aa r X i v : . [ m a t h . GN ] J u l Indecomposable Continuum with a Strong Non-Cut Point
Daron Anderson Trinity College Dublin. [email protected] Preprint September 2018
Abstract
We construct an indecomposable continuum with exactly one strong non-cut point. The methodis an adaptation of Bellamy [1]. We start with an ω -chain of indecomposable metric continuaand retractions. The inverse limit is an indecomposable continuum with exactly two composants. Our example is formed by identifying a point in each composant.
Every point p of an indecomposable metric continuum M is a weak cut point. That means thereare distinct x, y ∈ M − p such that each subcontinuum K ⊂ M with { x, y } ⊂ K has p ∈ K . The proof follows from M having more than one composant; and the composant-by-composant version of the result fails. Namely some q ∈ M might fail to weakly cut its composant κ ( q ) . In thatcase we call q a strong non-cut point of κ ( q ) . For example consider the endpoint c of the Knasterbuckethandle. It is easy to see κ ( c ) − c is even arcwise connected. Hence c has only trivial reasonsfor being a weak cut point.There exist indecomposable non-metric continua with exactly one composant − henceforth called Bellamy continua . Each Bellamy continuum is simultaneously an indecomposable contin-uum and a composant of an indecomposable continuum. Bellamy continua resemble indecom-posable metric continua in being compact. In this paper we show they resemble composants ofindecomposable metric continua in that they can have strong non-cut points.
Throughout X is a continuum. That means a nondegenerate compact connected Hausdorff space.For a, b ∈ X we say X is irreducible about { a, b } or irreducible from a to b to mean no proper subcontinuum of X contains { a, b } . The subspace A ⊂ X is called a semicontinuum to mean for Daron Anderson 1 Indecomposable Strong Non-Cut ach a, c ∈ A some subcontinuum K ⊂ A has { a, c } ⊂ K . Every subspace A ⊂ X is partitionedinto maximal semicontinua called the continuum components of A . For background on metriccontinua see [5] and [7]. The results cited here have analagous proofs for non-metric continua.For a subset S ⊂ X denote by S ◦ and S the interior and closure of S respectively. By boundarybumping we mean the principle that, for each closed E ⊂ X , each component C of E meets ∂E = E ∩ X − E . For the non-metric proof see §
47, III Theorem 2 of [5]. One corollary ofboundary bumping is that the point p ∈ X is in the closure of each continuum component of X − p .For b ∈ X we omit the curly braces and write X − b instead of X −{ b } . For distinct a, b, c ∈ X wesay b weakly cuts a from c and write [ a, b, c ] X to mean a and c have different continuum componentsin X − b . When there is no confusion we just write [ a, b, c ] . We say b ∈ X weakly cuts the semicontinuum A ⊂ X to mean [ a, b, c ] for some a, c ∈ A and call b a weak cut point to mean itweakly cuts X and a strong non-cut point otherwise. We define the interval [ a, c ] X = (cid:8) b ∈ X : [ a, b, c ] X (cid:9) . Again we often write [ a, c ] without con- fusion. Note [ a, c ] is not in general connected as the interval notation suggests. In case [ a, c ] isconnected and b ∈ [ a, c ] we have [ a, b ] ∪ [ b, c ] = [ a, c ] . Moreover [ a, b, c ] [ a,c ] for each b ∈ [ a, c ] − { a, c } .Clearly the weak cut structure is topologically invariant. That means [ a, b, c ] X ⇐⇒ [ h ( a ) , h ( b ) , h ( c )] Y for each a, b, c ∈ X and homeomorphism h : X → Y . We say X is indecomposable to mean it is not the union of two proper subcontinua. Equiva-lently each proper subcontinuum is nowhere dense. The composant κ ( x ) of the point x ∈ X isthe union of all proper subcontinua that have x as an element. Indecomposable metric continuaare partitioned into c many pairwise disjoint composants [6]. In case κ ( x ) = κ ( y ) then X is irre- ducible about { x, y } . There exist indecomposable non-metric continua [1, 2, 9] with exactly onecomposant, henceforth called Bellamy continua . We call X hereditarily unicoherent to mean it has some − and therefore all − of the equivalentproperties:(I) The intersection of any two subcontinua of X is connected.(II) [ a, b, c ] X ⇐⇒ [ a, b, c ] L for each subcontinuum L ⊂ X with a, b, c ∈ L .(III) [ a, c ] X = [ a, c ] L for each subcontinuum L ⊂ X with a, c ∈ L . (IV) Whenever a, c ∈ X the set [ a, c ] is a subcontinuum.The continuous function f : X → Y is called monotone to mean each f − ( y ) ⊂ X is connected for y ∈ Y . Theorem 6.1.28 of [4] says this implies f − ( K ) ⊂ X is a continuum for K ⊂ Y a Daron Anderson 2 Indecomposable Strong Non-Cut ontinuum. The function f is called proper to mean f ( L ) ⊂ Y is proper whenever L ⊂ X is aproper subcontinuum.The partition P of X into closed subsets is called upper semicontinuous to mean the following:For each P ∈ P and open U ⊂ X containing P there is open V ⊂ U with P ⊂ V and V a unionof elements of P . Upper semicontinuity of the partition is equivalent to the quotient space X/ P being a continuum.Throughout K ⊂ R is the quinary Cantor set . That means the points in [0 , × { } whose x -coordinate can be expressed in base- without the digits or . We write K for the middle third of K ; K for the leftmost two thirds of the portion of K right of K ; K for the leftmost two thirds ofthe portion of K right of K and so on. Formally K = K ∩ [2 / , / , K = K ∩ [4 / , − / and K n = K ∩ [1 − / n , − / n +1 ] for each n > . Let each k ( n ) = 1 − / n be the right endpoint of K n . Let G : K → K be the unique linear order-reversing isometry and define each P n = G ( K n +1 ) and p ( n ) = G ( k ( n + 1)) . Clearly K = { } ∪ (cid:0) S n P n (cid:1) ∪ (cid:0) S n K n (cid:1) ∪ { } is a disjoint union. We write Q ′ ⊂ R for the union of all semicircles in the upper half-plane with centre (1 − / n , for some n ∈ N and endpoints in K . We write R ′ ⊂ R for the union of all semicircles in the lower half-plane with centre (7 / n , for some n ∈ N and endpoints in K . We write Q for the set { ( x, y + 1) ∈ R : ( x, y ) ∈ Q ′ } and R for the set { ( x, y − ∈ R : ( x, y ) ∈ R ′ } .Throughout B = Q ′ ∪ R ′ is the quinary Knaster buckethandle . B is a metric indecompos-able hereditarily unicoherent continuum. It is easy to see [ x, y ] is an arc whenever x, y share acomposant and [ x, y ] = B otherwise. Let a be the point (3 / , / ∈ B . We write C forthe composant of the left endpoint c = (0 , and D for the composant of the right endpoint d = (1 , . Throughout ω = { , , , . . . } is the first infinite ordinal and ω the first uncountable ordinal. Every initial segment of ω is countable and every countable subset has an upper bound. For the ordered set Ω we say Ψ ⊂ Ω is cofinal to mean it has no upper bound in Ω . Every countable ordinalhas a cofinal subset order-isomorphic to ω . We say Ψ ⊂ Ω is terminal to mean Ω − Ψ has an upperbound.The poset Ω is said to be directed to mean for each γ, β ∈ Ω there is α ∈ Ω with γ, β ≤ α . Notemost authors require γ, β < α . This prohibits maximal elements. In this paper it is convenient toallow a directed set to have maximal elements. An inverse system over the directed set Ω consists of the following data: (1) a family of topo-logical spaces T ( α ) for each α ∈ Ω and (2) a family of continuous maps f αβ : T ( α ) → T ( β ) foreach β ≤ α such that (3) we have f βγ ◦ f αβ = f αγ whenever γ ≤ β ≤ α . The property (3) is called Daron Anderson 3 Indecomposable Strong Non-Cut k (1) k (2) . . .p (1) . . . . . . p (2) c d Figure 1: The quinary Knaster buckethandle commutativity of the diagram . The inverse limit T of the system is the space lim ←−{ T ( α ); f αβ : α, β ∈ Ω } = n ( x α ) ∈ Y α ∈ Ω T ( α ) : f αβ ( x α ) = x β ∀ β ≤ α o . We often suppress the index set and write for example lim ←−{ T ( α ); f αβ } . The functions f αβ are called the bonding maps . Write π β : T → T ( β ) for the restriction of the projection Q α T ( α ) → T ( β ) . The inverse limit X of a system { X ( α ); f αβ } of continua is itself a continuum. If moreovereach bonding map is surjective then so is each π β . In that case we call the inverse system (limit) surjective . For any subcontinuum L ⊂ X we have L = lim ←−{ π α ( L ); f αβ } where each f αβ is restricted to π α ( L ) . Note commutativity implies f αβ has range π β ( L ) hence the subsystem is well defined. Forcofinal Ψ ⊂ Ω the map ( x α ) α ∈ Ω ( x α ) α ∈ Ψ is an homeomorphism between X and the inverselimit lim ←−{ X ( α ); f αβ : α, β ∈ Ψ } over Ψ . We use transfinite recursion to construct the eponymous indecomposable continuum as the inverselimit of a system { X ( α ); f αβ : α, β < ω } of metric continua and retractions. This section shows how to construct each X ( β + 1) from X ( β ) . The following section deals with limit ordinals.To begin let X (0) = B be the quinary buckethandle. Let the composants C , D ⊂ B and points a , c ∈ C be as described in the Introduction. Define the following two sequences ( p n ) Daron Anderson 4 Indecomposable Strong Non-Cut nd ( q n ) in X (0) : Choose an homeomorphism [0 , [ a , c ] with a and c and leteach p n be the image of − /n . Let each q n be the point k ( n ) ∈ B as defined in the Introduction.Observe the pair of sequences (cid:0) ( p n ) , ( q n ) (cid:1) satisfies the following definition. Definition 3.1.
