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Topology and its Applications 126 (2002) 13– www.elsevier.com/locate/topol
Expansive homeomorphisms and indecomposable
subcontinua
Christopher Mouron
Department of Mathematics and Computer Science, Hendrix College, Conway, AR 72032, USA Received 13 February 2001; received in revised form 18 December 2001
Abstract
A continuum X is k- cyclic if given any ε > 0, there is a finite open cover U of X such that mesh(U) < ε and the nerve N(U) is has at most k distinct simple closed curves. A homeomorphism h : X → X is called expansive provided for some fixed ε > 0 and every x, y ∈ X there exists an integer n such that d(h n^ (x), h n^ (y)) > ε. We prove that if X is a k-cyclic continuum that admits an expansive homeomorphism, then X must contain an nondegenerate indecomposable subcontinuum. 2002 Elsevier Science B.V. All rights reserved.
MSC: primary 54H20, 54F15; secondary 54E40, 54F
Keywords: Expansive homeomorphism; Indecomposable continuum; Finitely cyclic; Tranche
1. Introduction
One important idea that is studied in a dynamical system is what happens to points that are close together over a period of iterations or time. In dynamical continuum theory, one is interested in how far points move apart under repeated homeomorphism of a continuum onto itself. A continuum is defined to be a compact, connected metric space. A homeomorphism h : X → X is called expansive provided for some fixed ε > 0 and
✩ (^) This work was supported in part by Texas Advanced Research Program Grant 003644-049, Texas Advanced Research Program Grant 0204-44-2413, and the 1997 Texas Tech Summer Graduate Research Award. The author would also like to thank his advisor Wayne Lewis for his guidance and many useful comments which helped make this paper possible. E-mail address: mouron@grendel.hendrix.edu (Ch. Mouron).
0166-8641/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved. PII: S 0 1 6 6 - 8 6 4 1 ( 0 2 ) 0 0 0 0 2 - 0
every x, y ∈ X there exists an integer n such that d(hn^ (x), hn^ (y)) > ε, where if n is a positive integer, then
hn^ (x) = h ◦ h ◦ · · · ◦ h(x) ︸ ︷︷ ︸ n
and if n is a negative integer, then
hn^ (x) = h−^1 ◦ h−^1 ◦ · · · ◦ h−^1 (x) ︸ ︷︷ ︸ |n|
Expansive homeomorphisms exhibit chaotic behavior in that no matter how close two points are, their images will eventually be a certain distance apart. Plykin’s attractors [6] and the dyadic solenoid [7] are examples of continua that admit an expansive homeomorphism. Both of these continua have the property of being indecomposable. A continuum is decomposable if it is the union of two proper subcontinuum. A continuum is indecomposable if it is not decomposable. Indecomposable continua are created by stretching and bending arcs an infinite number of times back and forth. Intuitively, it appears that in order to have expansiveness, the subset of the continuum between points that are close to each other would have to be continually stretched in order to move points away from each other. However, because of compactness, some folding or wrapping must also occur. Every known continuum that admits an expansive homeomorphism has an indecomposable subcontinuum. One way to describe continua is through sequences of finite open covers. Let U be a finite open cover. Since we are assuming that X is a one-dimensional continuum, we may also assume that each x ∈ X is in at most two elements of U. Define the mesh of U to be mesh(U) = sup{diam(U ) | U ∈ U} where diam(U ) = sup{d(x, y) | x, y ∈ U }. The nerve of U is a graph N(U) which has a vertex vi that corresponds to each element U (^) i of U and an edge between vi and vj if and only if U (^) i ∩ U (^) j = ∅. A continuum X is 1 arc-like , 2 tree-like , 3 circle-like , 4 G-like , 5 k-cyclic , if given any ε > 0, there is a finite open cover U of X such that mesh(U) < ε and the nerve N(U) is 1 an arc, 2 a tree, 3 a circle, 4 homeomorphic to a fixed graph G, 5 has exactly k distinct simple closed curves. X is finitely cyclic if it is k-cyclic for some k. The double Warsaw circle or a double Warsaw circle with any number of stickers attached to it are examples of finitely cyclic continua. The Sierpinski curve is an example of an one-dimensional continuum that is not finitely cyclic.
