On Topologically Finite Spaces
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1 saqartvelos mecierebata erovuli aademiis moambe, t 9, #, 05 BULLETIN OF THE GEORGIAN NATIONAL ACADEMY OF SCIENCES, vol 9, o, 05 Mathematics O Topologically Fiite Spaces Giorgi Vardosaidze St Adrew the First Called Georgia Uiversity of Patriarchate of Georgia, Tbilisi (Preseted by Academy Member Hvedri Iassaridze) ABSTRACT The otio of topologically fiite space was itroduced i 0 I the preset paper the questio whether the uio of fiite umber of topologically fiite spaces is topologically fiite is studied More precisely, it is show that the uio of two topologically fiite spaces (eve i the realm of separable, metrizable ad coected spaces) may ot be topologically fiite 05 Bull Georg Natl Acad Sci Key words: topologically fiite space, uio Itroductio The otio of topologically fiite space was itroduced i [] (see defiitio below), where, i particular, it was show that for ay {0,,,} { } there exists a separable ad metrizable space X with dim X ad X c (here, as usual, dim stads for the classical coverig dimesio fuctio, deotes cardiality of X ad c is the cardiality of cotiuum) I the same paper it was established also, that if a topological space is the uio of its two disjoit, topologically fiite, ope ad coected subspaces, the the space itself is topologically fiite The followig questio aturally rises, whether the uio of ay two topologically fiite spaces is topologically fiite Below we show that there exists a topologically ifiite locally compact ad path coected subspace of the real plae which is the uio of its two topologically fiite locally compact ad path coected subspaces Defiitios ad Notatios A topological space X is topologically fiite provided there is o proper subspace of X which is homeomorphic to the whole X Otherwise we say that the space X is topologically ifiite Evidetly, the otio of topological fiiteess is topological ivariat Basically, we accept otatios ad defiitios give i [] deotes the real plae with atural topology is the set of all itegers ad is the set of all atural umbers X 05 Bull Georg Natl Acad Sci
2 3 Giorgi Vardosaidze For ay ( a, b ), ( a, b ), ( a, b ) ( a, b ) ( a a ) ( b b ) odd (respectively, eve ) deotes the set of all odd (respectively, eve) itegers For ay deote: S ( x, y) : ( x, y), Thus, each S is the circumferece of the circle with the ceter at, ad radius By X we deote the uio of all S -s: X S It is obvious that X is the locally compact ad path coected subspace of Let, for each Thus, is the poit at which S touches S Deote: X { } ad X X \ X Evidetly, X X X ad X X We say, that i ad j are eighbor poits of X (or simply, eighbor poits), if i j Obviously, i ad j are eighbor poits of X if ad oly if there exists such that i, j S Let pr : be the first projectio, ie, for ay ( x, y), pr ( ) x It is clear that for each, pr ( ) For ay, X, where pr ( ) pr ( ) deote: X { pr ( ) pr ( )} Note that X, if ad oly if there exists i with, Si For ay ad ay distict poits, S there are exactly two (closed) arcs C ad C i S with ed-poits ad The sets C \{, } ad C \{, } are called below ope arcs i S with ed-poits ad Symbol Mai Costructio deotes the ed of proof We begi with the followig trivial assertio Assertio Let : A B be a ijective mappig from a set A to a set B Let also, A, A A ad A A cosists of exactly poits of A : A A { a,, a } The ( A ) ( A ) { ( a),, ( a )}, ie, the set ( A ) ( A ) cosists of exactly poits of B as well Propositio