ZERO-DIMENSIONAL SPACES HOMEOMORPHIC TO THEIR CARTESIAN SQUARES

Transcription

1 ZERO-DIMENSIONAL SPACES HOMEOMORPHIC TO THEIR CARTESIAN SQUARES W LODZIMIERZ J. CHARATONIK AND ŞAHİKA ŞAHAN Abstract. We show that there exists uncountably many zero-dimensional compact metric spaces homeomorphic to their cartesian squares as well as their n-fold symmetric products. 1. Introduction and Preliminaries In [4] Marjanović showed that there are exactly nine different zero-dimensional compact metric spaces X which are homeomorphic to 2 X which denotes the set of all non-empty, closed subsets of X. In this paper we look at this subject from a different perspective and show that there exists uncountably many zero-dimensional compact metric spaces homeomorphic to their cartesian products and furthermore to their n-fold symmetric products.as a tool we will use the Cantor-Bendixson rank of a zero-dimensional compact metric space. The derivative of a set X represents the set of all limit points of X and denoted by X. A compact space X is called zero-dimensional if every component of X is degenerate. 2. Cantor-Bendixson Rank in Cartesian Product Definition 2.1. The Cantor-Bendixson derivative of order α, or α-derivative of a compact space X is defined inductively as follows: (2.1.1) X (0) = X (2.1.2) X (α+1) = {x X (α) : x is a limit point in X (α) } (2.1.3) For limit ordinals γ: X (γ) = β<γ X(β) 2010 Mathematics Subject Classification. 54B20, 54D30, 54E45, 54G05. Key words and phrases. Cantor-Bendixson rank, Cartesian squares, zerodimensional spaces. 1

2 2 W LODZIMIERZ J. CHARATONIK AND ŞAHİKA ŞAHAN Then, the Cantor-Bendixson rank of a space X, denoted by rank(x) is defined as the least ordinal α such that X (α+1) =. Finally, the Cantor-Bendixson rank of a point x in a set X, denoted by CB(x) is defined, if x has a countable neighborhood, by CB(x) = min{ rank(u) : U is a countable compact neighborhood of x in X and x U}. Observation 2.2. For compact spaces X and Y we have, (X Y ) = X Y X Y. Any ordinal number α can be uniquely written as ω α 1 n 1 +ω α 2 n ω α kn k where α 1, α 2,..., α k are ordinals in a decreasing order and n 1, n 2,..., n k are integers. This is called as Cantor form. Let α = ω α 1 n 1 + ω α 2 n ω α kn k and β = ω α 1 m 1 + ω α 2 m ω α km k be two ordinals, the natural sum α β is defined by (see e.g. [3] pg.253): α β = ω α 1 (n 1 + m 1 ) + ω α 2 (n 2 + m 2 ) + + ω α k (n k + m k ). The following lemmas have been proved by J. J. Charatonik and W. J. Charatonik in [1]. Lemma 2.3. Let β be a limit ordinal, and for each α < β assign two ordinals α 1, α 2 such that α = α 1 α 2. Then,. sup{α 1 α < β} sup{α 2 α < β} β Lemma 2.4. Let α and β be ordinals such that α < β and β = β 1 β 2. Then there exists ordinals α 1 and α 2 such that, α 1 β 1, α 2 β 2, and α = α 1 α 2. Corollary 2.6 has been proposed by J. R. Prajs, but the proof to this Corollary has never been published. In order to present a proof to this corollary we first prove the following theorem. Theorem 2.5. Let X and Y be two compact spaces and let Z = X Y. Then Z (α) = {X (α X) Y (α Y ) α = α X α Y }. Proof. If we let α = 0, then it is clear that Z (0) = X (0) Y (0), so the first condition is satisfied.