d(g n^ (x (^) ε ), g n^ (y (^) ε )) = d(hnk^ (x (^) ε ), hnk^ (y (^) ε )) < ε for every integer n. Therefore, g = hk^ is not expansive. Now suppose that g = hk^ is not expansive. Pick any ε > 0, and choose δ > 0 such that d(x, y) < δ implies that d(hi^ (x), hi^ (y)) < ε for each i ∈ { 0 , 1 ,... , k}. Since g is not expansive there exists x (^) δ , y (^) δ ∈ X with x (^) δ = y (^) δ such that d(hnk^ (x (^) δ ), hnk^ (y (^) δ )) = d(g n^ (x (^) δ ), g n^ (y (^) δ )) < δ for every integer n. Thus, d(hnk+i^ (x (^) δ ), hnk+i^ (y (^) δ )) < ε for every integer n and every i ∈ { 1 , 2 ,... , k}. Since every integer can be represented by nk + i, h is not expansive. ✷
Proposition 4. No graph G admits an expansive homeomorphism.
Proof. We may assume that G is not an arc or simple closed curve. Let V be the set of vertices of degree 1 or greater than 2. Since G is not a simple closed curve, 2 |V | < ∞. Let E be the collection of edges between the vertices in V. Since 2 |V |, E is nonempty, and since V is finite, E is finite and each edge in E is periodic under any homeomorphism h. Thus, since each edge is homeomorphic to an arc, by Proposition 3, h cannot be expansive. ✷
2. Main result
In [2], Kato asked “If X admits an expansive homeomorphism, must X contain a nondegenerate indecomposable subcontinuum?”. Since every compact metric space of dimension greater than or equal to 2 contains a nondegenerate indecomposable continuum [5], only one-dimensional continua need to be considered. The next theorem is the main result of the paper and gives a partial answer in the affirmative to Kato’s question.
Theorem 5. If X is a k -cyclic continuum that admits an expansive homeomorphism, then X must contain a nondegenerate indecomposable subcontinuum.
The proof of the main result begins by assuming that X is a hereditarily decomposable k-cyclic continuum and h : X → X is an expansive homeomorphism. A series of assumptions on X are then shown to be valid until a contradiction is shown.
Note. It is assumed in each lemma, claim and theorem that the continuum X satisfies all the assumptions that were stated before each.
First it is shown that we may assume that X contains no proper, nondegenerate subcontinuum with period less than or equal to k + 1, where k is the cyclic bound for X. Then it is shown that if A is a proper subcontinuum of X, then A must be tree-like. Then under the previous assumption, it is shown that there exists a monotone map Φ from X to the simple closed curve S and a homeomorphism f : S → S such that f ◦ Φ = Φ ◦ h.
Since each proper subcontinuum of X is tree-like, Φ−^1 (y) is tree-like for each y in S. Thus, by Theorem 2, either X is a graph and h cannot be expansive, or X must contain an nondegenerate indecomposable subcontinuum.
Assumption 1. Suppose that X is a hereditarily decomposable k -cyclic continuum and that h : X → X is an expansive homeomorphism.
By Theorem 1, we may also assume that X is neither tree-like nor circle-like. Also, since X is not tree-like, we may assume that k 1.
Assumption 2. X contains no proper nondegenerate subcontinuum that has period less than or equal to max{ 2 , k + 1 }.
The next lemma gives the justification for Assumption 2 in that we may take X to be the minimal subcontinuum described.