For a ∈ X and c ∈ κ ( a ) we define a tail from a to c as an ordered pair T = (cid:0) ( p n ) , ( q n ) (cid:1) of sequences in κ ( a ) with the properties:(1) For each n ∈ N we have a ∈ [ p n , q n ] and a ∈ [ c, q n ] .(2) For each n ∈ N we have q n / ∈ [ c, a ] .(3) [ p , a ] [ p , a ] . . . .(4) [ a, q ] [ a, q ] . . . . (5) S (cid:8) [ p n , q n ] : n ∈ N (cid:9) = κ ( a ) − c . (6) S (cid:8) [ p n , a ] : n ∈ N (cid:9) = [ c, a ] − c .(7) For each n ∈ N and x ∈ X − [ c, a ] we have either [ c, q n ] ⊂ [ c, x ] or [ c, x ] ⊂ [ c, q n ] . The notion of a tail is pivotal to our example. Indeed as part of the construction we will at stage α < ω choose a tail T α = (cid:0) ( p nα ) , ( q nα ) (cid:1) on X ( α ) so that the tails behave nicely with respect to the bonding maps. This is made precise below.In the next definition and throughout when we write for example a β a γ the map in question is understood to be the bonding map f βγ . Similarily for subsets B ⊂ X ( β ) and C ⊂ X ( γ ) we write B → C to mean f βγ ( B ) = C . Definition 3.2.
Suppose { X ( β ); f βγ : γ, β < α } is an inverse system and each X ( β ) has a distin-guished pair of points ( a β , c β ) and pair of sequences (cid:0) ( p nβ ) , ( q nβ ) (cid:1) . We say the system is coherent to mean a β a γ and c β c γ and each p nβ p nγ , q nβ q nγ , [ p nβ , q nβ ] → [ p nγ , q nγ ] , [ p nβ , a β ] → [ p nγ , a γ ] and [ c β , a β ] → [ c γ , a γ ] for γ, β < α . In practice the distinguished points and sequences in Definition 3.2 will always come aboutfrom a tail. However in Section 4 we already have an inverse system, and most of the work goesinto showing a given pair of sequences is indeed a tail. Thus the definition is given in slightly more generality.
Daron Anderson 5 Indecomposable Strong Non-Cut t stage α < ω we have already constructed the coherent inverse system { X ( β ); f βγ : β, γ < α } of indecomposable hereditarily unicoherent metric continua and retractions. We assume the fol-lowing objects have been specified for each β < α :(i) Distinct composants C ( β ) , D ( β ) ⊂ X ( β ) (ii) Points a β , c β ∈ C ( β ) (iii) A tail T β = (cid:0) ( p nβ ) , ( q nβ ) (cid:1) from a β to c β We also assume for each γ, δ < α the three conditions hold:(a)
S (cid:8) X ( δ ) : δ < γ (cid:9) ⊂ D ( γ ) . (b) S (cid:8) f γδ (cid:0) X ( γ ) − X ( δ ) (cid:1) : γ > δ (cid:9) = C ( δ ) . (c) (cid:8) x ∈ [ c γ , a γ ] : f γδ ( x ) ∈ [ p nδ , a δ ] (cid:9) = [ p nγ , a γ ] for each n ∈ N .Conditions (a) and (b) come straight from [1] and will ensure the limit has exactly two com- posants. Condition (c) is needed to make the resulting point ( c β ) not weakly cut the composant.For an illustration of what Condition (c) means consider the system of retractions [0 , ← [0 , ← [0 , ← . . . where the bonding map [0 , n ] → [0 , m ] collapses [ m, n ] to the point m ∈ [0 , m ] . For X ( β ) = [0 , β ] and a β = β and each p nβ = 1 /n we see the system has Condition (c). Indeed our initial [ c , a ] ← [ c , a ] ← [ c , a ] ← . . . will turn out to be a copy of this simpler system, and sothe example should be kept in mind throughout.We are ready to begin the succesor step. Suppose α = β + 1 is a successor ordinal. We willconstruct an indecomposable hereditarily unicoherent metric continuum X ( β + 1) and retraction f β +1 β : X ( β +1) → X ( β ) . Then we can define the bonding maps f β +1 γ = f β +1 β ◦ f βγ . We will specifythe objects (i), (ii) and (iii) when β is replaced by β + 1 . Finally we will check the enlarged systemis coherent and Condition (a), (b) and (c) hold for all γ, δ ≤ β + 1 .We use terminology from the Introduction. Identify Q ∪ R ⊂ R with the subspace ( Q ∪ R ) ×{ a β } of R × X ( β ) . Define M ⊂ R × X ( β ) as follows. e P n = P n × {− , } × [ p nβ , a β ] e K n = K n × {− , } × [ a β , q nβ ] M = (cid:16) S n ∈ N e P n (cid:17) ∪ (cid:16) S n ∈ N e K n (cid:17) Daron Anderson 6 Indecomposable Strong Non-Cut roperties (4) and (5) of the tail imply M = e P ∞ ∪ M ∪ e K ∞ for e P ∞ = { } × {− , } × [ c β , a β ] e K ∞ = { } × {− , } × X ( β ) The set M ∪ ( Q ∪ R ) is compact since ( Q ∪ R ) is closed and bounded and M is contained inthe product K × {− , } × X ( β ) of compact sets. It is also metric since R , K , {− , } and byassumption X ( β ) are metric spaces. e K e K e K e K ∞ e P e P e P ∞ QR (0 , , c β ) ( / , , q β )( / , − , a β ) Figure 2: Schematic for M . For example the set e K is a family of copies of [ a β , q β ] attached to either halfof the bucket handle at the endpoint a β with the endpoints q β free. To obtain X ( β + 1) first make for each n ∈ N and k ∈ S P n the identification ( k, − , p nβ ) ∼ ( k, , p nβ ) and for each k ∈ S K n make the identification ( k, − , q nβ ) ∼ ( k, , q nβ ) . Then for each x ∈ [ c β , a β ] make the identification (0 , − , x ) ∼ (0 , , x ) and for each x ∈ X ( β ) make the identification (1 , − , x ) ∼ (1 , , x ) . It is straightforward to verify the decomposition is upper semicontinuous.Hence X ( β + 1) is a continuum and [7] Lemma 3.2 says X ( β + 1) is metric.For k ∈ K write B ( k ) for the quotient space of { ( x, y, z ) ∈ M : x = k } . For k / ∈ { , } clearly B ( k ) is homeomorphic to two copies of some [ p nβ , a β ] or [ a β , q nβ ] joined at the points correspondingto p nβ or q nβ respectively. Hence B ( k ) is a continuum irreducible from ( k, − , a β ) to ( k, , a β ) . B (1) is a copy of X ( β ) . Henceforth identify that copy with X ( β ) . B (0) is a copy of [ c β , a β ] . Denotethat copy by [[ c β , a β ]] and write e x ∈ [[ c β , a β ]] for the point corresponding to x ∈ [ c β , a β ] . Define a surjection g : X ( β + 1) → B onto the buckethandle. Daron Anderson 7 Indecomposable Strong Non-Cut β +1 a β X ( β ) q β q β q β +1 q β +1 q β +1 q β +1 c β +1 p β +1 p β +1 [[ c β , a β ]] B + ( / ) B − ( / ) k + ( / ) k − ( / ) Figure 3: The continuum X ( β + 1) is obtained by identifying the free endpoints of opposite pairs ofsubcontinua, and by identifying the two copies of [ c β , a β ] and X ( β ) that make up e P ∞ and e K ∞ respectively.Dashed lines indicate projection onto X ( β ) or [[ c β , a β ]] . g ( x, y, z ) = ( x, y − for x ∈ Q ( x, y + 1) for x ∈ R ( k, for x ∈ B ( k )(1 , for x ∈ X ( β )(0 , for x ∈ [[ c β , a β ]] For each x ∈ B the fibre g − ( x ) is either a singleton, X ( β ) , [ c β , a β ] or some B ( k ) . Thus all fibres are subcontinua and g is monotone. Therefore subcontinua pull back to subcontinua. Inparticular g − ( B ) = X ( β + 1) is a continuum. Define the points c β +1 = e c β and a β +1 = g − ( a ) and each p nβ +1 = e p nβ . The definition of q nβ +1 will be given later in the construction. Claim 1.