3 ZERO-DIMENSIONAL CARTESIAN SQUARES 3 Z (α+1) = (Z (α) ) = {X (αx) Y (α Y ) α = α X α Y } = {(X (α X ) ) Y (α Y ) X (αx) (Y (α Y ) ) α = α X α Y } = {X (α X +1) Y (α Y ) X (αx) Y (α Y +1) α = α X α Y } = {X (α X ) Y (α Y ) α + 1 = α X α Y }, as required. Here the third equation is a consequence of Observation 2.2. Now to show that, Z (γ) = {X (γx) Y (γ Y ) γ = γ X γ Y }, for each limit ordinals γ, first take z = x, y Z (γ). Then by Definition 2.1, x, y Z (γ) = β<γ Z(β), thus, by the inductive hypothesis we have x, y {X (βx) Y (β Y ) β = β X β Y }. Now, let γ X = sup{β X β < γ} and γ Y = sup{β Y β < γ}. By Lemma 2.3, γ X γ Y γ and by Lemma 2.4 there exists ordinals γ X and γ Y such that γ X < γ X, γ Y < γ Y and γ X γ Y = γ. Now since X(γX) = β<γ X(βX) and Y (γ Y ) = β<γ Y (β Y ) for all β < γ, we conclude that x, y X (γx) Y (γ Y ). Therefore x, y X (γx) Y (γ Y ) X (γ X ) Y (γ Y ) {X (γ X ) Y (γ Y ) γ = γ X γ Y }. In order to show the other inclusion let x, y {X (γx) Y (γ Y ) γ = γ X γ Y }. Then by Lemma 2.3 for any ordinal β, such that β < γ we can assign two ordinals β X and β Y such that β X < γ X, β Y < γ Y and β = β X β Y. Therefore, x, y X (γx) Y (γ Y ) X (βx) Y (β Y ) Z (β). Now by Definition 2.1 x, y Z (γ). So finally the equality holds. Corollary 2.6. Let X and Y be two compact spaces. Then, rank(x Y ) = rank(x) rank(y ). 3. Characterization I this section, a topological characterization of Z(α), an uncountable family {Z(α) : α < ω 1 } of zero-dimensional compact metric spaces will be provided. We will prove that uncountably many of them have the property that their Cartesian squares are homeomorphic to their factor spaces. Theorem 3.1. The following two conditions are equivalent for an ordinal α. (3.1.1) For every β, γ < α, we have β γ < α.

4 4 W LODZIMIERZ J. CHARATONIK AND ŞAHİKA ŞAHAN (3.1.2) α = 0 or there is an ordinal δ such that α = ω δ. Proof. (3.1.1) (3.1.2) Let α 0 and assume there is no δ such that α = ω δ. We may express α = ω α 1 k 1 +ω α 2 k 2 + +ω αn k n where α 1 > α 2 > > α n and k 1, k 2,... k n are nonzero natural numbers. If k 1 = 1, then n 2. Now, take β = γ = ω α 1, then we have β γ = ω α1 2 > α. If k 1 1, take β = ω α 1 (k 1 1)+ω α 2 k 2 + +ω αn k n and γ = ω α 1 (k 1 1), then β γ = ω α 1 (2k 1 2)+ω α 2 k 2 + +ω αn k n α and that is a contradiction. (3.1.2) (3.1.1) If α = 0 then the conclusion is true vacuously. Let α = ω δ for some ordinal δ. Then take β and γ less than α. We may express β = ω β 1 i 1 + ω β 2 i ω βn i n where β 1 > β 2 > > β n and i 1, i 2,... i n < ω, γ = ω γ 1 j 1 + ω γ 2 j ω γm j m where γ 1 > γ 2 >... γ m and j 1, j 2,..., j m < ω. Since β, γ < α, note that β 1, γ 1 < δ. Then β γ < ω max{β i,γ j }+1 < α. Corollary 3.2. There exist uncountably many ordinals α < ω 1 satisfying condition (3.1.1)( or equivalently (3.1.2)) of Theorem 3.1. To prove the next theorem we have copied a large part of the proof of Proposition 8.8 in [5] by adding necessary conditions on the ranks of points. Here S(X) represents the set of all points of X with no countable neighborhood. Theorem 3.3. Let α be a countable ordinal and let X 1 and X 2 be two compact metric spaces satisfying the following conditions for i {1, 2}: (3.3.1) For every x X i, if x has a countable neighborhood in X i, then CB(x) < α. (3.3.2) For every ordinal β such that β < α and for every uncountable, open subset U i of X i, there exist an x U i such that CB(x) = β. (3.3.3) S(X 1 ) is homeomorphic to S(X 2 ). Then the spaces X 1 and X 2 are homeomorphic. Proof. Let I(X i ) denote the set of all elements of X i with countable neighborhood. Thus X i = S(X i ) I(X i ) for i {1, 2} where S(X i ) is closed and I(X i ) is open. Since S(X 1 ) is homeomorphic to S(X 2 ), there exists a homeomorphism h : S(X 1 ) S(X 2 ). Let d 1 and d 2 denote the metrics for X 1 and X 2, respectively.