Lemma 6. If X admits an expansive homeomorphism h , then there exists a minimal nondegenerate subcontinuum of period less than or equal to k + 1 that contains no proper subcontinuum of period less than or equal to k + 1_._
Proof. Let Y be the collection of all subcontinua of X with period less than or equal to k + 1. Partially order Y by inclusion. Let P be any maximal chain in the ordering. There exists an m ∈ { 1 , 2 ,.. ., k + 1 } such that for every A (^) α ∈ P, there exists an E ∈ P such that E ⊂ A (^) α and E has period m. Let
Pm = {A (^) α ∈ P | A (^) α has period m},
and let A =
Aα ∈Pm A^ α^ =^
Aα ∈P A^ α^. Since^ Pm^ is ordered by inclusion,^ A^ must be a continuum. If A is degenerate, then for any ε > 0 there exists a nondegenerate continuum A (^) αm ∈ Pm such that diam(A (^) αm ) < ε. Therefore, hm^ is not expansive and thus, h is not expansive. Thus, A must be nondegenerate and have diameter at least as big as the expansive constant for hm. Notice that
hm^ (A) = hm
Aα ∈Pm
A (^) α
Aα ∈Pm
hm^ (A (^) α ) =
Aα ∈Pm
A (^) α = A.
Thus, A is a minimal periodic subcontinuum that contains no proper subcontinuum with period less than or equal to k + 1. ✷
Assumption 3. Every proper subcontinuum of X is tree-like.
Let x ∈ X and A ⊂ X. Define d(x, A) = inf{d(x, y) | y ∈ A} and d(A, B) = inf{d(x, y) | y ∈ A and x ∈ B}.
Theorem 7. If H and K are two subcontinua of X such that X = H ∪ K and H ∩ K has at least two components, then X is not tree-like.
must be connected and a tree. Let
U^ ˜K =
U | U ∈ UK and U ∩ (H ∩ K) = ∅
U ⊂U (^) H
U | U ∈ UK , U ∩ (H ∩ K) = ∅ and U (^) H ∈ UH
Claim. The nerve of U˜K is a tree.
Proof of Claim. Suppose not. Let {U˜ (^) i 1 , U˜ (^) i 2 ,... , U˜ (^) im , U˜ (^) i 1 } be elements of U˜ whose nerve is a simple closed curve. Let
U^ ˜H ∩K = {U˜ | U˜ ∈ U˜K and U˜ ∩ (H ∩ K) = ∅}.
Then, the nerve of U˜H ∩K is connected and by Lemma 10 is a tree. Suppose there exists a j such that U˜ (^) ij ∈/ U˜ (^) H ∩K. Then U˜ (^) ij ∩ (H ∩ K) = ∅, and therefore, U˜ (^) ij ∈ UK. If U˜ (^) ij ∈/ U˜ (^) H ∩K
for every j ∈ { 1 , 2 ,... , m}, then {U˜ (^) i 1 , U˜ (^) i 2 ,... , ˜U (^) im , U˜ (^) i 1 } ⊂ UK. However, that would imply that nerve of UK would contain a simple closed curve which is a contradiction. Thus, we may choose j, n such that {U˜ (^) ij ,... , U˜ (^) in } is a subchain of {U˜ (^) i 1 , U˜ (^) i 2 ,... , ˜U (^) im , U˜ (^) i 1 },
U^ ˜ (^) ip ∈/ U˜ (^) H ∩K for every U˜ (^) ip ∈ {U˜ (^) ij ,... , U˜ (^) in }, U˜ (^) i j′^ ∈^ U˜ (^) H ∩K , and U˜ (^) i n′^ ∈^ U˜ (^) H ∩K , where
U^ ˜ (^) i j′^ ,^ U˜ (^) i n′^ are the unique elements of^ { U˜ (^) i 1 , ˜U (^) i 2 ,... , U˜ (^) im , U˜ (^) i 1 } − {U˜ (^) ij ,... , U˜ (^) in } such that
U^ ˜ (^) ij ∩ U˜ (^) i j′^ = ∅^ and^ U˜ (^) in ∩ U˜ (^) i n′^ = ∅. Now, there exist^ U^ α^ , U^ β^ ∈^ UK^ such that^ U^ α^ ⊂ U^ ˜ (^) i j′^ ,^ U^ β^ ⊂^ U˜ (^) i n′^ ,^ U^ α^ ∩^ U˜ (^) ij = ∅, and U (^) β ∩ U˜ (^) in = ∅. Since H ∩ K is a continuum, there exists a chain {U (^) α , U (^) γ ,... , U (^) β } of elements of UH ∩K from U (^) α to U (^) β. Thus, {U˜ (^) ij ,... , U˜ (^) in , U (^) β ,... , U (^) α , U˜ (^) ij } ⊂ UK. However, the nerve of {U (^) ij ,... , U (^) in , U (^) β ,... , U (^) α , U (^) ij } is a simple closed curve. Thus, the nerve of UK is not a tree, which is a contradiction.
