Countable ordinal spaces and compact countable subsets of a metric space
aa r X i v : . [ m a t h . GN ] J u l COUNTABLE ORDINAL SPACES AND COMPACT COUNTABLESUBSETS OF A METRIC SPACE
BORYS ´ALVAREZ-SAMANIEGO
N´ucleo de Investigadores Cient´ıficosUniversidad Central del Ecuador (UCE)Quito, Ecuador
ANDR ´ES MERINO
Escuela de Ciencias F´ısicas y Matem´aticaFacultad de Ciencias Exactas y NaturalesPontificia Universidad Cat´olica del Ecuador (PUCE)Quito, Ecuador
Abstract.
We show in detail that every compact countable subset of a metricspace is homeomorphic to a countable ordinal number, which extends a resultgiven by Mazurkiewicz and Sierpinski for finite-dimensional Euclidean spaces. Inorder to achieve this goal, we use Transfinite Induction to construct a specifichomeomorphism. In addition, we prove that for all metric space (
E, d ), the car-dinality of the set of all the equivalence classes K E , up to homeomorphisms, ofcompact countable subsets of E is less than or equal to ℵ , i.e. | K E | ≤ ℵ . Wealso show that for all cardinal number κ smaller than or equal to ℵ , there existsa metric space ( E κ , d κ ) such that | K E κ | = κ . E-mail addresses : [email protected], borys [email protected], [email protected],[email protected] . Date : July 27, 2018.2010
Mathematics Subject Classification.
Key words and phrases.
Cantor-Bendixson’s derivative; ordinal numbers; ordinal topology. Introduction
The study of homeomorphisms between compact countable subsets of a topolog-ical space and countable ordinal numbers was began by S. Mazurkiewicz and W.Sierpinski in [8]. More precisely, they showed that for every compact countablesubset of an n -dimensional Euclidean space, there exists a homeomorphism betweenthis subset and some countable ordinal number. Moreover, a detailed proof of thislast result when the Euclidean space, under consideration, is the real line, was givenby the authors in [1]. Some related propositions can also be found in [3, 4]. Themain result of Section 2 is Theorem 2.1 below, which extends Lemma 3.6 in [1] foran arbitrary metric space. It is worth mentioning that Theorem 2 in [2] considerscompact, dispersed topological spaces with some additional properties, while The-orem 2.1 below regards the case of a metric space. Furthermore, it is stated in [7],without proof, that it is a known fact which can be proved by induction that Y is acountable locally compact space if and only if Y is homeomorphic to some countableordinal number (with the order topology). In this way, Lemmas 2.1, 2.3 and 2.4,proved in Section 2, are the comprehensive induction steps required in the Transfi-nite Induction used in the proof of Theorem 2.1. Section 3 is devoted to study thecardinality of the set of all the equivalence classes K M , up to homeomorphisms, ofcompact countable subsets of a metric space ( M, d ). Propositions 3.1 and 3.2 areused in the proof of Theorem 3.1, where it is shown that for all metric space (
E, d ),the cardinality of K E is less than or equal to ℵ . Propositions 3.3 to 3.5 shows thatfor all cardinal number κ ≤ ℵ , there exists a metric space ( E κ , d κ ) such that thecardinality of the set K E κ is equal to κ . Proposition 3.6 says that there exists acountable metric space ( F, d F ) such that | K F | = ℵ . Finally, Proposition 3.7 assertsthat there is an uncountable metric space ( G, d G ) such that | K G | = ℵ .We denote by OR , the class of all ordinal numbers. In addition, ω representsthe set of all natural numbers and ω is the set of all countable ordinal numbers.Further, we consider any ordinal number as being a topological space, endowed withits natural order topology. In order to describe this last topology, for all α, β ∈ OR such that α ≤ β , we write( α, β ) := { γ ∈ OR : α < γ < β } , [ α, β ) := { γ ∈ OR : α ≤ γ < β } . Thus, for any δ ∈ OR , the natural order topology for δ is given by the followingtopological basis { ( β, γ ) : β, γ ∈ OR , β < γ ≤ δ } ∪ { [0 , β ) : β ∈ OR , β ≤ δ } . OMPACT COUNTABLE SUBSETS OF A METRIC SPACE 3
Next definition was first introduced by G. Cantor in [5].
