Separable Indecomposable Continuum with Exactly One Composant
aa r X i v : . [ m a t h . GN ] J u l Separable Indecomposable Continuum withExactly One Composant
Daron Anderson Trinity College Dublin. [email protected] Preprint September 2018
Abstract
Indecomposable continua with one composant are large in the sense of being non-metrisable. We adapt the method of Smith [18] to construct an example which is small in the sense of being separable.
By a
Bellamy continuum we mean an indecomposable Hausdorff continuum with exactly one composant. To date there are only two known classes of Bellamy continua. Examples of the firstclass are Stone- ˇCech remainders of certain locally-compact spaces. The remainder is a continuum and under certain set-theoretic assumptions [3, 5, 8] it has exactly one composant.Examples of the second class are those obtained by Bellamy and Smith by carefully construct- ing an ω -chain of metric continua and retractions, so the inverse limit has exactly two composants. They then select one point from each composant, and identify the two points to get exactly onecomposant [4, 17, 18, 19].It is well-known that each Bellamy continuum is large in the sense of being non-metrisable[14]. This raises the question of whether a Bellamy continuum can be small in the sense ofbeing separable. The first class mentioned above provides no examples. Indeed Corollary 5.6of [20] says the Stone- ˇCech remainder of any well-behaved locally-compact Hausdorff space is ℵ -cellular hence non-separable.In this paper we modify the inverse-system of Smith [18] to produce a separable Bellamy continuum in the second class.Section 3 contains preliminary results about separability of certain inverse limits of metriccontinua. Section 4 applies the results to get an inverse system of metric continua and retractions. The modification itself is minor and only needed to make the limit separable − it is inessential Daron Anderson 1 Separable Bellamy Continuum o showing the limit has exactly two composants. We obtain a separable Bellamy continuum by spot-welding as usual.Section 5 shows both composants are non-metrisable. One composant is separable and theother is non-separable. Hence our example is separable but not hereditarily separable. The prob-lem is open whether there exists an hereditarily separable Bellamy continuum.Section 6 extends familiar results about Bellamy continua by showing each hereditarily uni-coherent metric continuum is a retract of a separable Bellamy continuum. The problem is openwhether the same holds for each separable continuum. Section 7 shows our modification is nec-essary, as the original limit of Smith is non separable.
Throughout X is a continuum. That means a nondegenerate compact and connected Hausdorff space. For background on metric continua see [11] and [14]. The results cited here have analagous proofs for non-metric continua. We call X separable to mean it has a dense countable subset, hereditarily separable to mean every subset is separable, and dense-hereditarily separable to mean every dense subset is separa- ble. Metric continua are separable but the converse fails in general. For each cardinal α we say X is α - cellular to mean it admits a family of α -many pairwise disjoint open subsets. Clearly each ℵ -cellular continuum is non-separable. 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 [11]. One corollary of boundary bumping is that the point p ∈ X is in the closure of each continuum component of X − p . Throughout all maps between continua are assumed to be continuous. We call f : Y → X a retraction to mean X is a subspace of the continuum Y and the restriction of f to X is the identity.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.For b ∈ X we omit the curly braces and write X − b instead of X − { b } without confusion.For a, b ∈ X we say X is irreducible about { a, b } to mean no proper subcontinuum of X contains { a, b } . For a, b ∈ X we write [ a, b ] for the intersection of all subcontinua that contain { a, b } . Note Daron Anderson 2 Separable Bellamy Continuum a, b ] is not in general connected as the interval notation suggests. Clearly h (cid:0) [ a, b ] (cid:1) = [ h ( a ) , h ( b )] for each a, b ∈ X and homeomorphism h : X → Y .We say X is indecomposable to mean it is not the union of two proper subcontinua. Equiv-alently each proper subcontinuum is nowhere dense. The composant κ ( x ) of the point x ∈ X is the union of all proper subcontinua that have x as an element. Indecomposable metric con-tinua are partitioned into c many pairwise disjoint composants [13]. In case κ ( x ) = κ ( y ) then X is irreducible about { x, y } . There exist indecomposable non-metric continua with exactly onecomposant [4] henceforth called Bellamy continua .We say X is hereditarily unicoherent to mean the intersection of any two subcontinua of X isempty or connected. Equivalently each interval [ a, b ] is a continuum. It then follows [ a, b ] is theunique subcontinuum of X irreducible about { a, b } . aG I Figure 1: The sin(1 /x ) continuum Throughout the sin(1 /x ) continuum is the metric continuum defined as the union of the graph G = (cid:8)(cid:0) x, sin(1 /x ) (cid:1) : − π/ ≤ x < (cid:9) and the arc I = { } × [ − , . The composant κ ( a ) of theendpoint a = ( − π/ , − is equal to G and its every nondegenerate subcontinuum is equal to the closure of its interior.The proper subcontinuum R ⊂ X is called a rung to mean each other subcontinuum K ⊂ X is either disjoint from, contained in, or contains R . For example the limiting arc I is a rung of the sin(1 /x ) continuum. By a ladder on X we mean a nested collection of rungs of X with Daron Anderson 3 Separable Bellamy Continuum ense union. Continua that admit ladders are rare. For example the sin(1 /x ) continuum admits noladders. Note what we call rungs are sometimes called terminal subcontinua , but that term alsohas several unrelated meanings across continuum theory [6].Throughout ω = { , , , . . . } is the first infinite ordinal and ω the first uncountable ordinal.Each proper initial segment of ω is countable and each countable subset has an upper bound.Each countable ordinal has a cofinal subset order-isomorphic to ω . For the ordered set Ω we say Ψ ⊂ Ω is cofinal to mean it has no upper bound in Ω .The poset Ω is called directed to mean for each γ, β ∈ Ω there is α ∈ Ω with γ, β ≤ α . Notemost authors require γ, β < α . The stronger condition prohibits Ω having a top element. In thispaper it is convenient to allow a directed set to have a top element.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 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 for all β ≤ α o .The functions f αβ are called the bonding maps . Write π β : T → T ( β ) for the restriction ofthe projection Q α T ( α ) → T ( β ) . If each bonding map is surjective so is each π β and we call the inverse system (limit) surjective . In case Ω has top element ∞ the inverse limit is a copy of T ( ∞ ) .The inverse limit of a system of continua is a continuum. We use transfinite recursion to construct the eponymous indecomposable continuum as the limitof a system { X ( α ); f αβ : α, β < ω } of metric continua and retractions. This section shows how toconstruct each X ( β + 1) from X ( β ) . The following section deals with limit ordinals.To begin let X (0) be the sin(1 /x ) continuum. Write a for the endpoint and select a sequence q n in X (0) with x -coordinates strictly increasing to from below. Observe q n satisfies the following definition of being a thick half-tail . Definition 3.1.
For X a continuum we define a half-tail at a ∈ X as a sequence q n ⊂ X with theproperties: Daron Anderson 4 Separable Bellamy Continuum [ a, q ] [ a, q ] . . . (2) S (cid:8) [ a, q n ] : n ∈ N (cid:9) = κ ( a ) (3) For each x ∈ X and n ∈ N either [ a, q n ] ⊂ [ a, x ] or [ a, x ] ⊂ [ a, q n ] .Moreover the half-tail q n is called thick to mean each [ a, q n ] is the closure of its interior.Next use induction to find a family D = { D , D , . . . } of pairwise disjoint subsets of κ ( a ) with each D n ⊂ [ a , q n ] dense. The fact that q n is thick implies that D is a tailing family asdefined below. Definition 3.2.