Thus, U˜ (^) ij ∈ {U˜ (^) i 1 , U˜ (^) i 2 ,... , U˜ (^) im , U˜ (^) i 1 } implies U˜ (^) ij ∈ ˜UH ∩K. Hence, {U˜ (^) i 1 , U˜ (^) i 2 ,... , U˜ (^) im , U˜ (^) i 1 } ⊂ U˜H ∩K. But that is impossible since the nerve of U˜H ∩K is a tree. Thus, the claim is proved.
Notice that if V ∈ ˜UK then either V ∈ UK or V ⊂ U (^) H where U (^) H ∈ UH. Thus, mesh(U˜K ) < ε. Let UH = {U (^) H 1 , U (^) H 2 ,... , U (^) Hn } and U˜K = {U (^) K 1 , U (^) K 2 ,... , U (^) K (^) m }. Now compose W of the following open sets: (1) WHi = U (^) Hi − K if U (^) Hi ∈ UH and U (^) Hi ∩ (H ∩ K) = ∅. (2) WK (^) j = U (^) K (^) j − H if U (^) K (^) j ∈ U˜K and U (^) K (^) j ∩ (H ∩ K) = ∅. (3) WHi = WK (^) j = (U (^) Hi − (K − (H ∩ K))) ∪ (U (^) K (^) j − (H − (H ∩ K))) if U (^) Hi ∈ UH ,
U (^) K (^) j ∈ U˜K and U (^) K (^) j ⊂ U (^) Hi. Notice that if WHi = WK (^) j and WHi′ = WK (^) j′ , then WK (^) j′ ∩ WK (^) j = ∅ if and only if U (^) K (^) j′ ∩ U (^) K (^) j = ∅ if and only if U (^) Hi′ ∩ U (^) Hi = ∅. Thus, by Lemma 10, the nerves of both WH = {WHi }ni= 1 and WK = {WK (^) j }mj = 1 are trees. Let
WH ∩K =
WHi | WHi ∩ (H ∩ K) = ∅
WK (^) j | WK (^) j ∩ (H ∩ K) = ∅
Then WH ∩K ⊂ WH (and likewise, WH ∩K ⊂ WK ). Thus, the nerve of WH ∩K is a tree. Also, if WHi ∈ WH and WHi ∩ (H ∩ K) = ∅, then WHi ∩ K = ∅. Likewise, if WK (^) j ∈ WK
and WK (^) j ∩ (H ∩ K) = ∅, then WK (^) j ∩ H = ∅. Thus if the nerve of W = WH ∪ WK is not a tree, then the nerve of
WH ∩K =
W | W ∩ (H ∩ K) = ∅
is not a tree, which is a contradiction. Thus, the nerve of W is a tree. Proof is similar when H ∩ K has k + 1 components. ✷
Suppose that U is a finite open cover of continuum A. Let B be a subcontinuum of A and let U(B) = {U ∈ U | U ∩ B = ∅}.
Lemma 12. Let X be a one-dimensional continuum and let A be a tree-like subcontinuum of X_. If_ U is a finite open cover of X , then there exists a finite open cover W of X such that W refines U , the nerve of W is one-dimensional, and the nerve of W(A) is a tree.