Definition 1.1 (Cantor-Bendixson’s derivative) . Let A be a subset of a topologicalspace. For a given ordinal number α ∈ OR , we define, using Transfinite Recursion,the α -th derivative of A , written A ( α ) , as follows: • A (0) = A , • A ( β +1) = ( A ( β ) ) ′ , for all ordinal number β , • A ( λ ) = \ γ<λ A ( γ ) , for all limit ordinal number λ = 0 ,where B ′ denotes the derived set of B , i.e., the set of all limit points (or accumulationpoints) of the subset B . Remark 1.1.
Given any subset of a T topological space, its derived set is closed.As a consequence of this last result, we have that if F is a closed subset of a T topological space, then ( F ( α ) ) α ∈ OR is a decreasing family of closed subsets. Moreover, P ( C ) and | C | denote, respectively, the power set and the cardinalityof the set C . We also write A ∼ B if there is a homeomorphism between thetopological spaces A and B . If ( X, τ ) is a topological space, then K X represents theset of all compact countable subsets of X , where a countable set is either a finite setor a countably infinite set, and K X := K X / ∼ denotes the set of all the equivalenceclasses, up to homeomorphisms, of elements of K X . If ( E, d ) is a metric space, x ∈ E and r >
0, we denote by B ( x, r ) and B [ x, r ] the open and closed balls, centered at x with radius r >
0, respectively. Furthermore, for all Y = ∅ , ρ Y is used to designatethe discrete metric on the set Y . We now give the following definition. Definition 1.2 (Cantor-Bendixson’s characteristic) . Let D be a subset of a topolog-ical space such that there exists an ordinal number β ∈ OR with the property that D ( β ) is finite. We say that ( α, p ) ∈ OR × ω is the Cantor-Bendixson characteristic of D if α is the smallest ordinal number such that D ( α ) is finite and | D ( α ) | = p . Inthis case, we write CB ( D ) = ( α, p ) . For the sake of completeness, we give here the proof of the following theorem,which was first introduced by G. Cantor in [6] for an n -dimensional Euclidean space.It deserves to point out that there are some known extensions, considering topolog-ical spaces, of the next result. Theorem 1.1.
Let ( X, τ ) be a Hausdorff space. For all K ∈ K X , there exists α ∈ ω such that K ( α ) is a finite set. B. ´ALVAREZ-SAMANIEGO AND A. MERINO
Proof.
Let K ∈ K X . We suppose, for a contradiction, that for all countable ordinalnumber γ , K ( γ ) is an infinite set. Let α ∈ ω . Since α + 1 ∈ ω , we have that K ( α +1) is an infinite set, and thus it is a nonempty set. By Remark 1.1, K ( α +1) ⊆ K . Then, K ( α +1) is a countable set. Using the fact that every nonempty, compact, perfect,Hausdorff space is uncountable, we obtain that K ( α +2) = K ( α +1) . Thus, by usingagain Remark 1.1, we get K ( α +2) K ( α +1) . We now define K α := K ( α +1) r K ( α +2) = ∅ . Then, { K γ : γ ∈ ω } is a family of nonempty sets. By the Axiom of Choice, thereexists a function f : ω → [ γ ∈ ω K γ such that for all γ ∈ ω , f ( γ ) ∈ K γ . We claim that f is injective. In fact, let β, δ ∈ ω be such that β < δ . Thus, β + 2 ≤ δ + 1. By Remark 1.1, K ( δ +1) ⊆ K ( β +2) . Then, K β ∩ K δ = ( K ( β +1) r K ( β +2) ) ∩ ( K ( δ +1) r K ( δ +2) ) = ∅ . Since f ( β ) ∈ K β and f ( δ ) ∈ K δ , it follows that f ( β ) = f ( δ ). Hence, f is a one-to-onefunction. Therefore, ℵ := | ω | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) [ γ ∈ ω K γ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | K | ≤ ℵ , giving a contradiction. This finishes the proof of the theorem. (cid:3) Remark 1.2.
