) we have
X1×Y1={[(Fem−1×Φen−1)∪(Fem−1×Φn)]\
\(Fm×Φen−1)}\(Fm×Φn).
By the assumption of induction and Theorem 1.3 from [3] one can prove thatX1×Y1 is quasiclosed inX×Y.
Case (−−). From (d) it follows that
X1×Y1= [(Fem−1×Φen−1)\(Fem−1×Φn)]\(Fm×Φen−1).
By the assumption the setsFem−1×Φen−1,Fem−1×ΦnandFm×Φen−1are quasiclosed inX×Y. Hence by Theorem 1.3 from [3]X1×Y1is quasiclosed inX×Y as well.
Proposition 1. For any pair of spaces X, Y ∈ T, if indX ≤ 0 and
indY ≤0, thenindX×Y ≤0.
Theproof is trivial.
Lemma 5. Let X, Y ∈ T and let either X 6= ∅ or Y 6= ∅. Then
Proof. If eitherd2(X) = 0 and d2(Y) =−1 ord2(X) =−1 andd2(Y) = 0, then the inequalityd2(X×Y)≤d2(X) +d2(Y) is obvious.
Assume that d2(X) = n ≥ 0 and d2(Y) = m ≥ 0. Then X =
∞
∪
t=1Xt,
Y = ∞∪
l=1Yl, whereH(Xt)≤0 andH(Yl)≤0 for anyt, l≥1, and, moreover, the equalities
X =n∪+1
i=1Xti, Y = m+1
∪
j=1Ylj
hold for any sequencest1, . . . , tn+1andl1, . . . , lm+1of natural numbers with pairwise disjoint numbers.
Introduce the notationZp ≡Xp×Yp for any p≥1. Since H(Xp) ≤0
and H(Yp) ≤ 0 for each p ∈ N, we have Xp =
∞
∪
i=1Xpi and Yp =
∞
∪
j=1Ypj, where eachXpi is quasiclosed inXp and eachYpj is quasiclosed inYp, and
for anyi, j∈Nwe have indXpi≤0, indYpj≤0.
Lemma 4 implies thatXpi×Ypj is quasiclosed inXp×Yp=Zp and by
Proposition 1 we have ind(Xpi×Ypj) ≤ 0. Moreover, it is obvious that
Zp=Xp×Yp=
∞
∪
i,j=1(Xpi×Ypj).
Let us now prove that if we are given n+m + 1 natural numbers p1, p2, . . . , pn+m+1 such that pi 6=pj for anyi6=j (1 ≤i, j ≤n+m+ 1),
thenn+∪m+1
i=1 Zpi=X×Y. (This, in particular, implies that
∞
∪
p=1Zp=X×Y.) The inclusion X ×Y ⊇ Zp1∪ · · · ∪Zpn+m+1 is obvious. Let us prove
the inverse inclusion. Assume that (x, y) ∈ X ×Y. It remains for us to show that if (x, y) does not belong to some m+n members of the sys-tem {Zp1, . . . , Zpm+n+1}, then (x, y) necessarily belongs to the remaining
member of this system.
Consider the case where (x, y)∈X×Y and (x, y)6∈Zp1∪ · · · ∪Zpm+n.
It will be shown that (x, y) ∈ Zpm+n+1. (All other cases are considered
analogously.) Let xnot belong to exactlyk (0 ≤k ≤m+n) members of the system{Xpi}
m+n
i=1 and belong to the remainingm+n−kmembers of this system. Then, since each subsystem of the system {Xt}∞t=1 consisting ofn+ 1 elements covers the spaceX, we havek≤n.
By the assumption (x, y)6∈Zp1∪ · · · ∪Zpm+n. Now if x∈Xpi, we shall
necessarily havey 6∈Ypi (1 ≤i ≤m+n). Hence y does not belong to at
least m+n−k members of the system {Ypi}mi=1+n. Since each subsystem of the system{Yl}∞l=1 consisting ofm+ 1 elements covers the spaceY, we havem+n−k≤m. Thereforen≤k.
From the inequalities k ≤ n and n ≤k we obtain the equality n = k. Thereforexdoes not belong to exactlynelements of the system{Xpi}
ele-ments, we have ( ∪n
j=1Xpij)∪Xpm+n+1 = X and, consequently, since x 6∈ n
∪
j=1Xpij, we havex∈Xpm+n+1.
Analogously,y does not belong to exactlym+n−k=m+n−n=m elements of the system{Ypi}
m+n
i=1 . Assume that they are setsYpj1, . . . , Ypjm.
(It is obvious that {pi1, . . . , pin} ∪ {pj1, . . . , pjm} = {p1, . . . , pm+n} and {pi1, . . . , pin} ∩ {pj1, . . . , pjm} =∅.) Consider the system {Ypj1, . . . , Ypjm,
Ypm+n+1}. Since this system consists ofm+ 1 members, we have ( m
∪
i=1Ypji)∪ Ypm+n+1 =Y. But y 6∈
m
∪
i=1Ypji and thus y ∈Ypm+n+1. Therefore (x, y)∈ Xpm+n+1×Ypm+n+1⊆Zp1∪ · · · ∪Zpm+n+1.
Lemma 6. LetX∈T andd2(X)≤n(where0≤n <+∞). Then there
exist n+ 1 subspaces X1, . . . , Xn+1 of the space X such that X = n+1
∪
i=1Xi
andd2(Xi)≤0 holds for any i= 1, . . . , n+ 1.
Proof. d2(X) ≤ n implies X =
∞
∪
t=1Xt where for each t ≥ 1 H(Xt) ≤ 0 (which in turn implies d2(Xt) ≤ 0) and for any pairwise disjoint natural
numberst1, . . . , tn+1we haveX= n+1
∪
i=1Xti, in particular,X = n+1
∪
k=1Xk where d2(Xk)≤0 for anyk= 1, . . . , n+ 1.
Applying Lemmas 2, 5, 6 and Corollaries 2, 3, we arrive at
Theorem 2. The GDF d2 satisfies the conditions T1T,T2T,T3T,T5T,T7T
simultaneously. In other words, the subsystem{TT
1 ,T2T,T3T,T5T,T7T}of the
system{TT
1 , . . . ,T8T} is realized.
References
1. L. G. Zambakhidze and I. G. Tsereteli, On the realizability of dimen-sion-like functions in the class of Tychonoff spaces. (Russian) Soobshch. Akad. Nauk Gruz. SSR 126(1987), No. 2, 265-268.
2. P. S. Alexandrov and B. A. Pasynkov, Introduction to the dimension theory. Introduction to the theory of topological spaces and the general dimension theory. (Russian)Nauka, Moscow,1973.
3. Y. Hayashi, On the dimension of topological spaces. Math. Japon. 3(1954), No. 2, 71-843.
5. P. A. Ostrand, Covering dimension in general spaces. Gen. Topol. and Appl. 1(1971), No. 3, 209-221.
(Received 21.09.1993)
Author’s address:
