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Topology Proceedings Topology Proceedings Web: http://topology.auburn.edu/tp/ Mail: Topology Proceedings Department of Mathematics & Statistics Auburn University, Alabama 36849, USA E-mail: [email protected] ISSN: 0146-4124 COPYRIGHT °c by Topology Proceedings. All rights reserved. Topology Proceedings Volume 23, Summer 1998, 141–153 SPLITTABILITY OVER LINEAR ORDERINGS A. J. Hanna∗ and T.B.M. McMaster† Abstract A partial order X is splittable over a partial or- der Y if for every subset A of X there exists an order preserving mapping f : X → Y such that f −1f(A)=A. We define a cardinal function sc(X) (the ‘splittability ceiling’ for X) to be the least cardinal β such that the disjoint sum of β copies of X fails to split over a single copy of X. We al- low sc(X)=∞ to cover the case where arbitrarily many disjoint copies may be split. We investigate this cardinal function with respect to (linear) par- tial orders. 1. Introduction A. V. Arhangel’skiˇı formulated and developed a range of defini- tions of splittability (or cleavability) in topology (see for example [1, 2]), of which the following are amongst the most basic. Definition 1.1. For topological spaces X and Y : − X is splittable over Y along the subset A of X if there exists continuous f : X → Y such that: ∗ The research of the first author was supported by a distinction award scholarship from the Department of Education for Northern Ireland. † The authors would like to express their gratitude to Steven Watson for his helpful comments and insight, especially regarding Theorem 1.8. Mathematics Subject Classification: 06A05, 06A06, 54A25, 54C99 Key words: splittability, partially ordered set, splittability ceiling 142 A. J. Hanna and T.B.M. McMaster (i) f(A) ∩ f(X \ A)=∅ or, equivalently, (ii) f −1f(A)=A. − X is splittable over Y if for every subset A of X there exists continuous f : X → Y such that f −1f(A)=A. It quickly becomes apparent that splittability is not exclu- sively a topological idea. Indeed, only a routine translation into the language of the appropriate category is required for an anal- ogous definition of splittability over other structures. (For ex- ample, splittability over semigroups is considered in [6].) Definition 1.2. Let X and Y be partially ordered sets (posets). − A map f between partial orders is increasing (or order pre- serving)ifx ≤ y implies f(x) ≤ f(y). − X is splittable over Y along the subset A of X if there exists increasing f : X → Y such that f −1f(A)=A. − X is splittable over Y if for every subset A of X there exists increasing f : X → Y such that f −1f(A)=A. The following result was obtained by D. J. Marron [4, 5]: Theorem 1.3. A poset X is splittable over the n-point chain if and only if: (i) X does not contain a chain of height greater than n, and (ii) X does not contain two disjoint chains of height n. Note 1.4. The previous result shows that it is not possible to split the (disjoint) sum of two copies of a finite chain over a single copy of the same finite chain. However, it is possible to disjointly embed two copies of ω (the positive integers with usual ordering) into a single copy of ω. Clearly, then, it is possible to split ‘two disjoint copies’ of ω over ω. SPLITTABILITY OVER LINEAR ORDERINGS 143 In general, suppose that α copies of a poset X can be dis- jointly embedded into a single copy of X. It is clear that the disjoint sum of α copies of X will split over a single copy of X. Indeed, if (X · α) can be embedded into X, we can split the sum of α copies of X over X. For notation and further information on linear orderings the interested reader is referred to [8]. Definition 1.5 (the ‘splittability ceiling’ for X). Let sc(X) be the least cardinal β such that the (disjoint) sum of β copies of X fails to split over a single copy of X. We allow sc(X)=∞ to cover the case where the sum of arbitrarily many disjoint copies may be split. Note 1.6. The critical case for deciding sc(X) is reached in attempting to split 2|X| copies. If we have more than 2|X| dis- joint copies of X and split along some subset of their sum, then there must be copies which we are splitting along the same