Bankston, Paul

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Marquette Universitye-Publications@MarquetteMSCS Faculty Research and Publications Mathematics, Statistics and Computer Science,Department of3-1-2013Mind the GapsPaul BankstonMarquette University, paul.bankston@marquette.eduPresented version. Mind the Gaps. 47th Spring Topology and Dynamics Conference. CentralConnecticut State University, March 23-25, 2013.Mind the GapsPaul Bankston, Marquette University47th Spring Topology and DynamicsConference23–25 March, 2013, Central ConnecticutState University0. Introduction1. Three Topological Interpretations2. Q-gap Freeness3. K-gap Freeness4. Strong K-gap Freeness5. Extra Strong K-gap Freeness10. Introduction.If [ , , ] is a ternary relation on a set Xinterpreting a notion of betweenness, wesay the structure 〈X, [ , , ]〉 is gap free ifeach two elements of X always have a thirdelement between them. This is the first-order sentence• Gap Freeness:∀ab(a 6= b → ∃x([a, x, b] ∧ x 6= a ∧ x 6= b))For example, if we start with a totally or-dered set 〈X,≤〉 and define [a, c, b] to mean(a ≤ c ≤ b) ∨ (b ≤ c ≤ a), then gap freenessin this interpretation means order-density.We take an “inclusive” view of between-ness; meaning that [a, c, b] automatically holdsif c ∈ {a, b}.In this talk we are interested in gap free be-tweenness relations naturally arising in thecontext of (Hausdorff) continua.21. Three Topological Interpretations.There are (at least) three interpretationsof betweenness in continua deserving men-tion; they’re all closely related.If X is a continuum, a, b, c ∈ X, and c 6∈{a, b}, we have:• [a, c, b]Q iff there’s a disconnection 〈A, B〉of X\{c} such that a ∈ A and b ∈ B; i.e.,a and b lie in different quasicomponentsof X \ {c}.• [a, c, b]C iff there’s no connected A ⊆X \ {c} with a, b ∈ A; i.e., a and b lie indifferent components of X \ {c}; and• [a, c, b]K iff there’s no continuum A ⊆X \ {c} with a, b ∈ A; i.e., a and b liein different continuum components ofX \ {c}.32. Q-gap Freeness.Clearly [ , , ]Q ⊆ [ , , ]C ⊆ [ , , ]K; henceQ-gap free =⇒ C-gap free =⇒ K-gap free.2.1 Proposition. If X is an aposyndeticcontinuum, then [ , , ]K = [ , , ]C. IfX is also locally connected, then [ , , ]K =[ , , ]Q. Q-gap freeness is a very strong property.2.2 Theorem (L. E. Ward). Q-gap free-ness in a continuum implies local connect-edness and hereditary decomposability. Itis equivalent to the connected intersectionproperty—the intersection of any two con-nected subsets is connected. 4Ward uses what we call Q-gap free as thedefining condition for a continuum to be atree. Less overloaded terminology is den-dron; indeed the metrizable dendrons arethe dendrites—locally connected and con-taining no simple closed curves.Currently we do not know of any literatureon the C-interpretation of betweenness, sohere is an opportunity to ask some ques-tions:• Does C-gap freeness imply Q-gap free-ness?• Failing this, are C-gap free continua lo-cally connected? Aposyndetic?• Or, is there some weakened form of theconnected intersection property that char-acterizes C-gap freeness?53. K-gap Freeness.Given a continuum X and a, b ∈ X, letK(a, b) constitute the subcontinua of X thatcontain both a and b. Then the K-interval[a, b]K bracketed by a and b is defined tobe⋂K(a, b). Hence [a, c, b]K holds iff c ∈[a, b]K.3.1 Proposition. A continuum is herediter-ily unicoherent iff each of its K-intervals isa subcontinuum. Hereditary unicoherence clearly implies K-gap freeness, and it is natural to ask whetherthis weakening of the connected intersec-tion property is actually a characterization.The answer turns out to be no.6A continuum X is a crooked annulus if ithas a decomposition X = M ∪N into sub-continua such that:• Both M and N are hereditarily inde-composable; and• M ∩ N = A ∪ B, where A and B aredisjoint nondegenerate subcontinua.3.2 Theorem. A crooked annulus is K-gap free without being even unicoherent,let alone hereditarily so. In a crooked annulus one can show thateach nondegenerate K-interval [a, b]K con-tains two nondegenerate subcontinua, onecontaining a and the other containing b.