Let, X be two distict poits of X The if, S for some, there exist two coected subsets C ad C of X such, that C C {, } Furthermore, let for ay i we have {, } S i Let also, pr ( ) pr ( ) ad C is a coected subset of X with, C The for ay X we shall have: a, ad C (as we have already oted above, i this case X ) Proof Suppose first,, S for some Let C ad C be two distict (closed) arcs i S with edpoits ad Clearly, C ad C are coected subsets of X ad C C {, } Suppose ow, for ay i we have {, } S i, pr ( ) pr ( ) ad C is a coected subset of X with, C ad let be ay elemet of X It is obvious that a ad, Show that C Ideed, assume, C Deote: U C Si \{ }, V C Si \{ } i i It is obvious that U ( U ), V ( V ), U V, U ad V are ope subsets of C ad Bull Georg Natl Acad Sci, vol 9, o, 05
3 O Topologically Fiite Spaces 33 (sice by the assumptio C ) C U V We get a cotradictio, because C is coected Propositio For ay homeomorphic embeddig h : X X we have: h( X) X Proof Suppose h( X) X The there has to be such that h( ) X Sice h( ) X, there is uique with h( ) S Let V X be a coected eighborhood of the poit i X The as h is homeomorphic embeddig, U h( V ) is a coected eighborhood of h( ) i h( X ) Two cases are possible: (a) S h( X ) ; (b) S h( X ) Case (a) ( S h( X ) ) Obviously, ay eighborhood of i S (ad i particular, V ) itersects S \{ } as well as S \{ } Let 0 V ( S \{ }) ad 0 V ( S \{ }) Deote: 0 h( 0) ad 0 h( 0) Clearly, 0, 0 U ad 0 0 Let C ad C be two distict arcs i S with ed-poits 0 ad 0 Evidetly, C, C h( X ) ad C C { 0, 0 } The (accordig to the Assertio above) we should have: h ( C ) h ( C ) { h ( ), h ( )} { h ( h( )), h ( h( ))} {, } O the other had, we have: 0 0, 0 0, pr ( 0 ) pr ( 0), there is o i with 0, 0 Si, 0 0 X { } ad h ( C ) ad h ( C ) are coected subsets of X Hece by Propositio, h ( C ) h ( C ) { 0,, 0}, which (taig ito the accout the equality h ( C ) h ( C ) { 0, 0} ) leads to a cotradictio Case (b) ( S h( X ) ) If S h( X ), there has to exist at least oe poit S with h( X ) U Obviously, ay eighborhood of i S (ad i particular, V ) itersects S \{ } Let 0 V ( S \{ }) Deote: 0 h( 0 ) Clearly, 0 U ad 0 h( ) Let C be a ope arc i S with ed-poits 0 ad h( ) such that C We shall ow show that ay coected subset C* h( X ) X, cotaiig the poits 0 ad h( ), cotais the whole C Ideed, assume, there is a poit C such that C * Note that sice h( X ) C * the C * Obviously, there are two ope subsets W ad W of the subspace X \{, } of X such that W W, W W X \{, }, 0 W ad h( ) W Deote: W W C * ad W W C * The W ad W are ope subsets of C *, both oempty ( 0 W ad h( ) W ), W W ad W W C * Thus we get a cotradictio (because the subspace C* h( X ) X is coected) Let ow C ad C be two distict arcs i S with ed-poits 0 ad Sice C C { 0, }, by the Assertio above we have: h( C ) h( C ) { h( 0), h( )} { 0, h( )} O the other had, cosider ay poit 0 C It is clear, that 0 0 ad 0 h( ) Furthermore, h( C ) ad h( C ) are coected subsets of h( X ) X both cotaiig the poits 0 ad h( ) Hece (as we have already see) 0 h( C ) ad 0 h( C ), ie, h( C ) h( C ) { 0, 0, h( )}, which cotradicts the equality h( C ) h( C) { 0, 0 } Propositio 3 Ay homeomorphic embeddig h : X X seds eighbor poits of X to the eighbor poits of X Proof Tae ay As (by Propositio ) h( X) X, there are, m with h( ) ad h( ) m We have to show, that either m or m Assume o the cotrary, m ad m Note that as h is ijectio, h( ) m, h( ) ad, the m Thus, oly two cases are possible: (a) m ; (b) m Case (a) ( m ) The poits ad belog to S The, by Propositio, there are two Bull Georg Natl Acad Sci, vol 9, o, 05