Suppose the continuum X has a half-tail q n at a ∈ X . By a tailing family on X we mean a pairwise disjoint collection D = { D , D , . . . } of countable subsets of X with each D n ⊂ [ a, q n ] and D n ∩ [ a, q m ] dense in [ a, q m ] for each m ≤ n .The notions of a half-tail and tailing family are pivotal to our example. Indeed as part of theconstruction we will at stage α < ω choose a half-tail q nα at a α ∈ X ( α ) and a tailing family D α on X ( α ) so both objectss behave nicely with respect to the bonding maps. This is made precisebelow. 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 3.3.
Suppose the continua X ( β ) and X ( γ ) have half-tails q nβ and q nγ at a β ∈ X ( β ) and a γ ∈ X ( γ ) and tailing families D β and D γ respectively. The map f : X ( β ) → X ( γ ) is called coherent to mean a β a γ and [ a β , q nβ ] → [ a γ , r nγ ] and f induces bijections D nβ → D nγ for each n ∈ N . The system { X ( β ); f βγ : γ, β < α } is called coherent to mean each bonding map iscoherent.At stage α < ω we have already constructed the coherent inverse system { X ( β ); f βγ : β, γ < α } of hereditarily unicoherent metric continua and retractions. We assume the following objects havebeen specified for each β < α : (i) Half-tails q nβ at a β ∈ X ( β ) (ii) Tailing families D β = { D β , D β , . . . } on X ( β ) We also assume for each γ, δ < α the two conditions hold:
Daron Anderson 5 Separable Bellamy Continuum a) S { X ( δ ) : δ < γ } ⊂ X ( γ ) − κ ( a γ ) (b) S (cid:8) f γδ (cid:0) X ( γ ) − X ( δ ) (cid:1) : α > γ > δ (cid:9) = κ ( a δ ) Conditions (a) and (b) come straight from [4] and will ensure the limit has exactly two com-posants. Coherence will ultimately be used to construct a tailing family on the inverse limit. Oncewe have shown the limit is a Bellamy continuum, the next lemma gives our main result.
Lemma 3.4.
Suppose X admits a half-tail q n and tailing family D = { D , D , . . . } . Then X isseparable. Proof.
Clearly S D = D ∪ D ∪ . . . has the cardinality of N × N which is well known to becountable. Now suppose U ⊂ X is open. Since κ ( a ) is dense it meets U . Since κ ( a ) = S n [ a, q n ] some [ a, q n ] meets U . Since [ a, q n ] ∩ U is open in [ a, q n ] it contains an element of D n by the definition of a tailing family. We conclude S D is dense in X .We are now ready to begin the successor step. Suppose α = β +1 is a successor ordinal. We willconstruct the hereditarily unicoherent 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 specify the objects (i) and (ii) when β is replaced by β + 1 . Finally we will check the enlarged system is coherent and Conditions (a) and (b) hold for all γ, δ ≤ β + 1 .To begin the construction of X ( β + 1) consider the following subset N of X ( β ) × [0 , . N = (cid:0) S (cid:8) [ a β , q nβ ] × { / (2 n − , / n } : n ∈ N (cid:9)(cid:1) ∪ (cid:0) X ( β ) × { } (cid:1) . Define the points b , b , b , . . . ∈ N . b n = (cid:0) a β , / (2 n + 1) (cid:1) b n +1 = (cid:0) q nβ , / (2 n + 1) (cid:1) b n +2 = (cid:0) q nβ , / (2 n + 2) (cid:1) b n +3 = (cid:0) a β , / (2 n + 2) (cid:1) To obtain X ( β + 1) make for each n ∈ N the identification b n +1 ∼ b n +2 . Call that point c n +1 ∈ X ( β + 1) . Then make the identification b n +3 ∼ b n +1) . Call that point c n +1) ∈ X ( β + 1) .Write J (2 n ) for the quotient space of each [ a β , q nβ ] × { / (2 n + 1) } and J (2 n + 1) for the quotientspace of [ a β , q nβ ] × { / n } . Observe each J ( n ) is irreducible from c n to c n +1 . Clearly the quotient space of S { J ( n ) : n ∈ N } is connected and its closure is the union with X ( β ) × { } . Therefore X ( β + 1) is connected. Identify X ( β ) with the subspace X ( β ) × { } . Theprojection X ( β ) × [0 , → X ( β ) respects the identifications and therefore induces a retraction f β +1 β : X ( β + 1) → X ( β ) . Daron Anderson 6 Separable Bellamy Continuum β +1 = ( a β , q β ,
1) ( a β , / q β , /
2) ( a β , / q β , / q β q β a β J (0) J (1) J (2) J (3) X ( β ) Figure 2: Schematic for X ( β + 1) . Dashed lines indicate identifications. The bonding map f β +1 β projectsto the right. Claim 1. X ( β + 1) is compact metric. Proof.
Let P be the partition induced on N by the identifications. Each partition element is either a singleton or doubleton hence closed. Let P ∈ P be contained in the open U ⊂ N . We claim some open neighborhood W of P is a union of partition elements. Then V = U ∩ W witnesseshow P is upper semicontinuous and X ( β + 1) is compact metric by [14] Lemma 3.2.For P a singleton of some J (2 n ) take W = J (2 n ) − { b n , b n +1 } . For P a singleton of some J (2 n + 1) take W = J (2 n + 1) − { b n +2 , b n +3 } . For P a singleton of X ( β ) observe U ∩ X ( β ) is anopen neighborhood of P in X ( β ) . Since F = f β +1 β respects P the open set W = F − (cid:0) U ∩ X ( β ) (cid:1) is a union of partition elements.For P a doubleton without loss of generality some n ∈ N has P = (cid:8) b n +1 , b n +2 (cid:9) . Since J (2 n ) − b n and J (2 n + 1) − b n +3 are open in N so is the union W = (cid:0) J (2 n ) − b n (cid:1) ∪ (cid:0) J (2 n +1) − b n +3 (cid:1) . The case for P = (cid:8) b n +3 , b n +1) (cid:9) is similar. Claim 2.
The composant κ ( a β +1 ) = X ( β +1) − X ( β ) . Hence Condition (a) holds for all γ, δ ≤ β +1 . Proof.
