The Shore Point Existence Problem is Equivalent to the Non-Block Point Existence Problem
aa r X i v : . [ m a t h . GN ] J u l The Shore Point Existence Problem is Equivalent to theNon-Block Point Existence Problem
Daron Anderson Trinity College Dublin. [email protected] Preprint February 2019
Abstract
We prove the three propositions are equivalent: ( a ) Every Hausdorff continuum has two or more shore points. ( b ) Every Hausdorff continuum has two or more non-block points. ( c ) Every Haus-dorff continuum is coastal at each point. Thus it is consistent that all three properties fail. We also give the following characterisation of shore points: The point p of the continuum X is a shorepoint if and only if there is a net of subcontinua in { K ∈ C ( X ) : K ⊂ κ ( p ) − p } tending to X in the Vietoris topology. This contrasts with the standard characterisation which only demands the net elements be contained in X − p . In addition we prove every point of an indecomposable continuum is a shore point. Leonel [11] has improved the classic non-cut point theorem of Moore [13] by showing every metric continuum has two or more shore points. Bobok, Pyrih and Vejnar [6] observed Leonel’stwo shore points have the stronger property of being non-block points.
In [2] the author proved it is consistent the result fails to generalise to Hausdorff continua.
Under Near Coherence of Filters (NCF) the Stone- ˇCech remainder H ∗ of the half-line lacks non-block points and hence lacks coastal points.This left open the question of whether there is a consistent example of a Hausdorff continuumwithout shore points. This paper gives a positive answer. Indeed we show the shore point andnon-block point existence problems are equivalent. They are also equivalent to a number of otherproblems involving shore, non-block, and coastal points of Hausdorff continua. We also prove every shore point p ∈ X has the stronger property of being a proper shore point .That means there is a net of subcontinua in the hyperspace { K ∈ C ( X ) : K ⊂ κ ( p ) − p } tending to X in the Vietoris topology. This is not apparent from the definition of a shore point, which only requires the net elements be contained in X − p . Daron Anderson 1 Shore and Non-Block Points
Terminology and Notation
For sets A and B define A − B = { a ∈ A : a / ∈ B } . For B = { b } we write A − b without confusion.For A ⊂ B we do not presume A is a proper subset of B . For a subset S ⊂ X denote by S ◦ and S the interior and closure of S respectively. The boundary of S means the set ∂S = S ∩ ( X − S ) .Throughout X is a continuum. That is to say a nondegenerate compact connected Hausdorffspace. For background on metric continua see [10] and [14]. The results cited here have analagousproofs for non-metric continua.Throughout all maps are assumed to be continuous. The map f : X → Y of continua is called monotone to mean f − ( y ) ⊂ X is connected for each y ∈ Y . Theorem 6.1.28 of [8] says moreover f − ( K ) ⊂ X is a continuum for each subcontinuum K ⊂ Y .For a, b ∈ X we call X irreducible about { a, b } to mean { a, b } is not contained in a propersubcontinuum of X . The subspace A ⊂ X is called a semicontinuum to mean for each a, c ∈ A some subcontinuum K ⊂ A has { a, c } ⊂ K . Every subspace A ⊂ X is partitioned into maximal semicontinua called the continuum components of A . For N ≥ we say the subcontinua X , . . . , X N ⊂ X form a decomposition and write X ⊕ . . . ⊕ X N to mean X ∪ . . . ∪ X N = X and no X n is contained in the union of the others. We call X decomposable to mean it admits a decomposition and indecomposable otherwise. The latter is equivalent to admitting no decomposition with N = 2 and equivalent to each proper subcontinuum being nowhere dense. We say X is hereditarily indecomposable to mean its every subcontinuumis indecomposable. Equivalently each pair of subcontinua are either disjoint or nested.The composant κ ( x ) of the point x ∈ X is the union of all proper subcontinua that have x as an element. Indecomposable metric continua are partitioned into c many pairwise disjointcomposants [12]. In case κ ( x ) = κ ( y ) then X is irreducible about { x, y } . By boundary bumping we mean the principle that, for each proper closed E ⊂ X , each compo-nent C of E meets the boundary ∂E = E ∩ X − E . For the non-metric proof see §
47, III Theorem 2of [10]. One corollary of boundary bumping is that any p ∈ X is in the closure of each continuumcomponent of X − p .Throughout C ( X ) is the set of subcontinua of X . We call p ∈ X a shore point to mean for eachfinite collection of open sets U , U , . . . U n ⊂ X some subcontinuum K ⊂ X − p meets each U m . This is equivalent to some net in { K ∈ C ( X ) : K ⊂ X − p } tending to X in the Vietoris topology.We do not need the full definition of the Vietoris topology here.We call p ∈ X a cut point to mean X − p is disconnected and a non-cut point otherwise. Clearly each shore point is non-cut. We call p ∈ X a non-block point to mean X − p has a dense continuum Daron Anderson 2 Shore and Non-Block Points omponent. Every non-block point is a shore point but the converse fails in general [6]. We call x ∈ X a coastal point to mean x is an element of some proper dense semicontinuum. Clearly X has a non-block point if and only if it has a coastal point.Theorem 5 of [3] says every point of a metric continuum is coastal. The result generalisesto separable continua [1] but not to Hausdorff continua. Under the set-theoretic axiom Near Co-herence of Filters the non-metric continuum H ∗ lacks non-block points and hence lacks coastalpoints [2]. Definition 3.1.