4 34 Giorgi Vardosaidze coected subsets C ad C of S (amely, the two (closed) arcs i S with ed poits ad ) such that C C cosists of exactly two poits: C C {, } Hece (sice h is ijectio), by the Assertio we shall have: h( C ) h( C) { h( ), h( )} { m, } That is, h( C ) h( C) must cosist of exactly two poits as well Furthermore, sice m, there is o i, with m, Si As h( C ) ad h( C ) are both coected, m, h( C) ad m, h( C), the by Propositio, there is at least oe j such that j m, j, j h( C ) ad j h( C ) ; hece j h( C ) h( C) Cosequetly, o the oe had, h( C ) h( C) { m, } ad, o the other had, we have: h( C ) h( C ) { m,, j} (where, ) Cotradictio j m j Cosideratio of Case (b) is carried out i the similar way Propositio 4 For ay homeomorphic embeddig h : X X we have: h( X) X Proof Cosider ay X By propositio there is m such that h( ) m By Propositio 3, two cases are possible: (a) h( ) m ad (b) h( ) m Case (a) h( ) m We shall show that for ay iteger i the followig equalities () h( i ) ad () h( ) hold i mi Proof of the equality () Let i Sice ad are eighbor poits of X ad h( ) m, by Propositio 3 we must have: h( ) m or h( ) m But, h( ) m ad as h, i particular, is ijective ad, the equality h( ) m is impossible So, h( ) m Assume ow, () is true for ay j i Sice i ad i are eighbor poits of X ad (by the assumptio) h( i ) mi, the by Propositio 3 we must have: h( i ) mi or h( i ) mi But, by the assumptio h( i ) h( ( i) ) h( ( i) ) mi ad as h is ijective ad i i, the equality h( i ) mi is impossible So, h( i ) mi The equality () is proved Proof of the equality () is carried out i the same way Case (b) h( ) m I the same way oe ca prove, that for ay iteger i the followig equalities tae place: (3) h( ) ad (4) h( ) i mi i mi Now show that h( X) X Accordig to the Propositio it suffices to prove that X h( X) Tae ay X The either m or there is i such that either m i or m i Suppose the first Case (a) taes place If m, the m h( ) If m i (where i ) the, by the equality () mi h( i ) If m i (where i ) the, by the equality () mi h( i ) Suppose ow, Case (b) taes place If m, the m h( ) If m i (where i ) the, by the equality (3) mi h( i ) If m i (where i ) the, by the equality (4) mi h( i ) Cosequetly, for ay X there exists X with h( ) ie, X h( X) The followig Corollary follows immediately from the propositio 4 Corollary For ay homeomorphic embeddig h : X X we have: h( X ) X Theorem The subspace X of is topologically fiite Proof Assume, X is ot topologically fiite The there exists a homeomorphic embeddig h : X X with h( X ) X Hece, there is a poit 0 X \ h( X ) Accordig to Propositio 4, X h( X) h( X ) Cosequetly, 0 X ad therefore 0 X Clearly, there is uique such that 0 S Let S (respectively, S ) be the upper (respectively, lower ) ope arc i S with ed-poits ad More precisely, S ( x, y) S y ad S ( x, y) S y mi Bull Georg Natl Acad Sci, vol 9, o, 05
5 O Topologically Fiite Spaces 35 Without loss of geerality, assume 0 S Now we shall show that ay poit of S belogs to h( X ) Ideed, suppose there is a poit S which does ot belog to h( X ) It is clear, that X \{ 0, } ca be represeted as the uio of two ope subsets W ad W of X with {,, } W, {,,} W, W W ad W W X \{, } As, 0 h( X ) ad (by the assumptio) h( X ), the 0 h( X ) ( W h( X )) ( W h( X )) Evidetly, W h( X ) ad W h( X ) are disjoit, ope subsets of h( X ) ad h( X ) ( W h( X )) ( W h( X )) Besides (as {,, } W h( X ) ad {,,} W h( X ) ), each of them