Last theorem implies that if ( X, τ ) is a Hausdorff space and K ∈ K X ,then CB ( K ) is well-defined and furthermore CB ( K ) ∈ ω × ω . Existence of homeomorphisms
Lemma 2.1. If ( E, d ) is a metric space, K ∈ K E , and CB ( K ) = (1 , , then K ∼ ω + 1 . Proof.
Since CB ( K ) = (1 , x ∈ E such that K ′ = { x } . Moreover, wesee that K = K (0) is infinite. Then, K r K ′ is a countably infinite set. Thus, thereis a bijection g from K r K ′ onto ω . We now define the following function f : K −→ ω + 1 z f ( z ) = g ( z ) , if z = x,ω, if z = x. OMPACT COUNTABLE SUBSETS OF A METRIC SPACE 5
From the definition of f , we obtain directly that f is a bijective function. We willnow show that f is continuous. Since every point belonging to K r K ′ is an isolatedpoint of K , it follows that f is continuous at every point of K r K ′ . Thus, it remainsto show the continuity of f at the point x . We take an open basic neighborhood V of f ( x ) = ω with regard to the order topology of ω + 1. We will now show that f − ( V )is a neighborhood of x . If V = [0 , β ), we have that β = ω + 1. Thus, V = ω + 1 and f − ( V ) = K is a neighborhood of x . On the other hand, if V = ( n, α ), then n < ω < α ≤ ω + 1 . Therefore, n ∈ ω and α = ω + 1. We now define the following set A := { z ∈ K : f ( z ) ≤ n } . Thus, x A . Moreover, since f is an injective function, we see that A is a finiteset. Let us take r := min { d ( z, x ) : z ∈ A } >
0. Then, K ∩ B ( x, r ) ⊆ f − (( n, ω + 1)) . In fact, if z ∈ K satisfies d ( z, x ) < r , then z A . Hence, f ( z ) > n . In addition,using the definition of function f , we see directly that f ( z ) < ω + 1. Thus, f ( z ) ∈ ( n, ω + 1) = V . Consequently, f − (( n, ω + 1)) is a neighborhood of x . Therefore, f is continuous at the point x . We thus conclude that f is continuous at every pointof its domain. Then, f is a continuous function. Finally, since f is a continuousbijective function, K is compact and ω + 1 is a Hausdorff space, it follows that f isa homeomorphism. In conclusion, K ∼ ω + 1. (cid:3) The next lemma extends Lemma 3.4 in [1] to the case of an arbitrary T topologicalspace. Lemma 2.2.
Let K and F be closed subsets of a T topological space such that K ∩ F = K ∩ int( F ) , where int( F ) is the set of all interior points of F . Then, forall α ∈ OR , we have that ( K ∩ F ) ( α ) = K ( α ) ∩ F. (2.1) Proof.