Let n ∈ N be arbitrary and U ⊂ J ( n ) − { c n , c n +1 } open. Since J ( n ) is irreducible from Daron Anderson 7 Separable Bellamy Continuum n to c n +1 boundary bumping implies J ( n ) − U = A ∪ B is the disjoint union of two nonemptyclopen sets that include c n and c n +1 respectively. It follows X ( β + 1) − U = (cid:0) J (1) ∪ . . . J ( n − ∪ A (cid:1) ∪ (cid:0) B ∪ J ( n + 1) ∪ . . . ∪ X ( β ) (cid:1) is a disjoint union of two clopen sets that include a β +1 and contain X ( β ) respectively.Now suppose the subcontinuum K connects a β +1 to X ( β ) . Observe for each n ∈ N the set O = J (0) ∪ J (1) ∪ . . . ∪ J ( n − − c n is clopen in X ( β + 1) − c n and has a β +1 ∈ O . We concludeall c n ∈ K .For K proper it excludes some open U ⊂ J ( n ) − { c n , c n +1 } . Then the two clopen sets from thefirst paragraph contradict how K is connected. We conclude κ ( a β +1 ) ⊂ X ( β + 1) − X ( β ) . Theother inclusion is witnessed by the subcontinua J (1) ∪ J (2) ∪ . . . ∪ J ( n ) .Recall each J ( n ) is hereditarily unicoherent. For n ≤ m it is straightforward to prove by induction each subcontinuum I ( n, m ) = J ( n ) ∪ . . . ∪ J ( m ) is hereditarily unicoherent. Claim 3.
Each subcontinuum of κ ( a β +1 ) is contained in some I ( n, m ) . Proof.
Suppose K ⊂ κ ( a β +1 ) is a proper subcontinuum. Without loss of generality assume a β +1 ∈ K . We claim K meets only finitely many J ( n ) . For otherwise there is a sequence n (1) , n (2) , . . . with n ( i ) → ∞ and elements x i ∈ K ∩ J ( n ( i )) . Consider the sequence f β +1 β ( x ) , f β +1 β ( x ) , . . . in X ( β ) . Since X ( β ) is compact metric f β +1 β ( x i ) has a subsequence tending to some x ∈ X ( β ) . It follows x i has a subsequence tending to ( x, . Since K is closed it includes ( x, ∈ X ( β ) . But then Claim 2 says K = X ( β + 1) and so K κ ( a β +1 ) contrary to assumption. Claim 4. X ( β + 1) is hereditarily unicoherent. Proof.
Suppose K and L are proper subcontinua. For K and L contained in κ ( a β +1 ) Claim 3says K ∪ L ⊂ I ( n, m ) for some n ≤ m . Since I ( n, m ) is hereditarily unicoherent K ∩ L is emptyor connected. For K, L ⊂ X ( β ) then K ∩ L is empty or connected since X ( β ) is hereditarilyunicoherent.Now suppose K meets κ ( a β +1 ) and X ( β + 1) − κ ( a β +1 ) = X ( β ) . We claim K = K ′ ∪ J ( n + 1) ∪ . . . ∪ X ( β ) for some n ∈ N and subcontinuum K ′ ⊂ J ( n ) with c n +1 ∈ K ′ . To that end let n ∈ N beminimal with K ∩ J ( n ) = ∅ . Then K ∩ J ( n −
1) = ∅ and so c n / ∈ K . Observe the subcontinuum I (0 , n ) ∪ K connects a β +1 to X ( β ) . Hence I (0 , n ) ∪ K = X ( β + 1) by Claim 2 and K contains Daron Anderson 8 Separable Bellamy Continuum ( n + 1) ∪ . . . ∪ X ( β ) . Since O = J (0) ∪ J (1) ∪ . . . ∪ J ( n ) − c n +1 is clopen in X ( β + 1) − c n +1 anddisjoint from X ( β ) we must have c n +1 ∈ K .Suppose K ∩ J ( n ) = A ∪ B is the disjoint union of two nonempty closed sets with c n +1 ∈ A .Since K ∩ J ( n ) is closed in X ( β + 1) so are A and B . Observe C = J ( n + 1) ∪ J ( n + 2) ∪ . . . ∪ X ( β ) is closed and C ∩ J ( n ) = { c n +1 } . Thus K = (cid:0) ( C ∩ K ) ∪ A (cid:1) ∪ B is a disjoint union of nonemptyclosed sets. We conclude K ∩ J ( n ) is connected.Thus K = K ′ ∪ J ( n + 1) ∪ . . . ∪ X ( β ) and L = L ′ ∪ J ( m + 1) ∪ . . . ∪ X ( β ) for some m, n ∈ N andsubcontinua K ′ ⊂ J ( n ) and L ′ ⊂ J ( m ) . For m ≤ n we have K ∩ L = L is connected. For n ≤ m we have K ∩ L = K is connected. For n = m we have K ∩ L = ( L ′ ∩ K ′ ) ∪ J ( m + 1) ∪ . . . ∪ X ( β ) which is a subcontinuum by hereditary unicoherence of J ( n ) . Claim 5.
Condition (b) holds for all γ, δ ≤ β + 1 . Proof.
By induction each
S (cid:8) f γδ (cid:0) X ( γ ) − X ( δ ) (cid:1) : β + 1 > γ > δ (cid:9) = κ ( a δ ) . Thus we can factor theset S (cid:8) f γδ (cid:0) X ( γ ) − X ( δ ) (cid:1) : β + 1 ≥ γ > δ (cid:9) as the union f β +1 δ (cid:0) X ( β + 1) − X ( δ ) (cid:1) ∪ κ ( a δ ) . So it is enough to show the second factor is contained in κ ( a δ ) . To that end write f β +1 δ (cid:0) X ( β + 1) − X ( δ ) (cid:1) = f β +1 δ (cid:0) X ( β + 1) − X ( β ) (cid:1) ∪ f β +1 δ (cid:0) X ( β ) − X ( δ ) (cid:1) By definition f β +1 δ = f βδ ◦ f β +1 β . Since f β +1 β is a retraction the second term equals f βδ (cid:0) X ( β ) − X ( δ ) (cid:1) which is contained in κ ( a δ ) by Condition (b) for earlier stages.Claim 2 says X ( β + 1) − X ( β ) = κ ( a β +1 ) . Since q nβ +1 is a half-tail we have κ ( a β +1 ) = S n [ a β +1 , q nβ +1 ] by Property (2). Hence the first term can be written f β +1 δ (cid:0) κ ( a β +1 ) (cid:1) = f β +1 δ (cid:0) S n [ a β +1 , q nβ +1 ] (cid:1) = S n f β +1 δ (cid:0) [ a β +1 , q nβ +1 ] (cid:1) which equals S n [ a δ , q nδ ] = κ ( a δ ) by coherence of the bonding maps and Property (2) at stage δ respectively. Claim 6.
The sequence q nβ +1 = c n − is a half-tail at a β +1 . Proof.
It is straightforward to verify J (0) ∪ J (1) ∪ . . . ∪ J (2 n − is a subcontinuum irreduciblefrom a β +1 to c n − . By hereditary unicoherence that subcontinuum is [ a β +1 , c n − ] . Since each c n +1 / ∈ [ a β +1 , c n − ] we have q n +1 β +1 / ∈ [ a β +1 , q nβ +1 ] which is Property (1). To prove Property (2) observe S n [ a β +1 , q nβ +1 ] = J (1) ∪ J (2) ∪ . . . = κ ( a β +1 ) by Claim 2.To prove Property (3) suppose x ∈ X ( β + 1) − [ a β +1 , q nβ +1 ] . Observe [ a β +1 , c n − ] − c n − isclopen in X ( β + 1) − c n − . Thus the continuum [ a β +1 , x ] includes the point c n − . By Zorn’s Daron Anderson 9 Separable Bellamy Continuum emma the continuum [ a β +1 , x ] contains a subcontinuum irreducible from a β +1 to c n − . Since [ a β +1 , c n − ] is the only such subcontinuum we have [ a β +1 , c n − ] ⊂ [ a β +1 , x ] and so [ a β +1 , q nβ +1 ] ⊂ [ a β +1 , x ] .For the successor continua in our system we need to use how the half-tail is thick. This holdsonly for successor stages. We will later see the limit continua are indecomposable. Hence theirevery subcontinuum has void interior and they cannnot admit a thick half-tail. Claim 7.