The function g : X ( β + 1) → B is proper. Proof.
Identify the plane R with the subspace R × { a β } of R × X ( β ) . For each n ∈ N let K ( n ) be the n th stage of construction of the quinary Cantor set. Let Q ( n ) ⊂ R be the union of allupper semicircles with centre (1 − / m , for some m ∈ N and endpoints in K ( n ) × { } . Let R ( n ) ⊂ R be the union of all lower semicircles with centre (7 / m , − for some m ∈ N andendpoints in K ( n ) × {− } . Define the continuum Y ( n ) = Q ( n ) ∪ R ( n ) ∪ X ( β + 1) . Recall K ( n ) consists of n intervals of length / n . For any such interval I with , / ∈ I itis a straightforward exercise to verify the closure of Y ( n ) − S { B ( k ) : k ∈ I } has exactly twocomponents A containing I × { } × { a β } and A containing I × {− } × { a β } . Daron Anderson 8 Indecomposable Strong Non-Cut × UI × B I × B A A [[ c β , a β ]] X ( β ) Figure 4: Schematic of Claim 1 for I an interval of K (2) Suppose the subcontinuum L ⊂ X ( β + 1) has g ( L ) = B . Clearly L contains the interior of Q ∪ R . Now assume L is proper. Then X ( β + 1) − L contains a basic open subset of the quotientspace of some P m × {± } × [ p mβ , a β ] or K m × {± } × [ a β , q mβ ] . Without loss of generality that set is V = I × { } × U for some open I ⊂ K m and U ⊂ [ a β , q mβ ] . We can shrink I further to make itan interval of some K ( n ) and shrink U to get a β / ∈ U .The subcontinua B ( k ) are homeomorphic for k ∈ I and S { B ( k ) : k ∈ I } is just I × B ( k ) . Treat U as a subset of B ( k ) . Since B ( k ) is irreducible from ( k, , a β ) to ( k, − , a β ) we know B ( k ) − U is the disjoint union B ∪ B of two clopen sets that include ( k, , a β ) and ( k, − , a β ) respectivly. Hence S { B ( k ) : k ∈ I } − V = ( I × B ) ∪ ( I × B ) is the disjoint union of two clopen sets containing I × { } × { a β } and I × {− } × { a β } respec-tively. It follows that Y ( m ) − V = (cid:0) ( I × B ) ∪ A (cid:1) ∪ (cid:0) ( I × B ) ∪ A (cid:1) is the disjoint union of two clopen sets containing I × { } × { a β } and I × {− } × { a β } respec-tively. Recall L is contained X ( β + 1) − V ⊂ Y ( m ) − V . Since the closed set L contains the interiorof Q ∪ R it meets both clopen sets which contradicts how L is connected. We conclude g ( L ) = B if and only if L = X ( β + 1) . Daron Anderson 9 Indecomposable Strong Non-Cut laim 2.
The composants of X ( β + 1) are { g − ( E ) : E ⊂ B is a composant } . Hence X ( β + 1) isindecomposable. Proof.
Since g is monotone and proper each g − ( E ) is contained in some composant of X ( β + 1) .Now suppose the subcontinuum L ⊂ X ( β + 1) meets two distinct g − ( E ) and g − ( E ) . It follows g ( L ) = B . Since g is proper we must have L = X ( β + 1) . The result follows.Claim 2 tells us g − ( C ) and g − ( D ) are distinct composants of X ( β + 1) . Define C ( β + 1) = g − ( C ) and D ( β + 1) = g − ( D ) . From the construction X ( β ) ⊂ D ( β + 1) . The next claimfollows. Claim 3.
Condition (a) holds for all γ, δ ≤ β + 1 . Observe each subcontinuum of B is contained in some union J ∪ J ∪ . . . ∪ J N of arcs where J i = [ a i , b i ] are alternately contained in Q ′ or R ′ and J i ∩ J j = { b n } ⇐⇒ { i, j } = { n, n + 1 } and J i ∩ J j = ∅ otherwise. Let each J i be identified with the corresponding arc in Q or R . The monotone surjection g witnesses how each subcontinuum L ⊂ X ( β + 1) is contained insome J ∪ B ( x ) ∪ J ∪ B ( x ) ∪ . . . ∪ J N ∪ B ( x N +1 ) ∪ J N +1 (†)for each J n ∩ B ( x n ) = ( x n , ± , a β ) and B ( x n ) ∩ J n +1 = ( x n , ∓ , a β ) and all other intersections are empty.Observe each summand of ( † ) is hereditarily unicoherent and irreducible between two end-points − and meets the previous factor at exactly one endpoint and the next factor at the other endpoint. It follows by induction the union is hereditarily unicoherent. Claim 4. X ( β + 1) is hereditarily unicoherent Proof.
Suppose the proper subcontinua L , L ⊂ X ( β + 1) have nonempty intersection. Since X ( β + 1) is indecomposable L ∪ L = X . Hence L ∪ L is contained in a union of the form ( † ) .Since the union is an hereditarily unicoherent subcontinuum L ∩ L is connected.Let { y , y , y . . . } ⊂ B be the set C ∩ K linearly ordered by x ≤ y ⇐⇒ [ c, x ] ⊂ [ c, y ] . Wemust have y = c . For each n ∈ N write y n ± = ( y n , ± , a β ) . Note for n = 0 we have y n + = y n − as elements of X ( β + 1) . For each n > write B ( n ) instead of B ( y n ) . Write I ( n ) for the arc in Q or R corresponding to the arc [ y n , y n +1 ] ⊂ B . By ( † ) each subcontinuum of C ( β + 1) is contained insome [[ c β , a β ]] ∪ I (0) ∪ B (1) ∪ . . . ∪ I ( N − ∪ B ( N ) ∪ I ( N ) . (††) Daron Anderson 10 Indecomposable Strong Non-Cut or example let x ∈ B ( n ) and y ∈ I ( m ) be arbitrary with n < m . It follows from hereditaryunicoherence that [ x, y ] = [ x, y n ± ] B ( n ) ∪ I ( n ) ∪ B ( n + 1) ∪ . . . ∪ B ( m ) ∪ [ y m ± , y ] I ( m ) (‡)for exactly one of the four choices of ± indices. The other intervals of [ x, y ] have a similarform based on whether each endpoint is in some B ( n ) or I ( m ) .For each n ∈ N write k ± ( n ) = (cid:0) k ( n ) , ± , a β (cid:1) . By definition k ( n ) is the right endpoint of K n . Thus we see B (cid:0) k ( n ) (cid:1) is the quotient space of the set { k ( n ) }× {− , } × [ a β , q n ] . Write B ± (cid:0) k ( n ) (cid:1) for the quotient space of the set { k ( n ) }× {± } × [ a β , q n ] . Then the continua B + (cid:0) k ( n ) (cid:1) and B − (cid:0) k ( n ) (cid:1) meet at the point (cid:0) k ( n ) , − , q n (cid:1) ∼ (cid:0) k ( n ) , , q n (cid:1) . Define each q nβ +1 ∈ X ( β + 1) as that point. The expression ( ‡ ) gives the equalities. [ p nβ +1 , q nβ +1 ] = [[ p nβ , a β ]] ∪ I (0) ∪ B (1) ∪ . . . ∪ I ( k ( n ) − ∪ B + ( k ( n ))[ a β +1 , q nβ +1 ] = [ a β +1 , k + (1)] ∪ B (1) ∪ . . . ∪ I ( k ( n ) − ∪ B + ( k ( n ))[ c β +1 , q nβ +1 ] = [[ c β , a β ]] ∪ I (0) ∪ B (1) ∪ . . . ∪ I ( k ( n ) − ∪ B + ( k ( n ))[ p nβ +1 , a β +1 ] = [[ p nβ , a β ]] ∪ [ e a β , a β +1 ][ c β +1 , a β +1 ] = [[ c β , a β ]] ∪ [ e a β , a β +1 ] Claim 5. T β +1 = (cid:0) ( p nβ +1 ) , ( q nβ +1 ) (cid:1) is a tail from a β +1 to c β +1 Proof.