We will use Transfinite Induction. • The case α = 0 follows directly. • We assume that the result holds for a given α ∈ OR , i.e., ( K ∩ F ) ( α ) = K ( α ) ∩ F .Then,( K ∩ F ) ( α +1) = (cid:0) ( K ∩ F ) ( α ) (cid:1) ′ = ( K ( α ) ∩ F ) ′ ⊆ ( K ( α ) ) ′ ∩ F ′ ⊆ K ( α +1) ∩ F, B. ´ALVAREZ-SAMANIEGO AND A. MERINO where in the last expression we have used the fact that F is closed. To show theother inclusion, let x ∈ K ( α +1) ∩ F . Since K is a closed subset of a T topologicalspace, using Remark 1.1, it follows that x ∈ K ∩ F = K ∩ int( F ). Therefore, thereexists a neighborhood U of x such that U ⊆ F . Let V be a neighborhood of x .We now take W := U ∩ V . We see that W is also a neighborhood of x . Then, ∅ = (cid:0) W r { x } (cid:1) ∩ K ( α ) = (cid:0) W r { x } (cid:1) ∩ K ( α ) ∩ F = (cid:0) W r { x } (cid:1) ∩ ( K ∩ F ) ( α ) ⊆ (cid:0) V r { x } (cid:1) ∩ ( K ∩ F ) ( α ) . Hence, x ∈ ( K ∩ F ) ( α +1) . Thus, K ( α +1) ∩ F ⊆ ( K ∩ F ) ( α +1) . Therefore, ( K ∩ F ) ( α +1) = K ( α +1) ∩ F . • Lastly, let λ = 0 be a limit ordinal number. We assume that for all β ∈ OR suchthat β < λ , ( K ∩ F ) ( β ) = K ( β ) ∩ F . Hence,( K ∩ F ) ( λ ) = \ β<λ ( K ∩ F ) ( β ) = \ β<λ ( K ( β ) ∩ F ) = \ β<λ K ( β ) ∩ F = K ( λ ) ∩ F. This finishes the proof. (cid:3)
Lemma 2.3.
Let ( E, d ) be a metric space and let α > be a countable ordinalnumber. Suppose that for all ordinal number β ∈ OR such that < β < α and forall e K ∈ K E with CB ( e K ) = ( β, p ) ∈ OR × ω , we have that e K ∼ ω β · p + 1 . Then,for all K ∈ K E such that CB ( K ) = ( α, , we get K ∼ ω α + 1 . Proof.
Let K ∈ K E be such that CB ( K ) = ( α,
1) with α >
1. Then, there exists x ∈ K with K ( α ) = { x } . We see that x ∈ K ( α ) ⊆ K ′′ . Thus, x is an accumulationpoint of K ′ . Then, there is a sequence ( x n ) n ∈ ω in K ′ r { x } such that ( d ( x n , x )) n ∈ ω is astrictly decreasing sequence converging to 0. Moreover, since { d ( z, x ) ∈ R : z ∈ K } is a countable set, it follows that for all n ∈ ω , A n := { d ( z, x ) ∈ R : z ∈ K } c ∩ ( d ( x n +1 , x ) , d ( x n , x ))is a nonempty set. Therefore, { A n : n ∈ ω } is a nonempty family of nonempty sets.By the Axiom of Countable Choice, there exists a sequence ( r n ) n ∈ ω of real numberssuch that for all n ∈ ω , d ( x n +1 , x ) < r n < d ( x n , x ) OMPACT COUNTABLE SUBSETS OF A METRIC SPACE 7 and for all z ∈ K we have that d ( z, x ) = r n . Thus, for all n ∈ ω , we define thefollowing sets F := B ( x, r ) c ,F n +1 := B [ x, r n ] r B ( x, r n +1 )and K n := K ∩ F n . We claim that for all n ∈ ω , K ∩ F n = K ∩ int( F n ) . In fact, let n ∈ ω . We see immediately that K ∩ int( F n ) ⊆ K ∩ F n . Reciprocally,let z ∈ K ∩ F n . We first consider the case when n = 0. We obtain that z ∈ K and d ( z, x ) ≥ r . Since z ∈ K , we have that d ( z, x ) = r . Thus, ε := d ( z, x ) − r > B ( z, ε ) ⊆ F . Then, z ∈ int( F ). Hence, K ∩ F ⊆ K ∩ int( F ). We now consider the case n ∈ ω r { } . We have that z ∈ K and r n ≤ d ( z, x ) ≤ r n − . Since z ∈ K , we obtain that d ( z, x ) = r n and d ( z, x ) = r n − . We now take ε n :=min { d ( z, x ) − r n , r n − − d ( z, x ) } >