The continuum X is called partially coastal to mean there are points x, y ∈ X with x coastal and y non-coastal. Theorem 1.
The following propositions are equivalent.(1) There exists a continuum without coastal points.(2) There exists a continuum with exactly one non-coastal point.(3) There exists a partially coastal continuum.(4) There exists a continuum without shore points.(5) There exists a continuum with exactly one shore point(6) There exists a continuum without non-block points.(7) There exists a continuum with exactly one non-block point.
Proof. (6) = ⇒ (1) because X has a non-block point if and only if it has a coastal point. (2) = ⇒ (3) follows from how every continuum has more than one point. (4) = ⇒ (6) follows from howevery non-block point is a shore point. (1) = ⇒ (2) : Suppose X has no coastal points. Let [0 , be an arc and the continuum Y beobtained by gluing ∈ [0 , to any fixed p ∈ X . Identify X and [0 , with their images in Y . Thefact each x ∈ Y − is coastal is witnessed by the dense semicontinuum S = (cid:0) S { (1 /n,
1] : n ∈ N } (cid:1) ∪ X = Y − .To see ∈ Y is non-coastal suppose otherwise. That means ∈ S ⊂ Y − q for some denseproper semicontinuum S ⊂ Y and point q ∈ Y . It is easy to see [0 , ⊂ S . Hence q / ∈ [0 , and q ∈ X − p . Daron Anderson 3 Shore and Non-Block Points et Q : Y → X be the monotone map that collapses [0 , to the point p ∈ X and leaves thepoints of X fixed. Then Q ( S ) ⊂ X is a dense semicontinuum with p ∈ Q ( S ) ⊂ X − q contradictingthe assumption that p ∈ X is non-coastal. (1) = ⇒ (5) : Suppose X has no coastal points and define Y , Q and S as before. The densesemicontinuum S witnesses how ∈ Y is a non-block point hence a shore point.To see is the only shore point first observe each q ∈ (0 , is a cut point hence not a shore point.Now suppose some q ∈ Y − [0 , is a shore point. Fix the open set U = (0 , . By assumption, forevery collection of open sets U , . . . , U n ⊂ X − , some subcontinuum K ⊂ Y − q meets U and all U m . Since K meets U and U we have ∈ K and hence p ∈ Q ( K ) .Now allow { U , . . . , U n } to range over all finite collections of open subsets of X . The subcon-tinua Q ( K ) ⊂ X − q constitute a dense proper semicontinuum of X . Therefore p ∈ X is coastalcontrary to assumption. We conclude q is not a shore point as required. The proof of (1) = ⇒ (7) is identical. (3) = ⇒ (4) : Suppose some p ∈ X is non-coastal. Take two disjoint copies X and X of X . Let Y be the continuum obtained from X ⊔ X by identifying the points a ∈ X and b ∈ X corresponding to p ∈ X . Identify X and X with their images in Y and write p ∈ Y for the shared image of a and b . We claim Y has no shore points. Observe Y − p is homeomorphic to the disjoint union of X − a and X − b . Hence p is a cutpoint and not a shore point. Now suppose q ∈ Y − p is a shore point. Without loss of generality q ∈ X . Fix some open set U ⊂ X − p . For every collection of open sets U , . . . , U n ⊂ X − p some subcontinuum K ⊂ Y − q meets U and all U m . Since K meets U and U we must have p ∈ K . Moreover K − p is disconnected and each component lies in one of X or X .Let K be the family of components that lie in X . Boundary bumping says L ∪ { p } is a contin-uum for each L ∈ K . It follows K ∩ X = S (cid:8) { p } ∪ L : L ∈ K (cid:9) is a subcontinuum of X − q thatmeets all U , . . . , U n . Now allow { U , . . . , U n } to range over all finite collections of open subsetsof X . Then the union M of all K ∩ X is a dense semicontinuum of X − q . Let Q : Y → X bethat map that compresses X to the point b ∈ X . Then the semicontinuum Q ( M ) contradict how X is not coastal at b ∈ X . We conclude no q ∈ Y is a shore point as required. (5) = ⇒ (1) : Suppose p ∈ X is the unique shore point. We claim p is not coastal, For then p ∈ S ⊂ X − q for some q ∈ X and dense proper semicontinuum S . This implies q is a non-block point and hence a shore point contrary to assumption. This implies either (1) or (3) which wehave shown are equivalent. (7) = ⇒ (3) : Suppose p ∈ X is the unique non-block point. Since there is a non-block point there is also a coastal point. But p ∈ X cannot be coastal as this would imply some q ∈ X − p is Daron Anderson 4 Shore and Non-Block Points non-block point. hence X is partially coastal.Under the set-theoretic axiom Near Coherence of Filters (NCF) the Stone- ˇCech remainder H ∗ of the half-line is a continuum with no coastal points [2]. From this we get the corollary. Corollary 3.2.