Properties (1) − (4) follow from the equalities above. Property (6) follows from the equal-ities, the definition of [[ c β , a β ]] as a copy of [ c β , a β ] , and how T β has Property (6).For Property (5) first observe each c β +1 / ∈ [ p nβ +1 , q nβ +1 ] . Then recall κ ( a β +1 ) = C ( β + 1) = g − ( C ) . First suppose x ∈ B (0) = [[ c β , a β ]] ⊂ [ c β +1 , a β +1 ] . Then Property (6) implies x is some [ p Nβ +1 , a β +1 ] ⊂ [ p Nβ +1 , q Nβ +1 ] .Otherwise ( ‡‡ ) says x is an element of some B ( n ) or I ( n ) . It is easy to verify the set { k ( n ) : n ∈ N } is cofinal in { y , y , y , . . . } . Thus y n + 1 < k ( N ) for some N ∈ N . Then the above equalities say x ∈ [ p Nβ +1 , q Nβ +1 ] .To prove Property (7) for T β +1 observe each [ c β +1 , x ] has one of the forms [[ c β , a β ]] ∪ I (0) ∪ B (1) ∪ . . . ∪ I ( N − ∪ B ( N ) ∪ [ y N ± , x ] I ( N ) Daron Anderson 11 Indecomposable Strong Non-Cut [ c β , a β ]] ∪ I (0) ∪ B (1) ∪ . . . ∪ B ( N − ∪ I ( N − ∪ [ y N ± , x ] B ( N ) depending on whether x ∈ I ( N ) or B ( N ) for some N ∈ N . For k ( n ) < N the expression for [ c β , q nβ +1 ] says [ c β , q nβ +1 ] ⊂ [ c β , x ] . Likewise [ c β , x ] ⊂ [ c β , q nβ +1 ] for N < k ( n ) .Finally assume N = k ( n ) . For x ∈ I ( N ) compare the two expressions to see [ c β , x ] ⊂ [ c β , q nβ +1 ] .For x ∈ B ( N ) we need only compare the final two summands [ y N ± , x ] B ( N ) and B + (cid:0) k ( n ) (cid:1) . Observeboth summands are sub- continua of B (cid:0) k ( n ) (cid:1) and include the point k + ( n ) . For x ∈ B + (cid:0) k ( n ) (cid:1) clearly [ y N ± , x ] B ( N ) ⊂ B + (cid:0) k ( n ) (cid:1) and so [ c β , x ] ⊂ [ c β , q nβ +1 ] .Otherwise x ∈ B − (cid:0) k ( n ) (cid:1) . Recall that B + (cid:0) k ( n ) (cid:1) = [ k + ( n ) , q nβ +1 ] and B − (cid:0) k ( n ) (cid:1) = [ q nβ +1 , k − ( n )] and B + (cid:0) k ( n ) (cid:1) ∩ B − (cid:0) k ( n ) (cid:1) = { q nβ +1 } . It follows [ y N ± , x ] B ( N ) includes q nβ +1 hence contains B + (cid:0) k ( n ) (cid:1) .We conclude that [ c β , q nβ +1 ] ⊂ [ c β , x ] . Recall we built X ( β + 1) from the subset M ∪ ( Q ∪ R ) of R × X ( β ) by making some iden- tifications. Since the projection R × X ( β ) → X ( β ) onto the third coordinate respects those identifications, it induces a continuous map f β +1 β : X ( β + 1) → X ( β ) . Claim 6.
The map f β +1 β : X ( β + 1) → X ( β ) is a retraction. Proof.
Write π : R × X ( β ) → X ( β ) for the projection onto the third coordinate. Recall the subset M = (cid:8) ( x, y, z ) ∈ M : x = 1 (cid:9) is the disjoint union { } × {− , } × X ( β ) of two copies of X ( β ) . In forming X ( β + 1) from M ∪ ( Q ∪ R ) we identify (1 , − , x ) ∼ (1 , , x ) for each x ∈ X ( β ) . Hence the restriction of f β +1 β to the quotient space of M is an homeomorphism onto X ( β ) . But recallwe have identified X ( β ) with the quotient space of M . Claim 7.
Condition (b) holds for all γ, δ ≤ β + 1 . Proof.
Recall we have defined each f β +1 γ = f β +1 β ◦ f βγ . Hence by induction and commutativity ofthe diagram it is enough to show f β +1 β (cid:0) X ( β + 1) − X ( β ) (cid:1) = C ( β ) . To that end recall we have X ( β + 1) = Q ∪ R ∪ [[ c β , a β ]] ∪ (cid:16) S (cid:8) B ( k ) : k ∈ K − { , } (cid:9)(cid:17) ∪ X ( β ) . Consider the image of each factor under the projection π onto the third coordinate. Both Q and R map onto the singleton a β ∈ C ( β ) . By definition [[ c β , a β ]] is a copy of [ c β , a β ] and the restriction of π is a homeomorphism [[ c β , a β ]] → [ c β , a β ] ⊂ C ( β ) . Daron Anderson 12 Indecomposable Strong Non-Cut or k ∈ K − { , } each B ( k ) is homeomorphic to two copies of some [ p nβ , a β ] or [ a β , q nβ ] joinedat the points corresponding to p nβ or q nβ respectively; and the restriction of π projects each copyonto the subset [ p nβ , a β ] or [ a β , q nβ ] of C ( β ) respectively.Conversely the construction ensures that for any given n ∈ N there are k , k ∈ K − { , } with B ( k ) and B ( k ) formed from two copies of [ p nβ , a β ] and [ a β , q nβ ] respectively as described above.From this we see the set f β +1 β (cid:0) X ( β + 1) − X ( β ) (cid:1) equals { a β } ∪ [ c β , a β ] ∪ (cid:16) S n [ p nβ , a β ] (cid:17) ∪ (cid:16) S n [ a β , q nβ ] (cid:17) = [ c β , a β ] ∪ (cid:16) S n [ p nβ , q nβ ] (cid:17) = [ c β , a β ] ∪ (cid:0) κ ( a β ) − c β (cid:1) = κ ( a β ) = C ( β ) . The first equality follows from hereditary unicoherence and how each a β ∈ [ p nβ , q nβ ] ; the second from Property (5) of the tail; and the last from the definition of C ( β ) . Claim 8.
The system { X ( γ ); f γδ : γ, δ ≤ β + 1 } is coherent. Proof.