The half-tail q nβ +1 is thick. Proof.
We claim each J ( n ) ◦ contains the dense open subset J ( n ) −{ c n , c n +1 } . Hence [ a β +1 , q nβ +1 ] ◦ contains the dense subset J (1) ◦ ∪ J (2) ◦ ∪ . . . ∪ J (2 n − ◦ . Let x ∈ J ( n ) − { c n , c n +1 } be arbitrary. Recall J ( n ) is a copy of some [ a β , q mβ ] with x corresponding to some y ∈ [ a β , q mβ ] and each of c n , c n +1 corresponding to exactly one of a β or q mβ . Choose open U ⊂ X ( β ) with y ∈ U ⊂ X ( β ) − { a β , q mβ } and positive ε < min n (cid:12)(cid:12) (cid:12) m − m + 1 (cid:12)(cid:12) (cid:12) , (cid:12)(cid:12) (cid:12) m − m − (cid:12)(cid:12) (cid:12)o . Observe U × (cid:16) m − ε, m + ε (cid:17) is an open subset of N is disjoint from { b , b , b , . . . } . Hence it corresponds to an open neighborhood of x in X ( β + 1) . The choice of ε ensures it is contained in J ( n ) − { c n , c n +1 } . Claim 8.
There exists a tailing family D β +1 = { D β +1 , D β +1 , . . . } on X ( β + 1) that makes f β +1 β coherent. Proof.
We first construct the sets D Mβ +1 . For now let M ∈ N be fixed. Since X ( β +1) is metric there is a countable basis U , U . . . for [ a β +1 , q Mβ +1 ] . By the first paragraph U meets some J ( n ) ⊂ [ b, r M ] that maps homeomorphically onto [ a, q m ] for some m ≤ M . Since U ∩ J ( n ) is open in J ( n ) theimage f β +1 β ( U ) ∩ [ a β , q nβ ] is open in [ a β , q nβ ] . Since m ≤ M the definition of a tailing family says D Mβ ∩ [ a β , q mβ ] is dense in [ a β , q mβ ] . Thus f β +1 β ( U ) ∩ [ a β , q mβ ] contains infinitely many elements of D Mβ ∩ [ a β , q mβ ] . Choose one such d (1) ∈ D Mβ ∩ [ a β , q nβ ] and select c (1) ∈ U with c (1) d (1) .Proceed by induction. At stage r we have chosen distinct c ( i ) ∈ U i for i = 1 , , . . . , r − and d ( i ) = f (cid:0) ( c ( i ) (cid:1) are distinct. Just like before f β +1 β ( U r ) includes infinitely many elements of D Mβ .Select some d ( r ) ∈ f ( U r ) ∩ D Mβ with d (1) , d (2) , . . . , d ( r ) distinct. Then select c ( r ) ∈ U r with c ( r ) d ( r ) . Daron Anderson 10 Separable Bellamy Continuum y construction we get a countable dense subset E Mβ +1 = { c (1) , c (2) , . . . } of distinct elementsof [ a β +1 , q Mβ +1 ] . For each d ∈ D M − { d (1) , d (2) , . . . } use surjectivity to select c ∈ [ a β +1 , q Mβ +1 ] with c d . Adjoin all such d to E Mβ +1 to get the set D Mβ +1 . By construction f β +1 β induces a bijection D Mβ +1 → D Mβ .Now let M vary over N . We get a family D β +1 = { D β +1 , D β +1 , . . . } of countable subsets of X ( β + 1) with each f β +1 β : D Mβ +1 → D Mβ a bijection and D Mβ +1 ⊂ [ a β +1 , q Mβ +1 ] dense. Since theelements of D β are pairwise disjoint so are the elements of D β +1 . Claim 7 shows each [ a β +1 , q mβ +1 ] is the closure of its interior. From this it follows D Mβ +1 ∩ [ a β +1 , q mβ +1 ] is dense in [ a β +1 , q mβ +1 ] whenever m ≤ M . We conclude D β +1 is a tailing family.To show coherence first recall by construction J (0) is a copy of [ a β , q β ] and projects under f β +1 β onto that copy. Since a β +1 ∈ J (0) corresponds to the point a β ∈ [ a β , q β ] we have a β +1 a β . Recall we define q nβ +1 = c n − and by definition c n − ∼ (cid:0) q nβ , / (2 n − (cid:1) . Since f β +1 β isinduced by the projection X ( β ) × [0 , → X ( β ) we have q nβ +1 q nβ . Finally recall [ a β +1 , c n − ] = J (0) ∪ J (1) ∪ . . . ∪ J (2 n − . By cooherence this set maps onto [ a β , q β ] ∪ [ a β , q β ] ∪ . . . ∪ [ a β , q nβ ] .By Property (1) for β the image equals [ a β , q nβ ] as required. 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 hereditarily unicoherent metric continua.For all β, γ, δ < α we assume the objects (i) and (ii) from Section 4 have been specified andConditions (a) and (b) 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 ) .That X ( α ) is hereditarily unicoherent follows from a straightforward modification of [7] Corol- lary 1. To see X ( α ) is metric we observe by definition α is a countable ordinal. The product Q β<α X ( β ) of countably many metric spaces is itself a metric space. The inverse limit X ( α ) isby definition a subset of that product and therefore a metric space It remains to show the enlarged system is coherent; to check Conditions (a) and (b) hold for
Daron Anderson 11 Separable Bellamy Continuum he enlarged system; and to specify the data (i) and (ii) for X ( α ) . By coherence at earlier stageseach a β a γ and q nβ q nγ . Hence there are well defined points a α = ( a β ) β<α and q nα = ( q nβ ) β<α in the limit X ( α ) with a α a β and q nα q nβ . For ease of notation write a and q n instead of a α and q nα respectively.For the proof that q n is a half-tail we refer to [2] where we introduce the more complicatednotion of a tail and study inverse limits of tails. The proof for half-tails follows from a closereading of the proofs of [2] Claims ?? and ?? and Lemma ?? . Hence we have the next claim. Claim 9.
The sequence q n is a half-tail at a . Moreover for each n ∈ N we have [ a, q n ] =lim ←− (cid:8) [ a β , q nβ ]; f βγ (cid:9) . Recall the proper subcontinuum R ⊂ X is called a rung to mean each other subcontinuum K ⊂ X is either disjoint from, contained in, or contains R . By a ladder on X we mean a nestedcollection of rungs of X with dense union. The proof that X ( α ) is indecomposable will followfrom the existence of a ladder. Lemma 4.1.
Suppose X admits a ladder. Then X is indecomposable. Proof.