Proof. Let U be a finite open cover of X. There exists a finite open cover V of X such that V refines U and the nerve of V(A) is a tree. Since X is one-dimensional, there exists a finite open cover D of X such that D refines V and the nerve of D is one-dimensional. Define
W =
D − D(A)
U ⊂V
U | U ∈ D(A) and V ∈ V(A)
Then by Lemma 10, W satisfies the conclusion of the lemma. ✷
From here on out, we may assume that if W is an open cover of X that refines open cover U such that the nerve of W(A) is a tree, then the nerve of W is one-dimensional. If U is a collection of sets, then U∗^ is the union of the elements of U. If x ∈ U , then the star of x in U , denoted by st(x, U), is the collection of elements of U that contain x (i.e., st(x, U) = {U ∈ U | x ∈ U }). If A ⊂ U∗, then st(A, U) = {U ∈ U | A ∩ U = ∅}. Inductively, sti+^1 (x, U) = st(sti^ (x, U)∗, U).
Lemma 13. Let X be a k -cyclic continuum and let T be a collection of tree-like subcontinua of X such that T has p elements. If U 1 is a finite open cover of X such that the nerve of U has at most k simple closed curves, then there exists a finite open cover W of X such that W refines U 1 , the nerve of W contains at most k simple closed curves, and the nerve of W(T ) is a tree for each T ∈ T_._
Proof. First, we must verify a claim.
Claim. There exists a finite open cover V of X such that V refines U 1 and the nerve of V(T ) is a tree for each T ∈ T_._
Proof of Claim. Pick T 1 ∈ T. Since T 1 is tree-like, there exists a finite open cover V 1 that refines U 1 such that the nerve of V 1 (T 1 ) is a tree.
Proof of Claim. The proof is by induction. Base case. Suppose that there exists a subset M 1 of M such that each A ∈ M 1 is 1-cyclic, M =
A∈M A^ =^
A∈M 1 A, and that^ M^ is tree-like. Pick any^ A^1 ∈^ M^1. By Lemma 13, there exists a sequence of finite open covers, {Un} ∞ n= 1 , of A 1 such that mesh(Un ) → 0 as n → ∞, Un+ 1 refines Un for each n, the nerve of Un has exactly 1 simple closed curve, and the nerve of Un (M) is a tree. Since M =
A∈M 1 A, there exist an^ A^2 ∈^ M^1 which is a proper subcontinuum of^ A^1 such that the nerve of U 1 (A 2 ) is a tree. Since A 2 is not tree-like, there exists an integer N 2 such that for every n N 2 , the nerve of Un (A 2 ) is not a tree. Thus, the nerve of UN 2 (A 2 ) must contain a simple closed curve. Let
V 1 =
UN 2 − UN 2 (A 2 )
U ⊂V
U | U ∈ UN 2 (A 2 ) and V ∈ U 1 (A 2 )
V 1 is a finite open cover of A 1 , and by Lemma 10, the nerve of V 1 (A 2 ) is a tree. Since the nerve of UN 2 contains exactly 1 simple closed curve, the nerve of V 1 must be a tree. Notice that mesh(V 1 ) mesh(U 1 ). Suppose that UNi has been found. Since M =
A∈M 1 A, there exists an^ A^ i+^1 ∈^ M^1 which is a proper subcontinuum of A 1 such that the nerve of UNi (A (^) i+ 1 ) is a tree. Since A (^) i+ 1 is not tree-like, there exists an integer Ni+ 1 such that for every n Ni+ 1 , the nerve of Un (A (^) i+ 1 ) is not a tree. Thus, the nerve of UNi+ 1 (A (^) i+ 1 ) must contain a simple closed curve. Let
Vi =
UNi+ 1 − UNi+ 1 (A (^) i+ 1 )
U ⊂V
U | U ∈ UNi+ 1 (A (^) i+ 1 ) and V ∈ UNi (A (^) i+ 1 )
Vi is a finite open cover of A 1 , and the nerve of Vi (A 2 ) is a tree. Since the nerve of UNi+ 1 contains exactly 1 simple closed curve, the nerve of Vi must be a tree. Notice that mesh(Vi ) mesh(UNi ). Thus, mesh(Vi ) → 0 as i → ∞. Thus, A 1 must be tree-like. This is a contradiction. Thus, M is not tree-like. Induction step. