Title On Countably Metacompact Spaces Author(s) Hayashi, Yoshiaki Editor(s) Bulletin of University of Osaka Prefecture. Series A, Engineering and nat Citation ural sciences. 1960, 8(2), p.161-164 Issue Date 1960-03-30 URL http://hdl.handle.net/10466/7939 Rights http://repository.osakafu-u.ac.jp/dspace/ } 161' 'g On Countably Metacompact Spaces Yoshiaki HAyAsHi* (Received January 29, 1960) A space X is called countably metacompact if every countable covering has a point-finite refinement. In this paper, we･study some prope,rties of countably meta- compact spaces. ･ The topological product Xx Y of a countably metacompact space X and a compact space Y is countably metacompact. In order that a topological space is countably metacompact it is necessary and suthcient that for e.very decreasing sequence {Fi} of closed sets with empty intersection there is a sequence {Gi} of open sets with eenGi--O such that Gi )Ei. There is a T2-space which is not countably metacompact, and there is a countably metacompact and Regular T2-space which is not countably paracompact. 1 Let X be a topological space, that is, a space with open sets such that the union of any family of open sets is open and the intersection of any finite number'of open 'sets is open. A covering of X is a family of open sets whose union is XL The covering is called countable if it consists of a countable family of open sets or finite if it consists of a finite family of open sets. The lcovering is called 1ocally finite if every point of X is contained in some open set which intersects only a finite number ･of sets of the covering and point-finite if each point of X is only a finite number of sets of the covering. A covering 8 is called a refinement of a covering U if every open sets of 8 is contained in some open set of U. The space X is called countably paracompact if every countable covering has a locally finite refinement and countably metacompact if every countable open covering has a point-finite refinement. ' The purpose of this paper is to study the peoperties of countably metacompact ' ' spaces. A space X i$ called compact･, if every covering has a finite refinement, paracompact if every covering has a locally finite･ refinement, metacompact*** if every covering has a point-finite refinement. It is 'clear ,that every compact, paracompact, or countably paracompact space is countably metacompaet. Every closed subset of a countably metacompact space is countably metacompact. It is known that the topological product of a countably paracompact space X and a compact space Y is countably paracornpact. We shall prove the similar theorem on countably metacompact spaces. * Department of Mataematics, Col!ege of General Education. ** We denote the empty set by O. , *** Cf. [1] p. !71. (Numbers in brackets refer to the references at the end of the paper.) 162 Y. HAyAsHi Theorem 1. The toPological Product Xx Y of a countably metacomPact sPace X and comPact space Yis countably metacomPact. ' [Proof] Let {Ui} (i--lj 2, ･･･) be a countable covering of XX Y.' Let Vi be the set of all points x of X such that xxYCUUj. If xEVi every point (x,y) of xxY has a js neighbourhood NxM (N is an open set of .X; M is an open set of Y), which is con- tained in the open set H.iUj. A finite number of sets these open sets M cover Y; let IVle be the intersection of the corresponding finite number of sets AIL The xEAZle, Alle is an open set of X, and ATle×Y9Y.iUj, and hence IVleCVi. Therefore Vi iS an open set of .XL AIso, for arfy xEX; sincexxYis compact, xxY is contained in some finite goUvMebreinrgOoff SxtLtS Qf the COVering {Ui}; hence x is in some' Vi. Therefore,{v,} isa Since {vi} is countable and x is countably metaEompact, {Vi} 'has a point-finite refinement E!B. For each open set W of EEB let g(va) be the first Vi cotitaning VV and let Gi be the union of all W for which g(W)=Vi. Then Gi is open, GiCVi and {Gi} is a point-finite covering of X. For each ]'s;i, set Gij=(GixY)nUj; then Gij is an open set in Xx Y. If (x, y) is a point of (X) Y) then, for some i, xEGi and hence (x,y)EGxY. Also, since xEGiCVi. (x, y)ExxYCjY?.,Uh and hence, for some 1'f{gi, (x,y)EUj. Hence (x,y)EGij. Therefore {Gij} is a covering of Xx ]YL Since GijCUj, {Gij} is a refinement of {Ui}. Also, if (x, y)EXx ]V; x is in only a finite number of sets of Gi. And hence, (x, y)g' is in only a finite number of sets of {Gij]. Therefore Kx Y is countably metacompact. ' ' ' 2 ' Theorein 2.