We first show rungs have void interior. Suppose otherwise the rung R ⊂ X has R ◦ = ∅ . Let b ∈ X − R be arbitrary and C the component of b in X − R . Then b witnesses how C R . Atthe same time C ⊂ X − R which implies C ⊂ X − R ◦ which in turn implies R C . Since R is a rung the subcontinua C and R must be disjoint. But this contradicts boundary bumping which says says C meets R . We conclude each R ◦ = ∅ . Now suppose L is a ladder on X and the proper subcontinuum L ⊂ X has nonvoid interior.Since S L is dense some rung R ∈ L meets L ◦ and X − L hence contains L . Since R has void interior so does L . We conclude each subcontinuum of X has void interior. This is equivalent to X being indecomposable.Since our simplest example of a rung is the limiting arc of the sin(1 /x ) continuum, and each X ( β + 1) looks like a sin(1 /x ) continuum limiting to X ( β ) , the next claim should be unsurprising. Claim 10.
Each X ( γ ) is a rung of X ( α ) . Proof.
The proof uses transfinite induction. Suppose for some e α ≤ α and all γ ≤ β < e α that X ( γ ) ⊂ X ( β ) is a rung. Now suppose the continuum K ⊂ X ( e α ) meets X ( γ ) and X ( e α ) − X ( γ ) .For e α = e β + 1 a successor ordinal there are two possibilities. First that K ⊂ X ( e β ) . In that case K ⊂ X ( e β ) meets X ( γ ) and X ( e β ) − X ( γ ) and we have X ( γ ) ⊂ K since X ( γ ) ⊂ X ( e β ) is a rung by Daron Anderson 12 Separable Bellamy Continuum nduction. Second that K meets X ( e β + 1) − X ( e β ) . It follows from Claim 2 that X ( e β ) ⊂ X ( e β + 1) is a rung. Hence X ( e β ) ⊂ K and X ( γ ) ⊂ K by Property (a) at stage e β .For e α a limit ordinal there are again two possibilities. First that K ⊂ X ( β ) for some β < e α .In that case X ( γ ) ⊂ X ( β ) is a rung by induction hence X ( γ ) ⊂ K as required. Second that K iscontained in no X ( β ) . In that case recall K = lim ←−{ f e αδ ( K ); f βδ : β, δ < e α } .If there was β < e α with f e αδ ( K ) ⊂ X ( β ) for all δ > β we would have K ⊂ X ( β ) contrary toassumption. Thus for each β < e α there is δ > β with f e αδ ( K ) X ( β ) . By induction X ( β ) ⊂ X ( δ ) is a rung. Hence the subcontinuum f e αδ ( K ) of X ( δ ) contains X ( β ) and its subset X ( γ ) . Bycommutativity f e αδ ′ ( K ) = f δδ ′ ◦ f e αδ ( K ) also contains X ( γ ) whenever γ ≤ δ ′ ≤ δ . In particularwhenever γ ≤ δ ′ ≤ β . Since β < e α is arbitrary we get X ( γ ) ⊂ f e αδ ′ ( K ) for all γ ≤ δ ′ < e α . It follows X ( γ ) ⊂ K as required. Claim 11. X ( α ) is indecomposable. Proof.
Recall we define X ( γ ) = (cid:8) ( x β ) ∈ X ( α ) : x β = x γ ∀ β > γ (cid:9) . By Lemma 4.1 it is enough toshow { X ( β ) : β < α } is a ladder in X ( α ) . To that end let γ < α be arbitrary and U ⊂ X ( γ ) open. We must show π − γ ( U ) meets some X ( β ) . Let x γ ∈ U be arbitrary and consider the following ( x β ) ∈ X ( α ) . For β ≥ γ define x β = x γ . We have f βγ ( x β ) = x γ since the bonding map is a retraction. For β ≤ γ define x β = f γβ ( x γ ) . It follows ( x β ) is a well defined element of X ( α ) . By definition ( x α ) ∈ X ( γ ) ∩ π − γ ( U ) . Claim 12.
Condition (a) holds for all γ, δ ≤ α . Proof.
By induction we only need to prove the case γ = α . We must show κ ( a α ) is disjoint fromeach X ( β ) . Since q n is a half-tail at a each x ∈ κ ( a ) is an element of some [ a, q n ] = lim ←− [ a β , q nβ ] . Thatmeans x β +1 ∈ [ a β +1 , q nβ +1 ] for each β < α . Since q nβ +1 is a half-tail at a β +1 we have [ a β +1 , q nβ +1 ] ⊂ κ ( a β +1 ) which is disjoint from X ( β ) by Condition (a) for β + 1 . We conclude x β +1 ∈ X ( β + 1) − X ( β ) . Since x β ∈ X ( β ) we have x β +1 = x β hence x / ∈ X ( β ) by definition of the embedding X ( β ) → X ( α ) . Claim 13.
Condition (b) holds for all γ, δ ≤ α . Proof.
By induction and commutativity it is enough to show f αδ (cid:0) X ( α ) − X ( δ ) (cid:1) = κ ( a δ ) for each δ ≤ α . Claim 12 says each X ( β ) is disjoint from κ ( a α ) . Hence X ( α ) − X ( β ) contains κ ( a α ) which maps onto κ ( a β ) by coherence. Daron Anderson 13 Separable Bellamy Continuum laim 14.
There exists a tailing family D α = { D α , D α , . . . } on X ( α ) that makes f αβ coherent foreach β < α . Proof.
It is clear that a a β and q n q nβ . Claim 9 says each [ a, q n ] = lim ←− (cid:8) [ a β , q nβ ]; f βγ (cid:9) hence [ a, q n ] [ a β , q nβ ] .By induction each f βγ is coherent hence induces bijections D nβ → D nγ . That means the inverselimit D nα = lim ←− (cid:8) D nβ ; f βγ (cid:9) is a well defined countable subset of X ( α ) . Lemma 2.5.9 of [9] says therestrictions f αβ : D nα → D nβ are bijective. Claim 9 says each D nα ⊂ [ a, q n ] .To see D α , D α , . . . are pairwise disjoint take x ∈ D nα and y ∈ D mα for some m = n . By definition f α ( x ) ∈ D n and f α ( y ) ∈ D m . Since D is a tailing family D m and D n are disjoint and the result follows. To show D α = { D α , D α , . . . } is a tailing family it remains to prove each D nα ∩ [ a, q m ] is densein [ a, q m ] whenever m ≤ n . To that end Claim 7 says the half-tail q nβ is thick whenever β < α isa successor ordinal. Since α is a limit ordinal the set Γ = { β < α : β is a successor ordinal } is cofinal in α . Hence we can write X ( α ) = lim ←−{ X ( β ); f βγ : β, γ ∈ Γ } where each X ( β ) comes witha distinguished thick half-tail. Recall π β = f αβ and the open subsets of [ a, q m ] = lim ←− [ a β , q mβ ] have the form π − β ( U ) for open U ⊂ [ a β , q mβ ] . By thickness [ a β , q mβ ] ◦ is dense in [ a β , q mβ ] . Hence V = U ∩ [ a β , q mβ ] ◦ is a nonemptyopen subset of X ( β ) . Since [ a β , q mβ ] ⊂ [ a β , q nβ ] we see V is open in [ a β , q nβ ] as well. Since D nβ is dense in [ a β , q nβ ] there is d ∈ V ∩ D nβ . Since D nβ = π β ( D nα ) we have π β ( x ) = d for some x ∈ D nα .But then π β ( x ) ∈ V ⊂ U and so x ∈ π − β ( U ) as required.Finally we have the main example. Theorem 1.