Propositions (1) − (7) from Theorem 1 are consistent.For each continuum X with a non-coastal point p the author has shown [1] there is a sub-continuum M ⊂ X with p ∈ M ◦ such that X/M indecomposable and non- coastal at M . Thisraises the question whether (1) − (7) are equivalent if we demand the continuum in question beindecomposable.Section 5 shows (4) and (5) never hold for indecomposable continua. In particular Theorem 3 says every point of an indecomposable continua is a shore point. For X indecomposable the remaining propositions seem more difficult and the methods ofTheorem 1 no longer apply. For example if X is indecomposable with (1) we cannot simplyattach an arc to prove (2) as the quotient space is manifestly decomposable. In the other direction X having (7) does not imply the same for X/M . For example assume
NCF and glue any point p ∈ H ∗ to the endpoint ∈ [0 , . Let X be the quotient space. Clearlyfor X/M to be indecomposable M is the union of the arc and some proper subcontinuum K ⊂ H ∗ with p ∈ K . Since the map H ∗ → H ∗ /K is monotone it follows X/M ∼ = H ∗ /K lacks non-blockpoints and fails (7) . Before proceeding to Section 5 we take a diversion to prove all shore points are proper shore points. Some of the terminology and results from Section 4 will be needed for Section 5.
Definition 4.1.
Recall p ∈ X is called a shore point to mean for each finite collection of open sets U , U , . . . , U N ⊂ X some subcontinuum K ⊂ X − p meets each U n . We call p a proper shorepoint to mean K can always be chosen as a subset of κ ( p ) − p . Otherwise we call p a trivial shorepoint . Definition 4.2.
Recall p ∈ X is called a non-block point to mean X − p has a dense continuum component. We call p a proper non-block point to mean κ ( p ) − p has a dense continuum component.Otherwise we call p a trivial non-block point . Daron Anderson 5 Shore and Non-Block Points his section shows there exist both proper and trivial non-block points but there is no such thingas a trivial shore point. Thus being a proper non-block point is a meaningful notion but being aproper shore point is not. Hereditarily indecomposable metric continua provide an example ofthe former.
Lemma 4.3.
Suppose M is a hereditarily indecomposable metric continuum. Each p ∈ M is atrivial non-block point. Proof.
Recall M has c many pairwise disjoint composants each of which is a dense semicontinuum[12]. Each composant other than κ ( p ) witnesses how p is a non-block point. It remains to showeach continuum component C of κ ( p ) − p is nowhere dense.Let q ∈ C be arbitrary. We can write C = S C as a union of proper subcontinua with q ∈ D for each D ∈ C . Since q ∈ κ ( p ) there exists a proper subcontinuum K with { p, q } ⊂ K . Clearly K meets but is not contained in each D ∈ C . Hereditary indecomposability implies each D ⊂ K and therefore C = S C ⊂ K . Since X is indecomposable K and thus C is nowhere dense asrequired. Observe we only required M to have more than one composant. The problem is open whether there exists a hereditarily indecomposable Hausdorff continuum with exactly one composant. For obstructions to finding such a space see Smith [15, 16, 17, 18].For κ ( p ) = X clearly the shore point p is proper. Henceforth assume X is irreducible about some { p, q } . We first treat the case when X is decomposable. Lemma 4.4.
Suppose X is decomposable and irreducible about { p, q } . Then p and q are propernon-block (shore) points. Proof.
By assumption we can write X = A ∪ B as the union of two proper subcontinua. Since X is irreducible we have without loss of generality p ∈ A ⊂ X − q and and q ∈ B ⊂ X − p .Choose any x ∈ A ∩ B . It follows from boundary bumping that p is in the closure of thecontinuum component C of x in A − p . Likewise q is in the closure of the continuum component D of x in B − q . Then x ∈ C ∪ D ⊂ X − { p, q } and C ∪ D is continuumwise connected.Observe the subcontinuum C ∪ D contains { p, q } . By irreducibility we have C ∪ D = X hence C ∪ D is dense. To see C ∪ D ⊂ κ ( p ) recall p ∈ A ⊂ X − q hence A ∪ D ⊂ κ ( p ) . Therefore the subset C ∪ D of A ∪ D witnesses how p is a proper non-block point. By symmetry the same argumentapplies to q .We now deal with indecomposable X . The definition allows finer control over why a given p ∈ X fails to be a shore point. Daron Anderson 6 Shore and Non-Block Points efinition 4.5.
Suppose p ∈ X and U = { U , . . . , U n +1 } are open subsets of X . We say p disrupts U to mean no continuum component of κ ( p ) − p meets all U , . . . , U n +1 . We say p trivially disrupts U to mean there are distinct elements U , U , . . . , U n ∈ U and nonempty open sets V i ⊂ U i suchthat p disrupts { V , . . . V n } . Otherwise we say p properly disrupts U .For p to properly disrupt U = { U , U , . . . , U n } the definition requires U be pairwise disjoint.It is also clear the shore point p ∈ X being trivial is equivalent to disrupting some nondegeneratefamily. By boundary bumping no p ∈ X can disrupt a family with only one element. So if p disrupts U = { U , U } then p properly disrupts U .By induction it follows if p disrupts U = { U , U , . . . , U n } there exist m ≤ n and elements U , U , . . . , U m ∈ U and open sets V i ⊂ U i such that p properly disrupts { V , . . . V m } . The next lemmas take advantage of that fact. Lemma 4.6.
Suppose p ∈ X properly disrupts { U , . . . , U n } and let r ≤ n be fixed. There exists a family K ( r ) of subcontinua of κ ( p ) − p that satisfies both properties below.(1) Each K ∈ K ( r ) meets U m for each m = r . (2) S K ( r ) is dense in U m for each m = r . Proof.
Without loss of generality r = 1 . For each m = 1 let V m ⊂ U m be an arbitrary open subset.Since p properly disrupts { U , . . . , U n } it cannot disrupt the family { V , . . . , V n } . That means there is a subcontinuum K ⊂ κ ( p ) − p that meets each of V , . . . , V n . Now let V , . . . , V n range over the open subsets of U , . . . , U n . The union of all continua K is dense in each U m . Lemma 4.7.
Suppose X is indecomposable. Each shore point p ∈ X is a proper shore point. Proof.
Suppose to the contrary that p ∈ X is a trivial shore point. That means p disrupts somefamily U = { U , . . . U n } of open subsets. By discarding some elements of U if necessary we canassume p properly disrupts U . Let K (1) be a family of subcontinua of κ ( p ) − p as described inLemma 4.6.Each K ∈ K (1) has a continuum component C ( K ) in κ ( p ) − p . Since p disrupts U we know C ( K ) is disjoint from U . Recall C ( K ) is a subcontinuum. It follows from boundary bumping p ∈ C ( K ) ⊂ X − U .As K ranges over the elements of K (1) the set S = S (cid:8) C ( K ) : K ∈ K (1) (cid:9) constitutes asemicontinuum with p ∈ S ⊂ X − U . Therefore S ⊂ X − U is a proper subcontinuum. ButProperty (2) of Lemma 4.6 says U ⊂ S . Thus S ⊂ X is a proper subcontinuum with nonvoid interior. Since X is indecomposable this cannot occur. We conclude p is a proper shore point. Daron Anderson 7 Shore and Non-Block Points heorem 2 follows from Lemmas 4.4 and 4.7.