The choice of a β +1 and c β +1 and f β +1 β makes it clear a β +1 a β and c β +1 c β . To see f β +1 β (cid:0) [ c β +1 , a β +1 ] (cid:1) = [ c β , a β ] recall that [ c β +1 , a β +1 ] = [[ c β , a β ]] ∪ [ e a β , a β +1 ] . The bonding mapprojects [[ c β , a β ]] onto [ c β , a β ] and sends [ e a β , a β +1 ] ⊂ I (0) to the point a β ∈ [ c β , a β ] . Likewise f β +1 β projects each [ p nβ +1 , a β +1 ] = [[ p nβ , a β ]] ∪ [ e a β , a β +1 ] onto [ p nβ , a β ] . Since the subcontinuum B + (cid:0) k ( n ) (cid:1) ⊂ [ p nβ +1 , q nβ +1 ] projects onto [ p nβ , q nβ ] we have [ p nβ , q nβ ] ⊂ f β +1 β (cid:0) [ p nβ +1 , q nβ +1 ] (cid:1) . To get equality observe [ p nβ +1 , q nβ +1 ] = [ p nβ +1 , a β +1 ] ∪ [ a β +1 , q nβ +1 ] . The pre-vious paragraph shows [ p nβ +1 , a β +1 ] maps onto [ p nβ , a β ] ⊂ [ p nβ , q nβ ] . Now we treat the remainder [ a β +1 , q nβ +1 ] . Recall we define (cid:2) a , k ( n ) (cid:3) as the unique arc in B with endpoints a and k ( n ) . Now observe [ a β +1 , q nβ +1 ] ⊂ g − (cid:0)(cid:0) a , k ( n ) (cid:3)(cid:1) .Consider the intersection K ∩ (cid:2) a , k ( n ) (cid:3) of the arc with the Cantor set. Inspecting the buck-ethandle we see p ( n ) ≤ x ≤ k ( n ) for each x ∈ K ∩ (cid:2) a , k ( n ) (cid:3) . Therefore (cid:8) k ∈ K : [ a β +1 , q nβ +1 ] meets B ( k ) (cid:9) is contained in the interval (cid:2) p ( n ) , k ( n ) (cid:3) ⊂ K . By construction all f β +1 β (cid:0) B ( k ) (cid:1) for / ≤ k ≤ k ( n ) are contained in f β +1 β (cid:0) B ( k ( n ) (cid:1) = [ p nβ , q nβ ] and all f β +1 β (cid:0) B ( k ) (cid:1) for p ( n ) ≤ k ≤ / are contained in f β +1 β (cid:0) B ( p ( n ) (cid:1) = [ p nβ , a β ] ⊂ [ p nβ , q nβ ] . We conclude that f β +1 β (cid:0) [ p nβ +1 , q nβ +1 ] (cid:1) ⊂ [ p nβ , q nβ ] as required. Daron Anderson 13 Indecomposable Strong Non-Cut laim 9.
Condition (c) holds for all γ, δ ≤ β + 1 . Proof.
Let h : [ c β +1 , a β +1 ] → [ c β , a β ] be the restriction of f β +1 β . By induction we need only showeach h − (cid:0) [ p nβ , a nβ ] (cid:1) = [ p nβ +1 , a nβ +1 ] . To that end recall [ c β +1 , a β +1 ] = [[ c β , a β ]] ∪ [ e a β , a β +1 ] . Since [ e a β , a β +1 ] ⊂ Q the map h takes [ e a β , a β +1 ] to a β and sends each e x ∈ [[ c β , a β ]] to the corresponding x ∈ [ c β , a β ] . It follows h − (cid:0) [ x, a β ] (cid:1) = [ e x, a β +1 ] for each x ∈ [ c β , a β ] . Taking x = p nβ we see h − (cid:0) [ p nβ , a β ] (cid:1) = [ e p nβ , a β +1 ] = [ p nβ +1 , a β +1 ] as required.Claim 9 completes the discussion of α = β + 1 a successor ordinal. This section deals with the limit stage of our construction. Henceforth assume α ≤ ω is a limitordinal and { X ( β ); f βγ : β, γ < α } a coherent system of indecomposable hereditarily unicoherentmetric continua and retractions. For all β, γ, δ < α we assume the objects (i), (ii) and (iii) from Section 3 have been specified and Conditions (a), (b) and (c) hold.
We define X ( α ) = lim ←−{ X ( β ); f βγ } and each f αβ as the projection from the inverse limit onto itsfactors. For each γ < α we identify X ( γ ) with the subset (cid:8) x ∈ X ( α ) : x β = x γ for all β > γ (cid:9) of X ( α ) .Straightforward modifications of [8] Theorem 3.1 and [3] Corollary 1 show X ( α ) is both inde-composable and hereditarily unicoherent. To see X ( α ) is metric we observe by definition that α is a countable ordinal. The product Q β<α X ( β ) of countably many metric spaces is itself a metricspace. The inverse limit X ( α ) is by definition a subset of that product and therefore a metric space. It remains to show the enlarged system is coherent; to check Conditions (a), (b) and (c) hold for the enlarged system; and to specify the data (i), (ii) and (iii) for X ( α ) . Much of the effort willgo into proving the pair of sequences given for (iii) is indeed a tail. To that end we first provesome general facts about tails.Recall at stage we defined the tail (cid:0) ( p n ) , ( q n ) (cid:1) on the quinary buckethandle. Observe thesequence ( p n ) tends to c and the intervals [ p n , a ] increase to be dense in [ c , a ] while the sequence ( q n ) has no limit and the intervals [ a , q n ] increase to be dense in the whole space. The next lemmashow this holds for general tails. Lemma 1.
Suppose the indecomposable and hereditarily unicoherent continuum X has a tail T = (( p n ) , ( q n )) from a to c . Then S n [ p n , a ] is nowhere dense and S n [ a, q n ] is dense. Daron Anderson 14 Indecomposable Strong Non-Cut roof.
For the first statement Property (6) says S n [ p n , a ] ⊂ [ c, a ] . By hereditary unicoherence [ c, a ] is a subcontinuum, and it is nowhere dense by indecomposability. We conclude the firstunion is nowhere dense.For the second statement Property (1) says each a ∈ [ p n , q n ] . Hence [ p n , q n ] = [ p n , a ] ∪ [ a, q n ] .Now observe by Property (5) that κ ( a ) − c = [ n ∈ N [ p n , q n ] = [ n ∈ N (cid:16) [ p n , a ] ∪ [ a, q n ] (cid:17) = (cid:16) [ n ∈ N [ p n , a ] (cid:17) ∪ (cid:16) [ n ∈ N [ a, q n ] (cid:17) = (cid:16) [ c, a ] − c (cid:17) ∪ (cid:16) [ n ∈ N [ a, q n ] (cid:17) . The left-hand-side is dense in X . The right hand-side is the union of two sets. We have alreadyshowed the right-hand-side is nowhere dense. It follows the second set is dense. This completes the proof. One can observe Section 3 made no explicit reference to strong non-cut points. In fact these points were mentioned explicitly when we talked about tails. The next lemma is needed to obtain the strong non-cut point mentioned in the title.
Lemma 2.
Suppose the indecomposable and hereditarily unicoherent continuum X has a tail T = ( p n , q n ) from a to c . Then c is the only point of κ ( a ) to not weakly cut the composant. Proof.
Property (5) says κ ( a ) − c is the union of a chain of subcontinua hence is a semicontinuum.This is the definition of c not weakly cutting κ ( a ) . Now let b ∈ κ ( a ) − c be arbitrary. Property (6) says b is an element of some [ p n , q n ] . Define L = [ p n , q n ] and P = [ p n +1 , q n +1 ] First suppose b / ∈ { p n , q n } . Then [ p n , b, q n ] L and thus [ p n , b, q n ] X by hereditary unicoherence. Now suppose b = p n . Then Properties (2) and (3) imply p n = p n +1 and q n = p n +1 respectively.Thus b / ∈ { p n +1 , q n +1 } which implies [ p n +1 , b, q n +1 ] P which in turn implies [ p n +1 , b, q n +1 ] X byhereditary unicoherence. The case for b = q n is similar. We conclude each b ∈ κ ( a ) − c weaklycuts the composant.We have assumed { X ( β ); f βγ : β, γ < α } is coherent. Hence each a β a γ , c β c γ , p nβ p nγ , q nβ q nγ and it follows the inverse limit X ( α ) has points a α = ( a β ) β<α , c α = ( c β ) β<α , p nα =( p nβ ) β<α and q nα = ( q nβ ) β<α . For ease of notation we suppress the subscript and write a, c, p n , q n instead of a α , c α , p nα , q nα respectively. Daron Anderson 15 Indecomposable Strong Non-Cut o see T α = (cid:0) ( p n ) , ( q n ) (cid:1) is a tail from a to c we use Corollary 4.2 and the facts [ p nβ , q nβ ] → [ p nγ , q nγ ] , [ p nβ , a β ] → [ p nγ , a γ ] and [ c β , a β ] → [ c γ , a γ ] respectively, from the definition of coherence,to get the three expressions. [ p n , q n ] = lim ←− (cid:8) [ p nβ , q nβ ]; f βγ ; γ, β < α } (I) [ p n , a ] = lim ←− (cid:8) [ p nβ , a β ]; f βγ ; γ, β < α (cid:9) (II) [ c, a ] = lim ←− (cid:8) [ c β , a β ]; f βγ ; γ, β < α (cid:9) (III)The three expressions show the expanded system { X ( β ); f βγ : β, γ ≤ α } is coherent. Expres-sions (I), (III) and (II) respectively imply T α has Properties (1), (2) and (3) of being a tail. Property(4) is slightly more complicated. Claim 10. T α has Property (4) of being a tail from a to c . Proof.
We first show each q n +1 β / ∈ [ p nβ , q nβ ] . Property (1) for T β says a β ∈ [ p nβ , q nβ ] . Hence [ p nβ , q nβ ] =[ p nβ , a β ] ∪ [ a β , q nβ ] . Property (6) says [ p nβ , a β ] ⊂ [ c β , a β ] thus q n +1 β / ∈ [ p nβ , a β ] by Property (2). Property(4) says q n +1 β / ∈ [ a β , q nβ ] . We conclude q n +1 β / ∈ [ p nβ , q nβ ] . Now we show q n +1 / ∈ [ a, q n ] which proves Property (4) for T α . Just like before we have [ p n , q n ] = [ p n , a ] ∪ [ a, q n ] . In particular [ a, q n ] is contained in [ p n , q n ] . We have already shown q n +1 β / ∈ [ p nβ , q nβ ] . Thus the expression (I) implies q n +1 / ∈ [ p n , q n ] as required.To show T α has Properties (5) and (6) we introduce some notation to measure how far a given subcontinuum extends along the tail. Notation 4.1.
For each β < α and subcontinuum L ⊂ X ( β ) define k L k = max (cid:8) n ∈ N : [ c β , q nβ ] ⊂ L (cid:9) where we allow the value k L k = ∞ in case all [ c, q n ] ⊂ L Lemma 1 says each S n [ a β , q nβ ] is dense. Therefore k L k is well defined whenever L ⊂ X ( β ) is proper. In case L = X ( β ) we define k L k = ∞ . Clearly k L k ≤ k P k for all L ⊂ P and k L k = 0 when either c β / ∈ L or L ⊂ X ( β ) = κ ( a β ) = X ( β ) − C ( β ) .Claims 11 and 12 will be used to show T α has Property (5). Claim 11.
We have k L k ≤ k f βγ ( L ) k for each γ, β < α and subcontinuum L ⊂ X ( β ) . Daron Anderson 16 Indecomposable Strong Non-Cut roof.
In case k L k = 0 the result is obvious. Hence assume k L k > . That means [ c β , q nβ ] ⊂ L forsome n > . Property (1) for T β says a β ∈ [ c β , q nβ ] hence [ c β , q nβ ] = [ c β , a β ] ∪ [ a β , q nβ ] . Property (6) says each p mβ ∈ [ c β , a β ] hence [ c β , q nβ ] = [ c β , p mβ ] ∪ [ p mβ , a β ] ∪ [ a β , q nβ ] . We conclude each [ p mβ , a β ] ⊂ L .To show [ c γ , q nγ ] ⊂ f βγ ( L ) first observe c β ∈ L and so c γ ∈ f βγ ( L ) by coherence. Now suppose x ∈ [ c γ , a γ ] − c γ . Property (6) for X ( γ ) says x ∈ [ p mγ , a γ ] for some m > . We have already shown [ p mβ , a β ] ⊂ L . Therefore f βγ ( L ) contains f βγ (cid:0) [ p mβ , a β ] (cid:1) = [ p mγ , a γ ] by coherence hence x ∈ f βγ ( L ) .Now suppose x ∈ [ c γ , q nγ ] − [ c γ , a γ ] . By the first paragraph we have [ c γ , q nγ ] = [ c γ , a γ ] ∪ [ a γ , q nγ ] hence x ∈ [ a γ , q nγ ] . The first paragraph proves [ c β , q nβ ] = [ c β , p nβ ] ∪ [ p nβ , a β ] ∪ [ a β , q nβ ] which equals [ c β , p nβ ] ∪ [ p nβ , q nβ ] since a β ∈ [ p nβ , q nβ ] . Therefore [ p nβ , q nβ ] ⊂ [ c β , q nβ ] . Finally observe x ∈ [ a γ , q nγ ] ⊂ [ p nγ , q nγ ] = f βγ (cid:0) [ p nβ , q nβ ] (cid:1) ⊂ f βγ (cid:0) [ c β , q nβ ] (cid:1) ⊂ f βγ ( L ) . We conclude x ∈ f βγ ( L ) as required. Taking n = k L k gives the result. Claim 12.
Suppose x ∈ κ ( a ) − [ c, a ] . Then x ∈ [ p n , q n ] for some n ∈ N . Proof.
By Lemma 4 the set
Ψ = (cid:8) β < α : x β / ∈ [ c β , a β ] (cid:9) is terminal. Replace α with Ψ and henceassume all x β / ∈ [ c β , a β ] . This does not change X ( α ) or whether x ∈ [ p n , q n ] . We deal with two cases separately. First assume (cid:8)(cid:13)(cid:13) [ c β , x β ] (cid:13)(cid:13) : β ∈ Ω } is bounded for someterminal Ω ⊂ α . Like before we can assume without loss of generality α = Ω . Hence thereis N ∈ N with each [ c β , q Nβ ] [ c β , x β ] . Property (7) for β says each [ c β , x β ] ⊂ [ c β , q Nβ ] . In particular x β ∈ [ c β , q Nβ ] = [ c β , a β ] ∪ [ a β , q Nβ ] and so x β ∈ [ a β , q Nβ ] ⊂ [ p Nβ , q Nβ ] . By (I) we conclude ( x β ) ∈ [ p N , q N ] .Now assume no such Ω exists. By induction we can select an increasing sequence β (1) <β (2) < . . . with (cid:13)(cid:13) [ c β ( n ) , x β ( n ) ] (cid:13)(cid:13) > n for each n ∈ N . Lemma 5 says π (cid:0) [ c, x ] (cid:1) = S (cid:8) f β (cid:0) [ c β , x β ] : β < α ) (cid:9) .In particular π (cid:0) [ c, x ] (cid:1) contains each f β ( n )0 (cid:0) [ c β ( n ) , x β ( n ) ] (cid:1) . Thus we have (cid:13)(cid:13) π (cid:0) [ c, x ] (cid:1)(cid:13)(cid:13) ≥ (cid:13)(cid:13) f β ( n )0 (cid:0) [ c β ( n ) , x β ( n ) ] (cid:1)(cid:13)(cid:13) ≥ (cid:13)(cid:13) [ c β ( n ) , x β ( n ) ] (cid:13)(cid:13) by Claim 11 and so (cid:13)(cid:13) π (cid:0) [ c, x ] (cid:1)(cid:13)(cid:13) > n .Since n ∈ N was arbitrary we conclude (cid:13)(cid:13) π (cid:0) [ c, x ] (cid:1)(cid:13)(cid:13) = ∞ . This means π (cid:0) [ c, x ] (cid:1) = X (0) .Replacing with any γ < α we see each π γ (cid:0) [ c, x ] (cid:1) = X ( γ ) hence [ c, x ] = X ( α ) . That means X ( α ) is irreducible from c to x . In other words x / ∈ κ ( c ) . Since κ ( c ) = κ ( a ) this contradicts theassumption. Claim 13. T α has Property (7) of being a tail from a to c . Daron Anderson 17 Indecomposable Strong Non-Cut roof.
Suppose x / ∈ [ c, a ] . It follows from (III) and Lemma 4 the set (cid:8) β < α : x β / ∈ [ c β , a β ] (cid:9) is terminal. Without loss of generality assume all x β / ∈ [ c β , a β ] . Now suppose [ c, q n ] [ c, x ] . Itfollows from Lemma 5 we cannot have [ c β , q nβ ] ⊂ [ c β , x β ] cofinally. Therefore [ c β , q nβ ] [ c β , x β ] terminally. Property (6) for β says [ c β , x β ] ⊂ [ c β , q nβ ] terminally. Thus [ c, x ] ⊂ [ c, q n ] by Lemma5. To deal with Property (5) of being a tail, we use Condition (c) from the construction Claim 14. T α has Property (6) of being a tail from a to c . Proof.