There exists a separable Bellamy continuum.
Proof.
By transfinite recursion we have a coherent system of metric continua { X ( α ); f αβ : β, α <ω } . Let X be the inverse limit. Recall this section assumes α ≤ ω is a limit ordinal. For thespecial case α = ω we have X = X ( α ) and Claim 14 says X admits a tailing family. Lemma 3.4 then says X is separable.Once we have expressed X as the inverse limit of a system of indecomposable metric continua,Conditions (a) and (b) and Theorem 1 of [4] will say X is indecomposable with at most twocomposants. The trivial composant E ⊂ X is the set S { X ( α ) : α < ω } of eventually constant Daron Anderson 14 Separable Bellamy Continuum - sequences. The nontrivial composant X − E is equal to the limit lim ←−{ κ ( a α ); f αβ : β, α < ω } .In this case the trivial composant contains the point ( a α ) α<ω hence is nonempty by construction.Claim 11 says X ( α ) is indecomposable whenever α < ω is a limit ordinal. We claim Γ = { α < ω : α is a limit ordinal } is cofinal in ω . Thus we can write X = lim ←−{ X ( α ); f αβ : α, β ∈ Γ } as the inverse limit of a system of indecomposable metric continua.To that end let β < ω be arbitrary and observe β × ω is well-ordered under ( δ, m ) ≤ ( γ, n ) ⇐⇒ (cid:0) m < n or m = n and δ ≤ γ ) . Hence β × ω is isomorphic to some ordinal e β . It is clear β × ω hasno top element hence e β is a limit ordinal. Since the initial segment β × { } is a copy of β we have β ≤ e β and since β × ω is countable we have e β < α .We conclude X has exactly two composants. Let the Bellamy continuum e X be obtained by choosing any x ∈ E and y ∈ X − E and identifying x ∼ y . Since e X is the image of X under the quotient map it is separable. We make some observations on the global properties of the inverse limit X from Section 4. Thefirst is the tailing family D = { D , D , . . . } on X has each D n contained in the nontrivial composant lim ←−{ κ ( a α ); f αβ : β, α < ω } . Thus we have the following. Claim 15.
The nontrivial composant of X is separable. Thus the nontrivial composant can be considered small . On the other hand Lemma 5.1 shows the trivial composant is large . Lemma 5.1.
Suppose the topological space T is the union of an ω -chain of proper closed subsets.Then T is non-separable. Proof.
Let B = { B ( α ) : α < ω } be a chain of proper closed subsets of T . Suppose D = { d , d , . . . } ⊂ T is dense. For each n ∈ N there exists α ( n ) < ω with d n ∈ B ( α ( n )) . The countablesubset { α ( n ) : n ∈ N } has an upper bound α < ω . Since B is a chain we have { d , d , . . . } ⊂ B ( α ) .Since B ( α ) is closed and proper D is not dense. Hence T is non-separable. The trivial composant of X is the union of the ω -chain { X ( α ) : α < ω } of proper subcontinua.Thus Lemma 5.1 gives the next two claims. Claim 16.
The trivial composant of X is non-separable. Daron Anderson 15 Separable Bellamy Continuum laim 17.
The continuum X is separable but not hereditarily separable and not dense-hereditarilyseparable.This suggests two open problems. Question 1.
Is the nontrivial composant of X hereditarily separable or dense-hereditarily separa-ble? Question 2.
Does there exist an hereditarily separable or dense-hereditarily separable Bellamycontinuum?The Introduction mentions all known Bellamy continua are Stone- ˇCech remainders, or arisefrom inverse-limits constructions similar to Section 4. For a positive answer to Question 2 weimagine an entirely new class of examples would be needed to obviate Lemma 5.1.Our next result is that neither composant of X is metrisable. Lemma 5.2 is similar to Lemma Lemma 5.2.
No metric space is the union of an ω -chain of compact proper subsets. Proof.
First recall [15] Theorem IV.5 (F) says compactness and sequential compactness are equiv-alent for metric spaces. Now suppose the metric space M is the union of the chain K = { K ( α ) : α < ω } of compact proper subsets. We claim M is compact.Suppose x , x , . . . is an arbitrary sequence in M . For each n ∈ N there exists α ( n ) < ω with x n ∈ K ( α ( n )) . The countable subset { α ( n ) : n ∈ N } has an upper bound α < ω . Since K is a chain we have { x , x , . . . } ⊂ K ( α ) . Since K ( α ) is compact metric there is a point x ∈ K ( α ) anda subsequence ( y n ) of ( x n ) with y n → x . We conclude M is sequentially compact hence compact. Since M is compact metric it admits a countable base U , U , . . . of open sets. Since M = S K each U n ∈ N meets K ( β ( n )) for some β ( n ) < ω . The countable subset { β ( n ) : n ∈ N } has anupper bound β < ω . Since K ( β ) meets all U n it is dense. Since K ( β ) is compact and M metric K ( β ) is closed. Thus K ( β ) = M contradicting the assumption that each K ( β ) is proper.The trivial composant of X is the union of the ω -chain S { X ( α ) : α < ω } of proper subcon-tinua. Thus Lemma 5.2 gives the next claim. Claim 18.
The trivial composant of X is non-metrisable.Next we prove the same for the nontrivial composant. Claim 19.
The nontrivial composant of X is non-metrisable. Daron Anderson 16 Separable Bellamy Continuum roof.
Clearly the nontrivial composant X − E is connected. Claim 15 says X − E is separable.Lemma 3 of [12] says X − E is strongly indecomposable as defined in [12] Section 2. For acontradiction suppose X − E is metrisable.Let the X → e X be the quotient map from Theorem 1. Since X − E is homeomorphic to itsimage we know e X is a compactification of X − E . Then [12] Theorem 8 says e X is irreduciblebetween some pair of points { a, b } . But that means a and b have different composants contradictinghow e X is a Bellamy continuum.To close the section we remark that metrisability cannot be droped from the hypothesis ofLemma 5.2. Example 5.8 of [1] is the closed unit ball in the Hilbert space ℓ ( ω ) of square-summable functions ω → R under the weak topology. In fact ℓ ( ω ) is even a continuum and theelements of the ω -chain are nowhere dense subcontinua. Bellamy [4] has shown each metric continuum is a retract of a Bellamy continuum. We use a sim- ilar technique to show each hereditarily unicoherent metric continuum is a retract of a separable
Bellamy continuum.For our example we took X (0) the sin(1 /x ) continuum. There is no obstruction to using instead any hereditarily unicoherent metric continuum Y . In case Y admits thick half-tail q n at the point a ∈ Y we can simply take Y = X (0) as the bottom element of the system { X ( α ); f αβ : α, β < ω } from Section 3.Writing X for the limit we see the projection π : X → X (0) is a retraction. Therefore each π ( x ) is a point of the trivial composant. Let the separable Bellamy continuum e X be obtained by choosing some x in the nontrivial composant and identifying x ∼ π ( x ) . Clearly π induces aretraction e X → Y from a separable Bellamy continuum.In case Y has no such tail, we must first build an hereditarily unicoherent metric continuum X (0) with a thick half-tail q n at the point a ∈ X (0) and a retraction X (0) → Y . We then take X (0) as the bottom element of the inverse system and compose the retractions e X → X (0) and X (0) → Y to get the desired retraction onto e X → Y . It remains to construct such a continuum X (0) . Lemma 6.1.