Theorem 2.
Shore points are the same as proper shore points.
The proof of Theorem 1 says we can build a continuum X without shore points by assuming NCFand gluing two copies of H ∗ together at a single point. This section is about whether arbitrarycontinua X without shore points come about this way − by joining together continua withoutnon-block points. We first show X is decomposable. Lemma 5.1.
Suppose p ∈ X is not a shore point. There is a decomposition X = X ⊕ . . . ⊕ X n with p ∈ X ∩ . . . ∩ X n . Proof.
Since p is not a shore point it disrupts some family U = { U , . . . U n } of open subsets. Like before, we can assume p properly disrupts U . For each r ≤ n let K ( r ) be a family of subcontinua of κ ( p ) − p as described in Lemma 4.6.The proof of Lemma 4.7 shows each K ( r ) = S K ( r ) is a subcontinuum with p ∈ K ( r ) ⊂ X − U r and U m ⊂ K ( r ) for each m = r . Let L ( r ) be the family of all subcontinua with both properties.Without loss of generality we can assume K ( r ) = S L ( r ) is the largest subcontinuum with bothproperties.Define the subcontinuum Y = K (1) ∪ K (2) ∪ . . . ∪ K ( n ) and write C ( x ) for the continuumcomponent of each x ∈ X − Y . We claim each C ( x ) meets no U r . For suppose otherwise. Then without loss of generality some C ( x ) meets U . Since p disrupts { U , . . . U n } we know C ( x ) failsto meet at least one U m . Without loss of generality m = 2 . Then C ( x ) is a subcontinuum with x ∈ C ( x ) ⊂ X − U .Now consider the subcontinuum K (2) ∪ C ( x ) . We have p ∈ K (2) ∪ C ( x ) ⊂ X − U and U m ⊂ K (2) ∪ C ( x ) for each m = 2 . Since we chose K (2) to be maximal with both propertieswe must have K (2) ∪ C ( x ) = K (2) . Hence C ( x ) ⊂ K (2) and x ∈ K (2) ⊂ Y contrary to theassumption that x ∈ X − Y . We conclude C ( x ) meets no U r .It follows for each x ∈ X − Y that C ( x ) is a subcontinuum with { p, x } ⊂ C ( x ) ⊂ X − ( U ∪ . . . ∪ U n ) . Define the subcontinuum C = S { C ( x ) : x ∈ X − Y } . Then { p, x } ⊂ C ⊂ X − ( U ∪ . . . ∪ U n ) .It follows X = C ∪ K (1) ∪ K (2) ∪ . . . ∪ K ( n ) and p is an element of each element of thecovering. By discarding any covering element contained in the union of the others we get the required decomposition. Daron Anderson 8 Shore and Non-Block Points he next theorem follows from Theorem 2 and Lemma 5.1.
Theorem 3.
Every point of an indecomposable continuum is a proper shore point.The proof of Theorem 3 for the special case of H ∗ uses standard techniques from the study ofthe Stone- ˇCech remainder. An earlier paper of ours gave a lengthy argument for H ∗ , that consid-ered separately two types of composants of H ∗ and did not use Theorem 2. That paper was neverpublished because the anonymous referee was able to give the much simpler proof we includehere as Lemma 5.2. The second half of the proof is based on [5] Theorem 4.1.We briefly recall the terminology from [2]: The Stone- ˇCech remainder H ∗ is the set of non-principal ultrafilters of closed sets on the half open interval H = [0 , ∞ ) . The topology on H ∗ isgenerated by the sets U ∗ = {D ∈ H ∗ : D ⊂ U for some D ∈ D} as U ranges over all open subsets of H . Note for U bounded we have U ∗ = ∅ . It is known that H ∗ is an indecomposable Hausdorff continuum.Suppose we have a nonprincipal ultrafilter D on ω and sequence of intervals I n = [ a n , b n ] with each b n < a n +1 . For each subset D ⊂ ω we write I D = S { I n : n ∈ D } for the subset of H . Wewrite I D for the closure in β H . That means the collection of ultrafilters D on H with I D ∈ D . We write I D for the subset T (cid:8) H ∗ ∩ I D : D ∈ D (cid:9) of H ∗ . Sets of the form I D are known to beproper subcontinua of H ∗ and are called standard subcontinua . For background on H ∗ see [9].For background on Stone- ˇCech compactifications in general see [7] and [19]. Lemma 5.2.