Let x ∈ [ c, a ] − c be arbitrary. By (III) each x β ∈ [ c β , a β ] . Let γ < α be fixed and observe Ψ = { β < α : γ ≤ β } is cofinal. Property (6) for γ says x γ ∈ [ p n , a γ ] for some n ∈ N . Let g : [ c β , a β ] → [ c γ , a γ ] be the restriction of f βγ . Then x γ ∈ [ p n , a γ ] hence x β ∈ g − (cid:0) [ p nγ , a γ ] (cid:1) which equals [ p nβ , a β ] by Condition (c). Thus x β ∈ [ p nβ , a β ] and x ∈ lim ←− (cid:8) [ p nβ , a β ] : β ∈ Ψ (cid:9) . The expression (II) says the set on the right-hand- side equals [ p n , a ] and so x ∈ [ p n , a ] .Since x ∈ [ c, a ] − c is arbitrary we see S (cid:8) [ p n , a ] : n ∈ N (cid:9) contains [ c, a ] − c . Property (6) for β says each c β / ∈ [ p nβ , a β ] thus c / ∈ S (cid:8) [ p n , a ] : n ∈ N (cid:9) . We conclude S (cid:8) [ p n , a ] : n ∈ N (cid:9) =[ c, a ] − c Claim 15. T α is a tail from a to c . Proof.
The expressions (I), (III) and (II) imply T α has Properties (1), (2) and (3) respectively. Properties (4), (6) and (7) follow from Claims 10, 14 and 13 imply T α respectively. Claim 12 says that S n (cid:2) p n , q n ] contains each x ∈ κ ( a ) − [ c, a ] . Combined with Property (6) and how [ p n , q n ] = [ p n , a ] ∪ [ a, q n ] we see S n (cid:2) p n , q n ] contains κ ( a ) − c . Since it is disjoint from D ( α ) it is a proper semicontinuum. Hence the only possibilities are S n (cid:2) p n , q n ] = κ ( a ) − c or S n (cid:2) p n , q n ] = κ ( a ) . To see the former observe Property (5) for each β says c β / ∈ [ p nβ , q nβ ] thus c / ∈ S n (cid:2) p n , q n ] . This shows Property (5) for T α .We have already chosen the points a = a α and c = c α . Claim 15 shows the pair T α of sequences is indeed a tail from a α to c α . Thus we have specified the objects (ii) and (iii) for stage α . It remainsto select the composants (i) and prove Conditions (a), (b) and (c) hold for all γ, δ ≤ α . Finally wemust show each X ( β ) occurs as a subspace of X ( α ) and how f αβ is a retraction. Daron Anderson 18 Indecomposable Strong Non-Cut or (i) we must choose C ( α ) = κ ( a α ) . Hereditary unicoherence along with (III) shows C ( α ) = κ ( c α ) as required. For each γ < α the identification X ( γ ) = (cid:8) x ∈ X ( α ) : x β = x γ for all β > γ (cid:9) makes it clear the X ( γ ) are nested and the bonding maps are retractions. Hence S { X ( β ) : β < α } is a semicontinuum of X ( α ) and thus contained in some composant D ( α ) of X ( α ) . Then condition(a) follows from the definition of D ( α ) . Claim 16.
The composants D ( α ) and C ( α ) of X ( α ) are distinct. Thus at stage α the condition on(i) holds. Proof.
Since X is indecomposable it is enough to show it is irreducible from c ∈ C ( α ) to some x ∈ D ( α ) . Let γ < α and x ∈ X ( γ ) be arbitrary and suppose { c, x } ⊂ L for some subcontinuum L ⊂ X ( α ) . We claim π β ( L ) = X ( β ) for all β > γ . Thus by surjectivity L = X ( α ) . Clearly { c β , x β } ⊂ π β ( L ) . By definition of the embedding X ( γ ) → X ( α ) we have x β = x γ hence { c β , x γ } ⊂ π β ( L ) . Since x γ = π γ ( x ) we have x γ ∈ X ( γ ) . By (a) for stage β we have X ( γ ) ⊂ D ( β ) . Hence π β ( L ) meets the distinct composants C ( β ) and D ( β ) of X ( β ) and so π β ( L ) = X ( β ) as required.To prove Condition (b) for the expanded system we use the following claim. Claim 17.
Suppose x γ = x γ +1 for some x ∈ X ( α ) and γ < α . Then x ∈ X ( γ ) . Proof.
We claim x β = x γ +1 for each β > γ + 1 . If x β ∈ X ( γ + 1) then since f βγ +1 is a retraction we have x β = x γ +1 . Otherwise x β ∈ X ( β ) − X ( γ + 1) and Condition (b) says f βγ +1 ( x β ) ∈ C ( γ + 1) . But recall f βγ +1 ( x β ) = f βγ +1 ◦ f αβ ( x ) = f αγ +1 ( x ) = x γ +1 . By assumption x γ +1 = x γ and therefore x γ ∈ C ( γ + 1) . But x γ ∈ X ( γ ) and by Condition (a) we have X ( γ ) ⊂ D ( γ + 1) . This is acontradiction since the composants C ( γ + 1) and D ( γ + 1) are disjoint. Claim 18.
Condition (b) holds for all γ, δ ≤ α . Proof.
We want to show
S (cid:8) f γδ (cid:0) X ( γ ) − X ( δ ) (cid:1) : α ≥ γ > δ (cid:9) = C ( δ ) for each δ < α . First observethe above union can be written f αδ (cid:0) X ( α ) − X ( δ ) (cid:1) ∪ (cid:16) S (cid:8) f γδ (cid:0) X ( γ ) − X ( δ ) (cid:1) : α > γ > δ (cid:9)(cid:17) . By induction the second factor equals C ( δ ) . So it is enough to show the first factor equals C ( δ ) as well. Daron Anderson 19 Indecomposable Strong Non-Cut o see f αδ (cid:0) X ( α ) − X ( δ ) (cid:1) ⊂ C ( δ ) let x ∈ X ( α ) − X ( δ ) be arbitrary. Claim 17 implies x δ +1 = x δ .Since f δ +1 δ is a retraction we have x δ +1 ∈ X ( δ + 1) − X ( δ ) and so f δ +1 δ ( x δ +1 ) ∈ C ( δ ) by Condition(b) at stage δ + 1 . But f δ +1 δ ( x δ +1 ) is just x δ = f αδ ( x ) and so f αδ ( x ) ∈ C ( δ ) .To see f αδ (cid:0) X ( α ) − X ( δ ) (cid:1) = C ( δ ) recall T α is a tail from a to c . Property (5) says { c } ∪ (cid:16) S n [ p n , q n ] (cid:17) = C ( α ) . By coherence the image under f αγ is { c γ } ∪ (cid:16) S n [ p nγ , q nγ ] (cid:17) which equals C ( γ ) by Property (5) of T γ .Finally we prove (c) for the expanded system. Claim 19.
Condition (c) holds for all γ, δ ≤ α . Proof.
By commutativity and Condition (c) at earlier stages it is enough to consider the case γ = α . We must show (cid:8) x ∈ [ c, a ] : x δ ∈ [ p nδ , a δ ] (cid:9) = [ p n , a ] .Let γ > δ be arbitrary and consider the γ -coordinate of a point x of the left-hand-side. Since x ∈ [ c, a ] and [ c, a ] = lim ←− (cid:8) [ c δ , a δ ]; f γδ (cid:9) by (III) we have x γ ∈ [ c γ , a γ ] . Since x δ = f γδ ( x γ ) we see x γ is an element of (cid:8) y ∈ [ c γ , a γ ] : f γδ ( y ) ∈ [ p nδ , a δ ] (cid:9) . Property (c) at earlier stages says this set equals [ p nγ , a γ ] . Thus x γ ∈ [ p nγ , a γ ] . By (II) we know the set lim ←− (cid:8) [ p nδ , a δ ]; f γδ (cid:9) is well defined and equals [ p n , a ] . Since x γ ∈ [ p nγ , a γ ] for all γ > δ we conclude (cid:8) x ∈ [ c, a ] : x δ ∈ [ p nδ , a δ ] (cid:9) ⊂ [ p n , a ] .For the other inclusion let x ∈ [ p n , a ] be arbitrary. By (II) each x δ ∈ [ p nδ , a δ ] . By Property (6) at stage δ we have [ p nδ , a δ ] ⊂ [ c δ , a δ ] . Hence x δ ∈ [ c δ , a δ ] and by (III) we have x ∈ [ c, a ] . We conclude [ p n , a ] ⊂ (cid:8) x ∈ [ c, a ] : x δ ∈ [ p nδ , a δ ] (cid:9) .We are ready to prove the main theorem. Theorem 3.
There exists a Bellamy continuum with a strong non-cut point.
Proof.