0. We get B ( z, ε n ) ⊆ F n . Hence, z ∈ int( F n ).Therefore, K ∩ F n ⊆ K ∩ int( F n ).We can now see that the family { K n : n ∈ ω } has the following properties: • Since K is a closed subset of E , we obtain that for all n ∈ ω , x n ∈ K n . • For all n ∈ ω , K n ⊆ K . • Since the intersection of two closed subsets is also closed, we see that for all n ∈ ω , K n is a closed subset. • Since every closed subset of a compact space is compact, we have that forall n ∈ ω , K n is compact. • Since for all n ∈ ω , K n is a countable set, we obtain that for all n ∈ ω , K n ∈ K E . • For all n ∈ ω , K ′ n = ∅ . In fact, let n ∈ ω . By Lemma 2.2, we have that K ′ n = ( K ∩ F n ) ′ = K ′ ∩ F n . Moreover, since x n ∈ K ′ ∩ F n , we see that x n ∈ K ′ n . • Since { F n : n ∈ ω } is a pairwise disjoint family of sets, we obtain that thefamily of sets { K n : n ∈ ω } is also pairwise disjoint. • We have that K = ] n ∈ ω K n ⊎ { x } . B. ´ALVAREZ-SAMANIEGO AND A. MERINO
In fact, since the sequence ( r n ) n ∈ ω converges to 0, we see that U n ∈ ω F n ⊎{ x } = E . Then, ] n ∈ ω K n ⊎ { x } = ] n ∈ ω ( K ∩ F n ) ⊎ { x } = K ∩ ] n ∈ ω F n ! ⊎ { x } = K ∩ ] n ∈ ω F n ⊎ { x } ! = K ∩ E = K. • For all n ∈ ω , K ( α ) n = ∅ . In fact, by Lemma 2.2, we see that for all n ∈ ω , K ( α ) n = ( K ∩ F n ) ( α ) = K ( α ) ∩ F n = { x } ∩ F n = ∅ . • Using the fact that an infinite subset of a compact subset of a topologicalspace has at least a limit point in the compact subset, using also Remark1.1 and the Cantor intersection theorem in a Hausdorff topological space, wesee that the last assertion implies that for all n ∈ ω , if CB ( K n ) = ( β n , p n ) ∈ OR × ω , then 0 < β n < α and p n ∈ ω r { } .It follows from the hypothesis that for all n ∈ ω , K n ∼ ω β n · p n + 1. By the Axiomof Countable Choice, there is a sequence ( f n ) n ∈ ω of homeomorphisms, such that forall n ∈ ω , f n : K n → ω β n · p n + 1 is a homeomorphism of the topological space K n onto ω β n · p n + 1. We now define the following function f : K −→ τ + 1 z f ( z ) = f ( z ) , if z ∈ K , n − X k =0 ω β k · p k + 1 + f n ( z ) , if z ∈ K n , for some n ∈ ω r { } ,τ, if z = x, where τ := X k ∈ ω ω β k · p k := sup ( n X k =0 ω β k · p k : n ∈ ω ) . Proceeding in a similar way to the proof of Lemma 3.3 in [1], it is possible to showthat τ = ω α and f is a homeomorphism of K onto τ + 1. Hence, K ∼ ω α + 1. (cid:3) Lemma 2.4.
Let ( E, d ) be a metric space. Let α be a countable ordinal number suchthat α > and let p ∈ ω . We assume that for all e K ∈ K E such that CB ( e K ) = ( α, , OMPACT COUNTABLE SUBSETS OF A METRIC SPACE 9 we have that e K ∼ ω α + 1 . Then, for all K ∈ K E with CB ( K ) = ( α, p ) , we get K ∼ ω α · p + 1 . Proof.