(E.g., if a ∈ A and b ∈ B, then [a, b]K =A ∪ B.) This clearly gives us K-gap free-ness.74. Strong K-gap Freeness.Recall the first-order statement of gap free-ness from above.• Gap Freeness:∀ab(a 6= b → ∃x([a, x, b] ∧ x 6= a ∧ x 6= b))If we replace negations of equality in theconclusion with negations of betweenness,we obtain a stronger property (when be-tweenness is interpreted properly).• Strong Gap Freeness:∀ab(a 6= b → ∃x([a, x, b]∧¬[x, a, b]∧¬[a, b, x]))With the Q- and the C-interpretations, stronggap freeness is not really stronger than gapfreeness because these interpretations sat-isfy8• Antisymmetry:∀abc(([a, b, c] ∧ [a, c, b])→ b = c)To see this, suppose [a, c, b]C and b 6= c.If c = a then clearly ¬[a, b, c]C; so assumec 6∈ {a, b}. Then there is a component A ofX \ {c} with a ∈ A and b 6∈ A. Thus A ∪ {c}is a connected subset of X \ {b} containinga and c; so ¬[a, b, c]. The Q-interpretationis antisymmetric as well because it is finerthan the C-interpretation.The topologist’s sine curve is not K-antisym-metric: if a is any point on the graph ofsin(1/x), 0 < x ≤ 1, and b and c are anytwo points on the line segment {0}×[−1,1],then both [a, c, b]K and [a, b, c]K hold.By Proposition 2.1, aposyndetic continuaare K-antisymmetric. However, the combspace is K-antisymmetric without being apo-syndetic.9Recall Ward’s result that Q-gap freenessin continua is equivalent to the connectedintersection property. This property au-tomatically implies both local connected-ness and hereditary decomposability, but itsweaker cousin hereditary unicoherence doesnot. (E.g., any pseudo-arc is hereditarilyunicoherent.) And while it is an open prob-lem whether hereditary unicoherence has afirst-order characterization ever so slightlystronger than K-gap freeness, we have thefollowing.4.1 Theorem. Strong K-gap freeness ina continuum is equivalent to the contin-uum’s being both hereditarily unicoherentand hereditarily decomposable.10Proof. The easy direction is to assume theconjunction of hereditary unicoherence andhereditary decomposability. For then eachnondegenerate K-interval is connected, byProposition 3.1, hence it is a decomposablecontinuum. Any point in the intersection ofa decomposition of a K-interval witnessesstrong K-gap freeness.For the opposite direction, let X be a stronglyK-gap free continuum. If M and N are sub-continua with M ∩N = A∪B, where A andB are nonempty, closed, and disjoint, weuse Zorn’s lemma to find a ∈ A, b ∈ B suchthat if a′ ∈ A, b′ ∈ B, and [a′, b′]K ⊆ [a, b]K,then [a′, b′]K = [a, b]K.So we use strong K-gap freeness to findc ∈ [a, b]K such that both [a, c]K and [c, b]Kare proper subsets of [a, b]K. But eitherc ∈ A or c ∈ B, and this contradicts theminimality of [a, b]K.11Thus we infer that X is hereditarily unico-herent; next we tackle hereditary decom-posability.Suppose Y is a nondegenerate indecom-posable subcontinuum of X. Then, by aresult of D. Bellamy, Y contains an inde-composable subcontinuum with more thanone composant; hence we may (WLOG)assume Y itself is irreducible about somedoubleton set {a, b}. But [a, b]K is a sub-continuum, by hereditary unicoherence, soY = [a, b]K. Now, by strong K-gap free-ness, there is some c ∈ [a, b]K such thatboth [a, c]K and [c, b]K are proper subsets of[a, b]K. By hereditary unicoherence again,we infer that Y is decomposable, a contra-duction. 125. Extra Strong K-gap Freeness.By extra strong gap freeness in an interpre-tation of betweenness we mean that bothgap freeness and antisymmetry hold.5.1 Theorem. Extra Strong K-gap free-ness in a continuum is equivalent to sayingthat all the continuum’s nondegenerate K-intervals are (Hausdorff) arcs. 13

Mind the Gaps

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