Suppose the metric continuum Y is hereditarily unicoherent. There exists a retrac- tion X (0) → Y from an hereditarily unicoherent metric continuum X (0) with a thick half-tail q n Daron Anderson 17 Separable Bellamy Continuum t the point a ∈ X (0) . Proof.
Choose a countable dense subset D = { d (1) , d (2) , . . . } of Y . For the sequence s = (1 , , , , , , , , , , . . . ) define each p n = d ( s ( n )) . Define the closed subset N ⊂ Y × [0 , . N = (cid:16) [ n ∈ N [ p , p n ] × { /n } (cid:17) ∪ (cid:0) Y × { } (cid:1) .To obtain X (0) from N first make for each odd n ∈ N the identification (cid:0) p n , /n (cid:1) ∼ (cid:0) p n +1 , / ( n +1) (cid:1) . Then make for each even n ∈ N the ident- ification (cid:0) p , /n (cid:1) ∼ (cid:0) p , / ( n + 1) (cid:1) .The picture for X (0) is similar to Figure 2. Then X (0) is a metric space as a subset of the prod-uct Y × [0 , of metric spaces. It is straightforward to show the first summand of N is connected with closure X (0) . Since Y × [0 , is compact so is X (0) . Identify Y with the subset Y × { } of X (0) . A similar argument to Section 3 shows X (0) is hereditarily unicoherent and q n = ( p n , /n ) is a thick half-tail at a = ( p , Clearly the projection N → Y onto the first coordinate respects the partition. Hence it induces a retraction X (0) → Y .The theorem follows. Theorem 2.
Each hereditarily unicoherent metric continuum is a retract of a separable Bellamy continuum.
There are several directions one might generalise Theorem 2. The first is to drop the reference to hereditary unicoherence.
Question 3.
Is each metric continuum a retract of a separable Bellamy continuum?
The methods in Sections 3 and 4 rely heavily on hereditary unicoherence and do not generalise.One special case that seems approachable is when Y is arcwise connected.If we take the proof of Lemma 6.1 and replace each [ p , p n ] with some arc from p to p n theresulting space X (0) is a spiral over Y . That means the composant κ ( a ) of Y − X (0) of a is anopen ray. So while X (0) is not itself hereditarily unicoherent, we see any two subcontinua of κ ( a ) have connected intersection. Since the maps f α map X ( α ) − X (0) into κ ( a ) it seems likely one could adapt our construction while paying very close attention to where hereditary unicoherence isused, and thus get a retractions from separably Bellamy continua onto arcwise connected continua.In particular [14] Theorem 8.23 says this includes all locally connected continua. One might also try to replace the hypothesis of Y being metrisable with merely being separable. Daron Anderson 18 Separable Bellamy Continuum uestion 4.
Is each separable hereditarily unicoherent continuum a retract of a separable Bellamycontinuum?Secions 3 and 4 do not generalise to answer Question 4. This is because in the non-metricrealm we might encounter the following type of subset.
Definition 6.2.
The dense subset D of the topological space T is called resolvable to mean it isthe disjoint union of two dense subsets. Otherwise we call the subset irresolvable .Irresolvable sets are pathological objects that never occur in metric continua. Lemma 6.3.
Every dense subset of a metric continuum is resolvable.
Proof.
Suppose X is a metric continuum. We first show each nonvoid open subset U ⊂ X isuncountable. Corollary 5.5 of [14] says U contains a proper subcontinuum K . Then Urysohn’slemma (see [16] Theorem III.2) says K surjects onto [0 , hence is uncountable. We conclude U is uncountable.Now suppose X is metric and D ⊂ X dense. Since finite subsets are closed we see D is infinite.Let U , U , . . . be a countable basis for X . Since each U i is infinite we can use induction to selectdistinct elements a , b , a , b , . . . ∈ X with each a i , b i ∈ U i . By construction A = { a , a , . . . } and B = { b , b , . . . } are dense. Since D (2) = D − A contains B it is also dense. For D (1) = A we geta disjoint union D = D (1) ∪ D (2) into disjoint dense subsets.On the other hand Theorem 4.1 of [10] shows the construction of a countable irresolvable subset of the non-metric separable continuum [0 , c . The following lemma will be useful. Lemma 6.4.
Suppose the subsets D and F of the continuum X differ by only finitely many ele- ments. Then neither or both of D and F are resolvable. Proof.
We claim D is resolvable if and only if A = D ∪ F is resolvable. First suppose D = D (1) ∪ D (2) is a disjoint union of dense subsets. For A (1) = D (1) and A (2) = D (2) ∪ ( A − D ) theequality A = A (1) ∪ A (2) witnesses how A is resolvable.Now suppose A = A (1) ∪ A (2) is a disjoint union of dense subsets. Clearly D (1) = A (1) ∩ D and D (2) = A (2) ∩ D are disjoint with union D . By assumption there are finite subsets C (1) , C (2) ⊂ X with D (1) = A (1) − C (1) and D (2) = A (2) − C (2) . Since nonvoid open subsets of X are infinite we see C (1) ◦ = C (2) ◦ = ∅ . Thus D (1) = A (1) − C (1) = A (1) − C (1) ◦ = A (1) = X .Hence D (1) is dense and likewise for D (2) . We conclude D is resolvable. The same proofshows F is resolvable if and only if A = F ∪ D is resolvable. We conclude D is resolvable if and only if F is resolvable. Daron Anderson 19 Separable Bellamy Continuum enceforth suppose the Hausdorff continuum X ( β +1) has a thick half-tail q nβ +1 at a β +1 ∈ X ( β + 1) and D β +1 = { D β +1 , D β +1 , . . . } is a tailing family on X ( β + 1) . Let the Hausdorff continuum X ( β + 2) and thick half-tail q nβ +2 at a β +2 ∈ X ( β + 2) and bonding map f β +2 β +1 : X ( β + 2) → X ( β + 1) be as constructed in Section 3. The next claim shows the obstruction to choosing an appropriate D β +2 on X ( β + 2) . Claim 20.
Suppose X ( β +2) admits a tailing family that makes f β +2 β +1 coherent. Then [ a β +1 , q β +1 ] ∩ D β +1 is resolvable as a subset of [ a β +1 , q β +1 ] . Proof.