Let K ∈ K E be such that CB ( K ) = ( α, p ) with α >
0. As mentioned in theproof of the previous lemma, we can show that p ∈ ω r { } . Thus, K ( α ) = { x , x , . . . , x p − } , where for all i, j ∈ { , . . . , p − } such that i = j , x i = x j . For all m ∈ { , . . . , p − } ,we define d m := min { d ( x m , x j ) ∈ R : 0 ≤ j ≤ p − j = m } > . Let m ∈ { , . . . , p − } . Since { d ( z, x m ) ∈ R : z ∈ K } is a countable set, there exists r m > r m ∈ { d ( z, x m ) ∈ R : z ∈ K } c ∩ (0 , d m ) . Thus, for all z ∈ K , we have that d ( z, x m ) = r m . We now define F m := B [ x m , r m ]and F := E r p − [ j =1 B ( x j , r j ) . Moreover, for all n ∈ { , . . . , p − } , we also define K n := K ∩ F n . We observe that for all n ∈ { , . . . , p − } , K ∩ F n = K ∩ int( F n ) . In fact, let n ∈ { , . . . , p − } . Since int( F n ) ⊆ F n , we see that K ∩ int( F n ) ⊆ K ∩ F n .Reciprocally, given z ∈ K ∩ F n , we see that z ∈ K and we consider the followingtwo cases: • We first examine the situation when n ∈ { , . . . , p − } . Since F n := B [ x n , r n ],we have that d ( z, x n ) ≤ r n . In addition, z ∈ K implies that d ( z, x n ) = r n . By taking ε n := r n − d ( z, x n ) >
0, we obtain that B ( z, ε n ) ⊆ F n . Hence, z ∈ int( F n ). • We now assume that n = 0. Using the fact that z ∈ F := E r p − S j =1 B ( x j , r j ),we conclude that for all j ∈ { , . . . , p − } , d ( z, x j ) ≥ r j . Furthermore, since z ∈ K , we see that for all j ∈ { , . . . , p − } , d ( z, x j ) = r j . We now take ε := min { d ( z, x j ) − r j : 1 ≤ j ≤ p − } >
0. Then, B ( z, ε ) ⊆ F . In order to prove this last assertion, let w ∈ B ( z, ε ). Wenow suppose, to derive a contradiction, that there exists i ∈ { , . . . , p − } such that d ( w, x i ) < r i . Then, ε + r i ≤ d ( z, x i ) ≤ d ( z, w )+ d ( w, x i ) < ε + r i ,which is a contradiction. Thus, z ∈ int( F ).Therefore, K ∩ F n ⊆ K ∩ int( F n ).Proceeding in a similar manner as in the proof of Lemma 2.3, we can show that thefamily { K n : 0 ≤ n ≤ p − } satisfies the next properties: • Using Remark 1.1, we have that for all n ∈ { , . . . , p − } , x n ∈ K n . • For all n ∈ { , . . . , p − } , K n ⊆ K . • For all n ∈ { , . . . , p − } , K n is a closed subset of E . • Since every closed subset of a compact space is compact, we obtain that forall n ∈ { , . . . , p − } , K n is compact. • For all n ∈ { , . . . , p − } , K n ∈ K E . • { K n : 0 ≤ n ≤ p − } is a pairwise disjoint family of sets. • K = p − ] n =0 K n . • Using Lemma 2.2, we conclude that for all n ∈ { , . . . , p − } , K ( α ) n = ( K ∩ F n ) ( α ) = K ( α ) ∩ F n = { x n } . • It follows from the last assertion that for all n ∈ { , . . . , p − } , CB ( K n ) =( α, n ∈ { , . . . , p − } , there exists ahomeomorphism g n : K n → ω α + 1 from the topological space K n onto ω α + 1. Wenow consider the following function g : K −→ τ + 1 z g ( z ) = g ( z ) , if z ∈ K ,ω α · n + 1 + g n ( z ) , if z ∈ K n , for some n ∈ { , . . . , p − } , OMPACT COUNTABLE SUBSETS OF A METRIC SPACE 11 where τ := ω α · p . By a similar argument to the one used in the proof of Lemma 2.3above, we obtain that function g is a homeomorphism from K onto τ + 1. Hence, K ∼ ω α · p + 1. (cid:3) Theorem 2.1.
Suppose that ( E, d ) is a metric space. Let α be a countable ordinalnumber such that α > and let p ∈ ω . If K ∈ K E satisfies CB ( K ) = ( α, p ) , then K ∼ ω α · p + 1 . Proof.