5 ZERO-DIMENSIONAL CARTESIAN SQUARES 5 We will define, by induction, two sequences R k and S k of finite sets such that I(X 1 ) = {R k : k {0, 1,... }} and I(X 2 ) = {Sk : k {0, 1,... }} and homeomorphism h k : R k S k such that CB(x) = CB(h k (x)) for all x R k, then the required homeomorphism h : X 1 X 2 will be defined as follows: { h h(x) if x S(X1 ) (x) = h k (x) if x R k To start, define R 0 = S 0 = and h 0 =. And assume that the sets R l I(X 1 ) and S l I(X 2 ) have been defined inductively for all l such that 0 l k 1. Now, for any n N, let Y 1 be any infinite subset of I(X 1 ), and X 1 (n, Y 1 ) be the set of n distinct points of Y 1 with the following property: (3.3.4) d 1 (x 1, S(X 1 )) d 1 (x, S(X 1 )) for any x 1 Y 1 \ X 1 (n, Y 1 ) and x X 1 (n, Y 1 ). Define X 2 (n, Y 2 ), where n N and Y 2 is any infinite subset of I(X 2 ), analogously. Since S(X 1 ) and S(X 2 ) are compact; there exist nonempty, finite subsets A k and B k of S(X 1 ) and S(X 2 ), resp., defined as follows A k = {a k,1, a k,2,..., a k,n(k) } B k = {b k,1, b k,2,..., b k,m(k) } such that each point of S(X 1 ) (respectively, S(X 2 )) is within 1/k of a point of A k (respectively, B k ). Observing that I(X 1 ) \ k 1 l=0 R l is an infinite set, let For each point p P k, let P k = X 1 (n(k), I(X 1 ) \ k 1 l=0 R l). α(p) = min{j : d 1 (p, a k,j ) = d 1 (p, A k )}. We index the points of P k as p k,1, p k,2,..., p k,n(k) in such a way that if s < t, then either (3.3.5) or (3.3.6) holds: (3.3.5) α(p k,s ) < α(p k,t ), (3.3.6) α(p k,s ) = α(p k,t ) and d(p k,s, a k,α(pk,s)) d(p k,t, a k,α(pk,t)).

6 6 W LODZIMIERZ J. CHARATONIK AND ŞAHİKA ŞAHAN Since I(X 2 ) \ k 1 l=0 S l contains points of all CB-rank less than α, we can define the sets, P k and Q k, as follows: let p k,1, p k,2,..., p k,n(k) be n(k) distinct points of I(X 2 ) \ k 1 l=0 S l such that for each i, d 2 (p k,i, h(a k,α(p k,t ))) d 1 (p k,i, h(a k,α(pk,t ))); then let and let CB(p k,i ) = CB(p k,i ); P k = {p k,1, p k,2,..., p k,n(k)} Q k = X 2(m(k), I(X 2 ) \ [P k ( k 1 l=0 S l)]). For each point q Q k, let β(q ) = min{j : d 2 (q, b k,j ) = d 2 (q, B k )}. We index the points of Q k as q k,1, q k,2,..., q k,m(k) in such a way that if s < t, then, as is analogous to (3.3.5) and (3.3.6) above, either (3.3.7) or (3.3.8) holds: (3.3.7) β(q k,s ) < β(q k,t ), (3.3.8) β(q k,s ) = β(q k,t ) and d 2(q k,s, b k,β(q k,t ) ). Now, we define Q k in terms of Q k and B k with a similar way that we have defined P k in terms of P k and A k as follows: noting that I(X 1 )\[P k ( k 1 l=0 R l)] contains points of all CB-rank less than α, let q k,1, q k,2,..., q k,m(k) be m(k) distinct points of I(X 1 )\[P k ( k 1 l=0 R l) such that for each i, and and let d 1 (q k,i, h 1 (b k,β(q k,t ))) d 2 (q k,i, b k,β(q k,t ) ), CB(q k,i ) = CB(q k,i ),

7 ZERO-DIMENSIONAL CARTESIAN SQUARES 7 Q k = {q k,1, q k,2,..., q k,m(k) }. Finally, let R k = P k Q k and S k = P k Q k, and define h k : R k S k as follows: h k (p k,i ) = p k,i for all p k,i P k, h k (q k,i ) = q k,i for all q k,i Q k. Therefore, R k, S k and the one-to-one and onto function h k : R k S k for each k = 0, 1, 2,... have been defined by induction. To be able to complete the definiton of the homeomorphism h of X 1 onto X 2 we need to prove the following two facts: (3.3.9) k=0 R k = I(X 1 ); (3.3.10) k=0 S k = I(X 2 ). First suppose that (3.3.9) is false. Then there is a point x 0 X 1 \ k=0 R k. Hence, by the definition of R k and P k, x 0 / k=1 P k = k=1 X 1(n(k), I(X 1, k 1 l=0 R l)). Thus, since x 0 I(X 1 ) \ k 1 l=0 R l for each k, from (3.3.4) we have the following: (3.3.11) d 1 (x 0, S(X 1 )) d 1 (p, S(X 1 )) for all p k=1 P k. From the definitions of P k and R k, we see that the sets P 1, P 2,... are mutually disjoint and nonempty; hence, k=1 P k is an infinite set. Thus, since X 1 is compact, inf{d 1 (p, S(X 1 )) : p k=1 P k} = 0.