Suppose the tailing family D β +2 = { D β +2 , D β +2 , . . . } makes f β +2 β +1 densely coherent. De- fine the subset F = D β +2 −{ c , c } of X ( β + 2) . Recall the subcontinua J (0) , J (1) , J (2) ⊂ X ( β +2) have dense interior and J (0) ∩ J (1) = { c } J (1) ∩ J (2) = { c } J (0) ∩ J (2) = ∅ . Therefore F ∩ J (0) , F ∩ J (1) and F ∩ J (2) are pairwise disjoint and dense in J (0) , J (1) and J (2) respectively. Recall the bonding map f β +2 β +1 induces projections J (0) → [ a β +1 , q β +1 ] and J (1) → [ a β +1 , q β +1 ] and J (2) → [ a β +1 , q β +1 ] . Therefore A = f β +2 β +1 (cid:0) F ∩ J (0) (cid:1) and B = f β +2 β +1 (cid:0) F ∩ J (1) (cid:1) are dense in [ a β +1 , q β +1 ] and f β +2 β +1 (cid:0) F ∩ J (2) (cid:1) is dense in [ a β +1 , q β +1 ] . Since [ a β +1 , q β +1 ] ⊂ [ a β +1 , q β +1 ] has dense interior we see C = [ a β +1 , q β +1 ] ∩ f β +2 β +1 (cid:0) F ∩ J (2) (cid:1) is also dense in [ a β +1 , q β +1 ] . By assumption f β +2 β +1 induces a bijection D β +2 → D β +1 and therefore A , B and C are pairwisedisjoint. Since F is contained in J (0) ∪ J (1) ∪ J (2) we see A ∪ B ∪ C = [ a β +1 , q β +1 ] ∩ f β +2 β +1 ( F ) .We conclude [ a β +1 , q β +1 ] ∩ f β +2 β +1 ( F ) is a resolvable subset of [ a β +1 , q β +1 ] .By definition F differs from D β +2 by only finitely many elements. Thus f β +2 β +1 ( F ) differs from f β +2 β +1 (cid:0) D β +2 (cid:1) = D β +1 by only finitely many elements. The same holds taking intersections with [ a β +1 , q β +1 ] . The result then follows from Lemma 6.4.Claim 20 says there is no way in general to define a suitable tailing family on X ( β + 2) . For suppose X ( β + 1) and [ a β +1 , q β +1 ] are non-metric. There is no guarantee a given dense subsetof [ a β +1 , q β +1 ] is resolvable. Thus a positive answer to Question 3 would go beyond the methods here. Daron Anderson 20 Separable Bellamy Continuum e close with the remark that irresolvable sets can be generated by trying to extend the system { X ( α ); f αβ : α, β < ω } of continua from Sections 3 and 4 beyond ω .Suppose X = X ( ω ) and D ω are constructed and X ( ω + 1) is as defined in Section 3. Forbrevity write a = a ω and q n = q nω and D = D ω . Suppose all countable dense subsets of each [ a, q n ] are resolvable. We give only the construction of D ω +1 ⊂ [ a ω +1 , q ω +1 ] = J (0) ∪ J (1) ∪ J (2) as the construction of D nω +1 is analogous.First we split [ a, q ] ∩ D = A ∪ B ∪ C into three pairwise disjoint dense subsets. Recall J (0) and J (1) are copies of [ a, q ] and J (2) a copy of [ a, q ] . Let A ′ ⊂ J (0) and B ′ ⊂ J (1) be the densesets corresponding to A and B and C ′ ⊂ J (2) the dense set corresponding to D − ( A ∪ B ) . Define D ω +1 = A ′ ∪ B ′ ∪ C ′ . In case all relevant countable dense subsets of [ a ω +1 , q nω +1 ] are resolvable we define X ( ω + 2) and D ω +2 similarly. Proceeding under the same assumption we continue to extend the system. Theorem 2 says we need no further assumptions to extend to limit ordinals. We claim we cannotextend to { X ( α ); f αβ : α, β < Ω } for any | Ω | > c . For then Theorem 2 says the limit X (Ω) is separable. But at the same time Corollary 1.12 of[20] says every separable compactum is the image of β N hence has at most | β N | = 2 c points. But since the chain of subcontinua X ( α ) ⊂ X (Ω) is strictly increasing X (Ω) has cardinality at least | Ω | > c which is a contradiction. We conclude the process terminates at { X ( α ); f αβ : α, β < η } for some ordinal η with | η | ≤ c . Thus some relevant countable dense subset D of some [ a η , q nη ] is irresolvable.Closer inspection of the proposed construction of each D nη +1 reveals some such D is built outof D Nη for some n ≤ N by successively splitting sets into two dense disjoint halves or taking intersections with [ a η , q mη ] for some n ≤ m ≤ N . The fundamental difference between our construction and that of Smith [18] is the use of identi-fications. We obtain X ( β + 1) from the set N = (cid:16) [ n ∈ N [ p , p n ] × { /n } (cid:17) ∪ (cid:0) Y × { } (cid:1) . by making for each n ∈ N the identifications (cid:0) q nβ , / (2 n − (cid:1) ∼ (cid:0) q nβ , / n (cid:1) and (cid:0) a β , / n (cid:1) ∼ Daron Anderson 21 Separable Bellamy Continuum a β , / (2 n + 1) (cid:1) . Smith instead obtains X ( β + 1) from N by attaching horizontal arcs from (cid:0) q nβ , / (2 n − (cid:1) to (cid:0) q nβ , / n (cid:1) and from (cid:0) a β , / n (cid:1) to (cid:0) a β , / (2 n + 1) (cid:1) as in Figure 3. a β +1 = ( a β , q β ,
1) ( a β , / q β , /
2) ( a β , / q β , / q β q β a β J (0) J (1) J (2) J (3) X ( β ) Figure 3: Schematic of the new X ( β + 1) . The arcs I β +1 ( n ) are shown in red. Suppose as in Section 4 we define each limit X ( α ) as the inverse limit of all previous X ( β ) .However, unlike before, we define each successor X ( β + 1) as the union of the three subspaces of X ( β ) × [0 , . (cid:0) S (cid:8) [ a β , q nβ ] × { / (2 n − , / n } : n ∈ N (cid:9)(cid:1) ∪ (cid:0) X ( β ) × { } (cid:1)S (cid:8) { q nβ } × [1 / n, / (2 n − n ∈ N (cid:9)S (cid:8) { a β } × [1 / (2 n + 1) , / n ] : n ∈ N (cid:9) For each n ∈ N define the arcs I β +1 (2 n −
1) = { q nβ } × [1 / n, / (2 n − and I β +1 (2 n ) = { a β } × [1 / (2 n + 1) , / n ] . Claim 21.
The new limit X is ℵ -cellular hence non-separable. Proof.
Let each V α +1 ⊂ X ( α ) be the interior of I α +1 (2) . Observe f α +1 α ( V α +1 ) = { a α } thus each f α +1 β ( V α +1 ) = { a β } for β < α + 1 . We claim the open sets U α +1 = π − α +1 ( V α +1 ) of X are pairwise Daron Anderson 22 Separable Bellamy Continuum isjoint. For suppose x ∈ U α +1 . Then by definition x α +1 ∈ V α +1 and by commutativity x β +1 ∈ f α +1 β +1 ( V α +1 ) which equals { a β +1 } . Thus x β +1 = a β +1 . But since a β +1 / ∈ V β +1 we must have x / ∈ U β +1 for each β + 1 < α + 1 .Since ω is a limit ordinal the map by β β + 1 from ω to itself is well defined and injective.Thus { U α +1 : α < ω } has the same cardinality as ω namely ℵ . We conclude X is ℵ -cellular.The proof for the example of Smith [18] is similar, with the interiors of the sets a α × [1 / , (page 594) playing the role of V α +1 and the points α (page 595) playing the role of a α . Acknowledgements
This research was supported by the Irish Research Council Postgraduate Scholarship Scheme grant number 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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