We proceed by Strong Transfinite Induction on the ordinal number α > α = 1. Now, let α ∈ ω be such that α >
1. We suppose that the conclusion is true for all ordinal number β such that0 < β < α . By Lemmas 2.3 and 2.4, the result is also valid for α . Thus, the theoremholds for all countable ordinal numbers greater than zero. (cid:3) Remark 2.1.
The hypothesis about the countable cardinality of the ordinal number α in Lemma 2.3, Lemma 2.4 and Theorem 2.1 can be omitted. In fact, if ( E, d ) isa metric space, K ∈ K E , ( α, p ) ∈ OR × ( ω r { } ) and K ∼ ω α · p + 1 , then α ∈ ω . Cardinality of the partition
Let (
X, τ ) be a topological space. We consider the set K X of all compact countablesubsets of X . The set K X := K X / ∼ provides a partition of the set K X into disjointequivalence classes, more precisely, K X = { [ K ] ∈ P ( K X ) : K ∈ K X } , where, for all K ∈ K X [ K ] := { K ∈ K X : K ∼ K } . The following two propositions will be used in the proof of Theorem 3.1 below.
Proposition 3.1.
Let ( X, τ ) be a T topological space. For all K , K ∈ K X suchthat K ∼ K , we have that CB ( K ) = CB ( K ) .Proof. Let K , K ∈ K X be such that K ∼ K and let f : K → K be a homeo-morphism from K onto K . We will first show that for all ordinal number α ∈ OR , K ( α )1 ∼ K ( α )2 , where f | K ( α )1 is a homeomorphism between these two sets. In order toprove this last assertion, we use below Transfinite Induction. • In the case when α = 0, we see that K (0)1 = K ∼ K = K (0)2 and f = f | K (0)1 : K (0)1 → K (0)2 is a homeomorphism from K (0)1 onto K (0)2 . • We now suppose that the result holds for a given ordinal number α , i.e., K ( α )1 ∼ K ( α )2 and f | K ( α )1 is a homeomorphism between K ( α )1 and K ( α )2 . By Remark 1.1above, we have that K ( α +1)1 ⊆ K ( α )1 . Thus, f ( K ( α +1)1 ) = f | K ( α )1 ( K ( α +1)1 ) = f | K ( α )1 (( K ( α )1 ) ′ ) = ( K ( α )2 ) ′ = K ( α +1)2 . Then, f | K ( α +1)1 : K ( α +1)1 → K ( α +1)2 is a homeomorphism from K ( α +1)1 onto K ( α +1)2 .Therefore, K ( α +1)1 ∼ K ( α +1)2 . • Finally, let λ = 0 be a limit ordinal number. We presume that for all β ∈ OR such that β < λ , K ( β )1 ∼ K ( β )2 , where f | K ( β )1 is a homeomorphism from K ( β )1 onto K ( β )2 . Since f is an injection, we have that f ( K ( λ )1 ) = f \ β<λ K ( β )1 ! = \ β<λ f ( K ( β )1 ) = \ β<λ K ( β )2 = K ( λ )2 . Therefore, f | K ( λ )1 is a homeomorphism between K ( λ )1 and K ( λ )2 , i.e., K ( λ )1 ∼ K ( λ )2 .Then, for all α ∈ OR , | K ( α )1 | = | K ( α )2 | . We suppose that CB ( K ) = ( β, p ) ∈ OR × ω .Thus, β is the smallest ordinal number such that K ( β )1 is finite. Furthermore, since | K ( β )1 | = p , we obtain that | K ( β )2 | = | K ( β )1 | = p . With this, we conclude that CB ( K ) = ( β, p ) = CB ( K ). (cid:3) Proposition 3.2.
Let ( E, d ) be a metric space and let K , K ∈ K E . If CB ( K ) = CB ( K ) , then K ∼ K .Proof. Let CB ( K ) = CB ( K ) = ( α, p ), for some ordinal number α and some p ∈ ω . • If α = 0, we have that K and K are both finite sets with p elements,therefore K ∼ K . • If α > p ∈ ω , by Theorem 2.1, we have that K ∼ ω α · p + 1 and K ∼ ω α · p + 1, thus K ∼ K . (cid:3) Propositions 3.1 and 3.2 imply that for a metric space (
E, d ), the partition of K E is fully characterized by the Cantor-Bendixson characteristic. Theorem 3.1.