subset (since X has precisely 2|X| subsets) and hence the ‘same’ map will do. In other words, if sc(X) ≥ 2|X| then sc(X)=∞. Definition 1.7. [8] A cardinal number ℵα is said to be regular if it is not the sum of fewer than ℵα cardinal numbers smaller than ℵα. Theorem 1.8. For any partial order X,ifsc(X) =6 ∞ then sc(X) is a regular cardinal. Proof. Suppose sc(X)=λ<∞ is not a regular cardinal; then λ can be expressed as the sum of α cardinals βi each less than λ, where α is less than λ. Let Y = [ Xi be the disjoint union i∈λ of λ copies of X. We can write Y = [( [ Xj ). For each i ∈ α i∈α j∈βi we can split [ Xj over a single copy Xβi of X, since βi <λ. j∈βi 144 A. J. Hanna and T.B.M. McMaster Likewise we can split [ Xβi over a single copy of X, since α<λ. i∈α Hence we can split λ copies of X (along any subset) over X -a contradiction. Proposition 1.9. The splittability ceiling for the chain of pos- itive integers ω is infinity (i.e. sc(ω)=∞). Proof. Given X, the (disjoint) sum of copies of ω and a subset A of X, we define a map f : X → ω as follows: x (if x ∈ A and x is odd) or (x/∈ A and x is even), f(x)= x + 1 otherwise. It is clear that f is increasing and that f(A) is a subset of the odds while f(X \ A) is a subset of the evens. It follows that f splits X along A over ω as required. The corresponding result holds for the negative integers ω∗ and for the integers ω∗ + ω. Proposition 1.10. Let α be an ordinal (considered as a linear order). Then ∞ if α is a limit ordinal sc(α)= 2 if α is a non-limit ordinal Proof. Note that each element in a limit ordinal has an imme- diate successor. The first part of the result follows from similar methods as employed for ω.Ifα is a non-limit ordinal we specify the subset A to contain the ‘odd’ ordinals less than α. Similarly we specify the subset B to contain the ‘even’ ordinals less than α. We can express α = ξ + n where ξ is a limit ordinal and n is finite. Now f(x) ≥ x for all x ∈ α and f(x)=x for x = ξ + i (0 ≤ i<n) whenever f is a map splitting α along A or B over α. Clearly it will not be possible to split the sum of two copies of α along A and B respectively over a single copy of α. SPLITTABILITY OVER LINEAR ORDERINGS 145 Proposition 1.11. The splittability ceiling for the chain of ra- tionals η is infinity (i.e. sc(η)=∞). Proof. Decompose η into two disjoint subsets C and D, each of which is dense in η. Enumerate both C and D in an arbitrary fashion. Given disjoint copies of η and a subset A to split along, define a map for each copy. Begin by enumerating the copy X1 = {x1,x2,x3,...}.Ifx1 ∈ A (∈/ A) map x1 to the first point in the enumeration of C (D). The process continues inductively (using a method similar to that devised by Cantor to show that every countable linear order can be embedded into η). Proposition 1.12. The splittability ceiling for the chain of the real numbers λ is c+ (i.e. sc(λ)=c+). Proof. We first note that it is possible to disjointly embed continuum-many copies of λ into λ. To prove the result we show that there are only continuum-many increasing maps from the reals into the reals. We know that there are only continuum- many maps from the rationals into the reals. Given increasing f : R → R consider its restriction to the rationals f|Q. For how many increasing maps g : R → R do we have f|Q = g|Q? We can show that f and g can only differ at countably many points: for each irrational x select both a strictly decreasing sequence (an) and a strictly increasing sequence (bn) of rationals, each converging to x.Now(f(an)) converges to some limit l 0 0 while (f(bn)) converges to some limit l .Ifl = l then f(x)= g(x)=l, otherwise f(x),g(x) ∈ [l0,l] and f(X)∩[l0,l]={f(x)}. Since there can only be countably many disjoint intervals in the reals, there can only be countably many points x where f(x) =6 g(x). It follows that there can only be continuum-many maps within each equivalence class. Hence there are at most continuum-many increasing maps from the reals to the reals. Clearly if we have more than continuum-many disjoint copies of λ and pick differ- ent subsets in them, then the union of these copies cannot be 146 A.
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