is oempty We get a cotradictio, because h( X ) is coected Furtheremore, prove that every coected saubspace C of h( X ) cotaiig two distict poits, S, cotais also ay poit of the ope arc i S with ed-poits ad Ideed, suppose there is a poit of the ope arc i S with ed-poits ad such that C It is clear, that h( X ) \{ 0, } ca be represeted as the uio of two ope i X subsets W ad W with W, W, W W ad W W h( X ) \{ 0, } Evidetly, W C ad W C are disjoit, ope subsets of C ad C ( W C) ( W C) Besides (as W C ad W C ), each of them is oempty We get a cotradictio, because C is coected Now tae ay poit S As X, the (by Propositio 4) h ( ) X Clearly, there is (uique) with h ( ) S Sice S is a ope subset of h( X ) cotaiig the poit ad h( h ( )), the by cotiuity of h there exists a eighborhood V of the poit h ( ) i X such that h ( V ) S Tae ay poit V S differet from the poit h ( ) ad let ( ) h S Obviously, Sice h ( ), S, by Propositio, there exist two coected subsets C ad C of X such, that C C { h ( ), } Hece, accordig to the Assertio above, we should have: h( C ) h( C) { h( h ( )), h( )} {, } O the other had, let be ay poit of the ope arc i S with ed-poits ad Obviously, ad Sice the coected subsets h( C ) ad h( C ) of h( X ) both cotai the poits ad, the (as it is already show above) h( C ) h( C ) {,, } which cotradicts the equality h( C ) h( C ) {, } fiite Cosequetly, our assumptio that h( X ) X is ot true Hece, h( X ) X, ie, X is topologically Example Below a topologically ifiite locally compact ad path coected subspace Y of the real plae, which is the uio of its two topologically fiite locally compact ad path coected subspaces Y ad Y, is costructed For ay For ay eve deote: S ( x, y) :, ( x, y) ad odd deote: 5 S ( x, y) :, ( x, y) R ( x, y) x ; y 0 ( x, y) x ; 0 y 3 ( x, y) x ; y 3 ( x, y) x ; 0 y 3 Thus S (respectively, S ) is the circumferece of the circle with the ceter at the poit, (respec- Bull Georg Natl Acad Sci, vol 9, o, 05
6 36 Giorgi Vardosaidze 5 tively,, ) ad radius As to R, it is the boudary of the rectagle with vertexis,3,,3 ad,0 Let, Y S S R eve eve odd Y S R eve odd ; Y S R a d eve odd,0, Clearly, Y, Y ad Y are locally compact ad path coected subspaces of Oe ca easily see that both Y ad Y are homeomorphic to the space X Hece, by Theorem, Y ad Y are topologically fiite Besides, Y Y Y However, the space Y is topologically ifiite Ideed, deote: S0 ( x, y) S0 : y ad 5 S0 ( x, y) S0 : y It is ot difficult to chec that the space Y is homeomorphic to the followig its proper subspace: 5 5 Y \ S 0 S0 ( x, y) : x ; y ( x, y) : x ; y Acowledgemet The author thas Professor I Tsereteli for his guidace ad assistace Bull Georg Natl Acad Sci, vol 9, o, 05
7 O Topologically Fiite Spaces 37 matematia topologiurad sasruli sivrceebis Sesaxeb g vardosaize saqartvelos sapatriarqos wmida adria pirvelwodebulis qartuli uiversiteti, Tbilisi (warmodgeilia aademiis wevris x iasarizis mier) topologiurad sasruli sivrcis ceba SemoRebuli iyo 0 wels warmodgeil asromsi gamovleulia saitxi sasruli raodeobis topologiurad sasrul sivrceta gaertiaebis topologiurad sasrulobis Taobaze ufro zustad, acveebia, rom ori topologiurad sasruli sivrcis gaertiaeba (separabelur, metrizebad da bmul sivrceta lassic i) SeiZleba ar iyos topologiurad sasruli REFERENCES Tsereteli I (0) Topology ad its Applicatios, 59, 6: Egelig R (989) Geeral Topology, Berli, Helderma Received February, 05 Bull Georg Natl Acad Sci, vol 9, o, 05
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