Let p ∈ H ∗ be arbitrary and U = { U , . . . , U n } a family of open sets. For each standard subcontinuum I D with p ∈ I D some other standard subcontinuum J D ⊂ κ ( p ) has p / ∈ J D but J D meets each U m . In particular each point of H ∗ is a shore point. Proof.
We only consider n = 2 and U = { U , U } as the general case is similar. First choosedisjoint open U, V ⊂ H with the basic open sets U ∗ ⊂ U and V ∗ ⊂ U . Then observe U = A ∪ A ∪ . . . is a disjoint union of open intervals and likewise for V = B ∪ B ∪ . . . .Choose an increasing sequence c < c < . . . in H so each [ c n , c n +1 ] contains at least one A i and B j . It follows c n → ∞ . Define sequences R n = [ c n , c n +1 ] and L n = [ c n +2 , c n +3 ] . By construction R D and L D are disjoint and each meets U ∗ and V ∗ . Therefore p is an element of atmost one of R D and L D . Without loss of generality p / ∈ R D .Taking J D = R D it remains to show J D ⊂ κ ( p ) . In case J D meets I D the fact that H ∗ isindecomposable says J D ∪ I D = H ∗ hence J D ⊂ κ ( p ) as required. Otherwise I D and J D are Daron Anderson 9 Shore and Non-Block Points isjoint. Recall I D = T (cid:8) H ∗ ∩ I D : D ∈ D (cid:9) and J D = T (cid:8) H ∗ ∩ J D : D ∈ D (cid:9) are intersections ofclosed subsets of the compact H ∗ . Since they are disjoint the Cantor property says H ∗ ∩ I A and H ∗ ∩ J B are disjoint for some A, B ∈ D .It follows from the definition of I A and I B that I A ∩ J B ⊂ H is compact. Since A ∩ B ∈ D and I A ∩ B ∩ J A ∩ B ⊂ I A ∩ J B we also see I A ∩ B ∩ J A ∩ B is compact. That means the set C = { n ∈ A ∩ B : I n meets J m for some m ∈ A ∩ B } is finite. Since D is nonprincipal C / ∈ D and so D = ( A ∩ B − C ) ∈ D .According to the definiton of I D as in intersection, the subcontinuum I D does not change if weredefine the intervals I n for all n ∈ D c . Likewise for J D . Hence we can redefine I n , J n for n ∈ D c without changing I D and J D and hence assume I ω ∩ J ω is empty. Then we can combine the two sequences into some { K n : n ∈ ω } where each interval K n is some I m or J m and each K n +1 is to the right of K n .Define the maps α : ω → ω by letting each α ( n ) be the unique i ∈ ω with I n = K i . In other words I n = K α ( n ) . Likewise define β : ω → ω by J n = K β ( n ) . It follows from the definitions that I D = K α ( D ) and J D = K β ( D ) where we define the ultrafilter α ( D ) = { E ⊂ ω : α − ( E ) ∈ D} and likewise for β ( D ) . Define a finite-to-one function g separately over the disjoint sets A = { α ( n ) : n ∈ ω } and B = { β ( n ) : n ∈ ω } by g ( α ( n )) = α ( n ) and g ( β ( n )) = α ( n ) . The definition g ( β ( n )) = α ( n ) makessense because β is injective. From here it quickly follows g ( α ( D )) = α ( D ) = g ( β ( D )) . From [4]Lemma 10 and the paragraph before that lemma, if there exists a finite-to-one function g : ω → ω with g ( α ( D )) = g ( β ( D )) then there also exists a finite-to-one monotone function f : ω → ω with f ( α ( D )) = f ( β ( D )) . Since f is finite-to-one each f − ( n ) is finite. Thus the convex hull L n of { K m : m ∈ f − ( n ) } is a closed interval. Since f is monotone each L n +1 is to the right of L n . So we can define astandard subcontinuum L f ( α ( D )) . We claim K α ( D ) , K β ( D ) ⊂ L f ( α ( D )) . Since J D = K β ( D ) thisshows J D ⊂ κ ( p ) .To that end write α ( D ) = U . First observe each K n ⊂ L f ( n ) since L f ( n ) is the hull of { K m : m ∈ f − ( f ( n )) } and we have m ∈ f − ( f ( n )) for m = n . It follows K U ⊂ L f ( U ) for each U ∈ U . Hence we can write K U = \ U ∈U H ∗ ∩ K U ⊂ \ U ∈U H ∗ ∩ L f ( U ) . It follows from the definition of f ( U ) and U being an ultrafilter that f ( U ) ∈ f ( U ) for each Daron Anderson 10 Shore and Non-Block Points ∈ U . In the other direction, for each E ∈ f ( U ) we have f − ( E ) ∈ U and so E contains the set f ( U ) for U = f − ( E ) . It follows { f ( U ) : U ∈ U } is a cofinal subset of f ( U ) . Hence the intersectionon the right-hand-side equals \ W ∈ f ( U ) H ∗ ∩ L W = L f ( U ) . We conclude K α ( D ) ⊂ L f ( α ( D )) . By symmetry the same holds for K β ( D ) . This completes theproof.We would like to show our example of spot-welding two copies of H ∗ is generic in the followingsense. Conjecture 4.