Let ( E, d ) be a metric space. The cardinality of K E is less than orequal to ℵ .Proof. We define the function f CB : K E −→ ω × ω [ K ] f CB ([ K ]) = CB ( K ) . OMPACT COUNTABLE SUBSETS OF A METRIC SPACE 13
By Theorem 1.1 and Proposition 3.1, we see that function f CB is well-defined. More-over, Proposition 3.2 shows that f CB is an injective function. Thus, | K E | ≤ | ω × ω | = | ω | =: ℵ . (cid:3) In general, we cannot strengthen the last result. To show this, we give the follow-ing three propositions.
Proposition 3.3.
For all n ∈ ω , there exists a metric space ( E n , d n ) such that | K E n | = n .Proof. Let n ∈ ω . We now consider the space ( n, ρ n ), where ρ n is the discrete metricon the set n := { , . . . , n − } . Since n is a finite set, we have that every subset of n is compact, i.e., K n = P ( n ) . Moreover, for all K ∈ K n , we see that K (0) = K is a finite set. Therefore, CB ( K ) =(0 , | K | ). Thus, f CB ( K n ) ⊆ { } × n. On the other hand, for all k ∈ { , . . . n − } , there exists F ⊆ n such that | F | = k ,i.e., CB ( F ) = (0 , k ). Thus, f CB ( K n ) = { } × n. Hence, | K n | = | f CB ( K n ) | = n. (cid:3) Proposition 3.4.
There exists a metric space ( E, d E ) such that | K E | = ℵ .Proof. We consider the metric space ( ω, ρ ω ). Since ρ ω is the discrete metric on theset ω , we see that a subset of ω is compact if and only if it is a finite set. Then,for all K ∈ K ω , K (0) = K is a finite set. Hence, for all K ∈ K ω , CB ( K ) = (0 , | K | ).Thus, f CB ( K ω ) ⊆ { } × ω . On the other hand, since for all k ∈ ω , there exists K ⊆ ω such that | K | = k , it follows that f CB ( K ω ) = { } × ω . Therefore, | K ω | = | f CB ( K ω ) | = ℵ . (cid:3) Proposition 3.5.
There exists a metric space ( F, d F ) such that | K F | = ℵ .Proof. We take the metric space ( R , d ), where d is the usual metric on the set R .By Theorem 3.4 in [1], we obtain that | K R | = ℵ . (cid:3) Finally, it is worth mentioning that the last two results do not depend on thecardinality of the underlying metric spaces considered there, as it can be seen in thenext two propositions.
Proposition 3.6.
There exists a countable metric space ( G, d G ) such that | K G | = ℵ .Proof. Proceeding in a similar way as in the proof of Theorem 2.1 in [1] and consid-ering the density of the rational numbers, Q , in R , we can see that for all countableordinal number α ∈ ω , and for all a, b ∈ Q such that a < b , there exists a set K ∈ K Q such that K ⊆ ( a, b ] and K ( α ) = { b } . By using this last statement, we canprove an analogous result to Corollary 2.1 in [1], more precisely, we have that forall countable ordinal number α ∈ ω , and for all p ∈ ω , there is a set K ∈ K Q suchthat | K ( α ) | = p . Then, f CB ( K Q ) = (cid:0) ω × ( ω r { } ) (cid:1) ∪ (0 , . Hence, | K Q | = | f CB ( K Q ) | = | ω × ω | = | ω | =: ℵ . (cid:3) Proposition 3.7.
There exists an uncountable metric space ( H, d H ) such that | K H | = ℵ .Proof. We take the uncountable metric space ( R , ρ R ), where ρ R is the discrete metricon the real line. Proceeding in a similar fashion as in the proof of Proposition 3.4,we obtain | K R | = ℵ . (cid:3) References [1]
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OMPACT COUNTABLE SUBSETS OF A METRIC SPACE 15 [8]