* in order that a topological space X is countably metacompact it is necessaf:y and szaficient that .ICbr every decreasing sequence {Flr} of closed sets with fhMaet:GY ,/;ntRtr.SeCtiOn there iS a Sequence {Gi} of oPen sets with empty .{intersection such A topological space X is called regular if for every pair (A,x) such that A is a closed set.,and x is a point ¢A :there is a pair of disioint open sets U and V with U )A and VEx. A 'topolQgical space X is called normal if for every pair of disjoint closed sets A ahd 'B there is a pair of disjoint open sets U and V with UDA and V)B. The fellowing two properties of a normal space X are equivalent**: (1) The space K is countably para,compact. (2) The space Xis countably metacompact. - ' However, these properties (1), (2) of some regular space are not equivalent, that is the fol!owing theorem holds. On Countably Mbtacompact SPaces 163 cTohuenOtraebMly p3bracTollf;laictZ.S a COUntably metacompact and reguiar space which is not [Proof] Let Ri= {O, 1, 2, 3, ･･･, to, ･･･, 2} Where 2 is the first odinal number in all 3rd- class ordinals, and R2={O,.1, 2. 3, ･.･, tu} where to is the first ordinal in all 2nd-class ordinals. For each of Ri and R2,.we define its topology by the,limit of ordinals as usual.* Set R=RixR2 and give the weak topology of the product space to R. And let X==R-(2, tu). ' ' Since both Ri and R2 are regular, R is regular therefore .X is regular. We shall prove X is not countably paracompact. Let Fi={(9, n)ln=i, i+1,i+2. ･･･}(i=1,2, 3, ･"). Then {E} is a decreasing sequence of closed sets of X wigh-empty intersection. Suppose that there is a sequence {Gn}' of open sets of X with ,nn,Gi**=O such that GDE. Since (9, n) (n2i) is contained in Gi, there is an ordinal ai,n<2 such that {(e,n)lai,n<es;;9} ]Gi. ' Set' ai=supai,n then ai<2 and {(e,n)Iai<e:{l;9, n= i, i+1, ･･･}CGi. Set a=supai then a<2 and n {(a,n)i i a<6f{;2, n==i,' i+1, ･･･}CGi therefore, for every at'(a<a'<9). (a', tu)cGi (i=1, 2, 3, ･･･). This is contradictoty to the assumption nGi=O. Therefore, by a theorem in [3], X is i not countably paracompact. We shall prove that X is countably metacompact, Let {Fi} be a decreasing sequence of closed sets with empty intersection. Denote Ri-coordinate of each point x of X by Ri[x] and R2-coordinate of x by R2[x]. We shall prove that there is an integer m such that Ri[x]=2 VxE,Pla. Suppose there iS some xEEn with Ri[x] =2 for･each m(lf{l;m<oo). Let xi (i=1, 2, ･･･) be a point of Fi with Ri[xi]42. SetB==lim***Ri[xi], then BL<2. And let {yjll'--1,2,･･･} be a sub- bco sequence of {xi} whith limyj==B. Set l=limR2[yj] and let {2icIle==1,2,･･･} be a sub- seqdenc of [yj] with limR2[zic]=l. j-ee As heo B<2 and IS:tu, (B,l)EX And lim2ic=(B,l). Therefore, since {R} is decreasing and {2ic} ic-).ee is a subsequence ic-oo of {xi} and xiER (i=1,2, ･･･), UE)(3,l), that is nR40. This is contradictory to the.assumption 'nR =O. Therefore there is an integer m such that Ri[x] =9 VxejF;n. Let Ri[x]==9 VxE"F;n. Since {Fi} is decreasing, Ri[x]=2 vxEFi. for every l)}lm. And also R2[x]<di VxEFL for every l)}lm. We define Gi (i=1, 2, 3, ･･･) as follows: (a) Gi=X (i=1, 2, ･･･, m-1), (b) Gi-- {xl there is a point yEFI with R2[y]==R2[x]} (i----m, m+1, ･･･). Then Gi (i=1,2,･･･) is open and GiDE. We shall prove that fiGi=O. Suppose nGi=O and xEfiGi. From the definition, there are ym, ym+i, ･･･ such that yiEF; and R2[yi] tt =R2[x] i (i=m, m+1, ･･･). As .Ri[yi]==9 (i=mm+1, ･･･),y.=ym+i=･･･. Hence ?F}=¥O. This * We define neighbourhood of a point p as follow$: for each q<p, {p'lq<p'E{;p} is a neighbourhood of p. ** A is the closure of A. *** lim is the upper limit of ordinals. 164 Y. HAyAsfii is contradictory to the assumption nn=:O.