Suppose X has no shore points and p ∈ X . There is a decomposition X = X ⊕ . . . ⊕ X N with p ∈ X ∩ . . . ∩ X N and p non-coastal when treated as a point of each X n . Conjecture 4 asks for a particularly nice decomposition of X . One variant of the conjecture is that every decomposition with p ∈ X ∩ . . . ∩ X N is nice . The stronger conjecture however is false.For a counterexample assume NCF and take two copies H and H of H ∗ and identify points x ∈ H and x ∈ H respectively with the endpoints and of the arc. Denote by X the quotientspace.Observe for X = H ∪ [0 , and X = [0 , ∪ H we have the decomposition X = X ⊕ X butthe semicontinua H ∪ [0 , and (0 , ∪ H witness how p = 1 / is a coastal point of each element.Thus the stronger conjecture fails. To see the weaker conjecture holds consider the second choice of decomposition X = H ∪ [0 , / and X = [1 / , ∪ H . By the same reasoning as in Theorem 1 we see p = 1 / is a coastalpoint of neither element. The second decomposition above is minimal in the following sense.
Definition 5.3.
Suppose X = X ⊕ . . . ⊕ X N and X = Y ⊕ . . . ⊕ Y N are decompositions. Wewrite Y ⊕ . . . ⊕ Y N ≤ X ⊕ . . . ⊕ X N to mean there is a permutation σ of { , , . . . , N } with each Y n ⊂ X σ ( n ) . We call a decomposition minimal to mean it is minimal with respect to this partialorder. For example, each minimal decomposition X ⊕ X of the arc has X ∩ X a singleton. Eachminimal decomposition X ⊕ X of the circle has X ∩ X a doubleton. For X formed by spot-welding finitely many indecomposable continua the natural decomposition is minimal. It may prove useful that minimal decompositions always exist. Daron Anderson 11 Shore and Non-Block Points emma 5.4.
Each decomposition is ≤ -above some minimal decomposition. Proof.
The proof uses Zorn’s lemma. Suppose { X i (1) ⊕ . . . ⊕ X i ( N ) : i ∈ I} is a chain ofdecompositions. Without loss of generality I has top element ∈ I and no bottom element.Since X (1) ⊕ . . . ⊕ X ( N ) is a decomposition there are points x n ∈ X ( n ) − S { X ( m ) : m = n } .Let X i (1) ⊕ . . . ⊕ X i ( N ) be arbitrary. Since i ≤ some permutation σ i has each X i (cid:0) σ i ( n ) (cid:1) ⊂ X ( n ) . Thus X i (cid:0) σ i ( n ) (cid:1) includes at most the element x n of { x , x , . . . , x N } . But since X i (1) ∪ . . . ∪ X i ( N ) = X each x n is an element of some X i ( m ) . We conclude each x n ∈ X i (cid:0) σ i ( n ) (cid:1) − S (cid:8) X i (cid:0) σ i ( m ) (cid:1) : m = n (cid:9) .For j ≤ i we know X j (cid:0) σ j ( n ) (cid:1) is contained in some X i (cid:0) σ i ( m ) (cid:1) . The point x n witnesses how m = n . We conclude each X j (cid:0) σ j ( m ) (cid:1) ⊂ X i (cid:0) σ i ( m ) (cid:1) . Then [14] Proposition 1.7 says the intersec- tion X n = T (cid:8) X i (cid:0) σ i ( n ) (cid:1) : i ∈ I (cid:9) is a subcontinuum. We claim X , . . . , X N form a decompositionwhich is clearly a lower bound for the chain. First observe each x n witnesses how X n is notcontained in S { X ( m ) : m = n } .To show X ∪ . . . ∪ X N = X let x ∈ X be arbitrary. For each i ∈ I there is n i ≤ N with x ∈ X i (cid:0) σ i ( n i ) (cid:1) . Write A ( n ) = (cid:8) i ∈ I : x ∈ X i (cid:0) σ i ( n ) (cid:1)(cid:9) for n = 1 , , . . . , N . Since I has no bottom element one of A ( n ) is cofinal. Thus x ∈ T (cid:8) X i (cid:0) σ i ( n ) (cid:1) : i ∈ A ( n ) (cid:9) which equals T (cid:8) X i (cid:0) σ i ( n i ) (cid:1) : i ∈ I (cid:9) by cofinality and thus x ∈ X n . Since x is arbitrary we see X n cover X .We conclude the chain { X i (1) ⊕ . . . ⊕ X i ( N ) : i ∈ I} has a lower bound X ⊕ . . . ⊕ X N . Zorn’slemma then implies each decomposition is above a minimal decomposition. Conjecture 5.