By Theorem 1 of [1] the limit X = lim ←−{ X ( α ); f αβ : β, α < ω } is indecomposable with atmost two composants. The trivial composant E ⊂ X is the set S { X ( α ) : α < ω } of eventuallyconstant ω -sequences. The nontrivial composant X − E is equal to lim ←−{ C ( α ); f αβ : β, α < ω } .The points a = ( a β ) and c = ( c β ) witness how X − E is nonempty. Therefore X has exactly twocomposants. Theorem 15 applied to { X ( α ); f αβ : β, α < ω } says there is a tail from a = ( a β ) to c = ( c β ) .Lemma 2 says c does not weakly cut X − E . Daron Anderson 20 Indecomposable Strong Non-Cut hoose any two points x ∈ E and y ∈ X − E both distinct from c . Let e X be obtained bytreating { x, y } as a single point. It follows e X is an indecomposable continuum with exactly onecomposant and c ∈ e X is not a weak cut point.Finally we prove the lemmas cited throughout. Lemma 4.
Suppose { Y ( β ); f βγ : γ, β ∈ Ω } is an inverse system whose limit Y has points p = ( p β ) and q = ( q β ) and a = ( a β ) . Suppose the set (cid:8) β ∈ Ω : [ p β , a β , q β ] (cid:9) is cofinal. Then [ p, a, q ] . Proof.
First replace Ω with (cid:8) β ∈ Ω : a β ∈ [ p β , a β , q β ] (cid:9) hence assume each [ p β , a β , q β ] . Nowsuppose L ⊂ Y is a subcontinuum with { p, q } ⊂ L . Then each { p β , q β } ⊂ π β ( L ) and so a β ∈ π β ( L ) since π β ( L ) is a subcontinuum. Now recall L = lim ←−{ π β ( L ); f βγ : γ, β ∈ Ω } . Since each a β ∈ π β ( L ) we have a ∈ L . Since L ⊂ Y is arbitrary we conclude [ p, a, q ] . Corollary 4.2.
Suppose { Y ( β ); f βγ : γ, β ∈ Ω } is an inverse system whose limit Y has points p = ( p β ) and q = ( q β ) . Suppose (cid:8) β ∈ Ω : Y ( β ) = [ p β , q β ] (cid:9) is cofinal. Then Y = [ p, q ] . Lemma 5.
Suppose { X ( β ); f βγ : γ, β ∈ Ω } is an inverse system with points a = ( a β ) and b = ( b β ) . Each π γ (cid:0) [ a, b ] (cid:1) = S (cid:8) f βγ (cid:0) [ a β , b β ] (cid:1) : β > γ (cid:9) . Proof.
Write J γ = S (cid:8) f βγ (cid:0) [ a β , b β ] (cid:1) : β > γ (cid:9) . We know [ a, b ] has the form lim ←−{ I β ; f βγ : γ, β ∈ Ω } for some subcontinua I β ⊂ X ( β ) . Since each I β contains { a β , b β } it contains [ a β , b β ] . Bycommutativity each I γ = f βγ ( I β ) contains f βγ (cid:0) [ a β , b β ] (cid:1) for β > γ . hence I γ contains J γ and by closure contains J γ .By commutativity each f βγ maps J β onto J γ . By continuity f βγ maps J β onto J γ . Hence J =lim ←−{ J β ; f βγ : γ, β ∈ Ω } is a well defined subcontinuum containing { a, b } . We have shown it isminimal with respect to containing { a, b } . We conclude J = [ a, b ] and each I β = J β as required. We have already determined the weak cut structure of the nontrivial composant. Namely c ∈ X − E is the only point to not weakly cut its composant. Daron Anderson 21 Indecomposable Strong Non-Cut his section examines the trivial composant E ⊂ X . We show E is weakly cut by its everypoint. Hence the continuum e X from Theorem 1 has exactly one strong non-cut point.To that end recall E is the set of eventually constant ω -sequences. The first claim is that thesubcontinua of E share the property of being eventually constant . Claim 20.
Suppose the subcontinuum L ⊂ E meets both X ( β ) and X − X ( β + 1) for some β < ω .Then X ( β + 1) ⊂ L . Proof.
By assumption L meets X ( α ) − X ( β + 1) for some α > β + 1 . By hereditary unicoherence X ( α ) ∩ L is a subcontinuum. Since X ( α ) ∩ L = (cid:0) ( X ( α ) ∩ L ) ∩ X ( β + 1) (cid:1) ∪ (cid:0) ( X ( α ) ∩ L ) − X ( β + 1) (cid:1) we know f αβ +1 (cid:0) X ( α ) ∩ L (cid:1) equals f αβ +1 (cid:0) ( X ( α ) ∩ L ) ∩ X ( β + 1) (cid:1) ∪ f αβ +1 (cid:0) ( X ( α ) ∩ L ) − X ( β + 1) (cid:1) . Since the map is a retraction the first summand equals L ∩ X ( β + 1) which meets X ( β ) by assumption and hence meets D ( β + 1) . By (b) the second summand is contained in C ( β + 1) . Thus the subcontinuum f αβ +1 ( X ( α ) ∩ L ) ⊂ X ( β + 1) meets the distinct composants C ( β + 1) and D ( β + 1) hence equals X ( β + 1) . Since the second summand is contained in C ( β + 1) the firstsummand L ∩ X ( β + 1) must contain D ( β + 1) . Since L ∩ X ( β + 1) is closed and D ( β + 1) dense we have L ∩ X ( β + 1) = X ( β + 1) and so X ( β + 1) ⊂ L . One consequence of Claim 20 is any subcontinuum that meets all X ( α ) must contain all X ( α ) and hence equal X . The next claim follows. Claim 21.
Each subcontinuum of E is contained in some X ( α ) . Claim 22.
Each point of E weakly cuts its composant. Proof.
Let b ∈ E be arbitrary. Then b ∈ X ( β ) for some β < ω . Choose any a ∈ X ( β ) − b and c ∈ X ( β + 2) − X ( β + 1) . By Claim 20 each subcontinuum that includes a and c must contain X ( β + 1) . Hence the subcontinuum contains X ( β ) and includes b . We conclude [ a, b, c ] . Theorem 6.
There exists a Bellamy continuum with exactly one strong non-cut point.
Proof.
Let X be the continuum from Theorem 1 and X → e X the quotient map that treats { x, y } as the single point z ∈ e X . We have already shown e c ∈ e X is a strong non-cut point. Now suppose b ∈ e X − e c . For b = z we have [ r, b, s ] for each r ∈ f κ ( x ) − z and s ∈ f κ ( y ) − z . Daron Anderson 22 Indecomposable Strong Non-Cut ow assume b = z . Then b = e d for some unique d ∈ X − { x, y } . The composant κ ( d ) is oneof E or X − E . In the first case Claim 22 says [ r, d, s ] for some pair r, s ∈ κ ( d ) . In the second caseLemma 2 says the same.Boundary bumping implies each continuum component of X − d contains a nondegeneratesubcontinuum hence has infinitely many points. That means we can reselect r and s outside { x, y } if necessary. Hence e r, e s = z . We claim [ e r, b, e s ] thus b weakly cuts e X .Suppose to reach a contradiction { e r, e s } ⊂ L ⊂ e X − b for some sub- continuum L ⊂ e X . Clearly z ∈ L as otherwise L = e P for some subcontinuum P ⊂ X − { x, y } . Then P contradicts how [ r, d, s ] .Let C r and C s be the continuum components of e r and e s in L − z respectively. Then C r = e D r and C s = e D s for semicontinua D r , D d ⊂ X − { x, y } . It follows D r , D s ⊂ κ ( d ) − { x, y } . Boundarybumping says z ∈ C r and z ∈ C s . The definition of the quotient topology implies D r and D s meet { x, y } . Without loss of generality κ ( d ) = κ ( x ) = E . Recall D r and D s are mapped into L which is proper. Continuity of the quotient says D r and D s are mapped into L hence proper. We cannot have y ∈ D r as then the proper subcontinuum D r meets both composants of X . Likewise y / ∈ D s . We conclude both x ∈ D r and x ∈ D s . Hence D r ∪ D s is a subcontinuum of X . Since X is indecomposable each summand is nowhere dense, and the same follows for the union. Thus the subcontinuum D r ∪ D s contradicts how [ r, d, s ] .We conclude no such subcontinuum L ⊂ e X exists and therefore [ e r, b, e s ] . Acknowledgements
This research was supported by the Irish Research Council Postgraduate Scholarship Scheme grantnumber GOIPG/2015/2744. The author would like to thank Professor Paul Bankston and DoctorAisling McCluskey for their help in preparing the manuscript.
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