8 8 W LODZIMIERZ J. CHARATONIK AND ŞAHİKA ŞAHAN Hence, by (3.3.11), d 1 (x 0, S(X 1 )) = 0; however, since x 0 I(X 1 ), this is impossible and this completes the proof of (3.3.9). The proof of (3.3.10) is similar using the sets Q k. Finally, we complete the definition of h : X 1 X 2. First of all, note that h is well defined since the sets S(X 1 ), R 1, R 2,... are mutually disjoint. Also, by (3.3.9) h is defined on all X 1 and by (3.3.10) h maps onto X 2 since h k (R k ) = S k for each k and h[s(x 1 )] = S(X 2 ). Due to the fact that the sets S(X 2 ), S 1, S 2,... are mutually disjoint, h and each h k is one-to-one h is one-to-one. Finally,the continuity of h follows from the uniform continuity of h, the properties of the sets A k and B k and (3.3.5)-(3.3.8). Therefore, since X 1 is compact and X 2 is Hausdorff, h is a homeomorphism of X 1 onto X 2. Theorem 3.4. Let α be a countable ordinal and let X 1 and X 2 be two zero-dimensional compact metric spaces satisfying the following conditions for i {1, 2}: (3.4.1) For every x X i, if x has a countable neighborhood in X i, then CB(x) < α. (3.4.2) For every ordinal β such that β < α and for every uncountable, open subset U i of X i, there exist an x U i such that CB(x) = β. Then the spaces X 1 and X 2 are homeomorphic. Proof. In this case S(X 1 ) and S(X 2 ) are homeomorphic to the Cantor set, so the condition (3.3.3) of Theorem 3.3 is satisfied. Denote by Z(α) the (topologically unique) metric spaces satisfying the assumptions of Theorem 3.4. In particular Z(0) is the Cantor set and Z(1) is the Pe lczyński space described in p.70 in [5]. Theorem 3.5. If α is an ordinal satisfying the condition (3.1.1) (or equivalently (3.1.2)) of Theorem 3.1, then Z(α) Z(α) is homeomorphic to Z(α) and for every natural number n, the hyperspace F n (Z(α)) is homeomorphic to Z(α). In particular there are uncountably many compact metric spaces X homeomorphic to their Cartesian products X n and to their hyperspaces F n (X). Proof. Note that if α satisfies condition (3.1.1) of Theorem 3.1 then Z(α) Z(α) satisfies the conditions of Theorem 3.4, so it is homeomorphic to Z(α).

9 ZERO-DIMENSIONAL CARTESIAN SQUARES 9 Note that the spaces Z(α) are not the only ones that are homeomorphic to their Cartesian squares. For example, the disjoint union Z(0) Z(1) also has this property. Problem 3.6. Characterize all zero-dimensional compact metric spaces homeomorphic to their Cartesian squares. Problems 3.7. Is there a zero-dimensional compact metric space X such that X is homeomorphic to X X, but not homeomorphic to F 2 (X) nor to F n (X), for some n? Similarly, is there a zero-dimensional compact metric space X such that X is homeomorphic to F n (X), for some n, but not to X X? Is there a zero-dimensional compact metric space and two natural numbers n, m such that X is homemomorphic to F n (X), but not to F m (X)? References 1. J. J. Charatonik and W. J. Charatonik, A Degree of Non-Local Connectedness, Rocky Mountain J. Math., 31 (2001), no.4, T. Jech, Set Theory: The Third Millenium Edition, Springer, K. Kuratowski and A. Mostowski, Set Theory, North-Holland and PWN, Amsterdam, Warszawa, M. M. Marjanović, Exponentially Complete Spaces III, Publ. Inst. Math. (Beograd) (N.S.) 14(28) (1972), A. Illanes and S. B. Nadler, Jr., Hyperspaces. Fundementals and Recent Advances, Monographs and Textbooks in Pure and Applied Mathematics, 216. Marcel Dekker, Inc., New York, Department of Mathematics and Statistics, Missouri University of Science and Technology, 400 West 12th St., Rolla, MO, address: wjcharat@mst.edu address: ssxx4@mst.edu