Suppose X has no shore points and X = X ⊕ . . . ⊕ X N is a minimal decomposition with p ∈ X ∩ . . . ∩ X N . Then p is non-coastal when treated as a point of each X n .Thus far we have only the partial results Lemmas 5.5, 5.8 and 5.9. Henceforth assume X hasno shore points. Fix p ∈ X and let X = X ⊕ X be a decomposition with p ∈ X ∩ X . The casefor n > is similar. Lemma 5.5.
Each dense semicontinuum of X (resp. X ) at p contains X − X (resp. X − X ). Proof.
Suppose for example p ∈ S ⊂ X − q for some dense semicontinuum S ⊂ X and q ∈ X − X . Then X ∪ S ⊂ X − q is a dense semicontinuum of X . This implies q ∈ X is a non-blockpoint hence a shore point contrary to assumption. Corollary 5.6.
Suppose X ∩ X = { p } . Then p is non-coastal as an element of X and X . Daron Anderson 12 Shore and Non-Block Points he following notation is part of Lemma 5.8
Notation 5.7.
Suppose X = X ⊕ X . Write S ( X ) (resp. S ( X ) ) for the collections of properdense semicontinua of X (resp. X ) that meet both X ∩ X and X − X (resp. X − X ). Definetwo subsets of X ∩ X . C = { x ∈ X ∩ X : x ∈ S for each S ∈ S ( X ) } . C = { x ∈ X ∩ X : x ∈ S for each S ∈ S ( X ) } . Lemma 5.8.
Suppose p is coastal as an element of both X and X . Then one of C ∩ C = ∅ or C ∪ C = X ∩ X holds. Proof.
Since p ∈ X is coastal S ( X ) is nonempty and likewise for S ( X ) . Suppose C ∩ C = ∅ and C ∪ C = X ∩ X . That means there are x ∈ C ∩ C and y ∈ X ∩ X − C ∪ C . Select S ∈ S ( X ) and S ∈ S ( X ) with y / ∈ S and y / ∈ S .By definition we have x ∈ S and x ∈ S . Thus S ∪ S ⊂ X is a dense semicontinuum that excludes the point y ∈ X . This contradicts how X has no shore points. We conclude C ∩ C = ∅ and so C ∪ C = X ∩ X .In the first case of Lemma 5.8 we can say more. Lemma 5.9.
Suppose p is coastal as an element of both X and X and C ∩ C = ∅ . Then eachsubcontinuum of X that meets C and C also contains X ∩ X − C ∪ C . Proof.
Suppose the subcontinuum K ⊂ X meets C and C . Let x ∈ X ∩ X − C ∪ C bearbitrary. Select S ∈ S ( X ) and S ∈ S ( X ) with x / ∈ S and x / ∈ S . It follows S ∪ K ∪ S ⊂ X is a dense semicontinuum. Since X has no shore points S ∪ K ∪ S = X and so x ∈ K . Since x ∈ X ∩ X − C ∪ C is arbitrary we conclude X ∩ X − C ∪ C ⊂ K . 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 Doctor
Aisling McCluskey for their help in preparing the manuscript, and the anonymous referee for theirattention and suggestions.
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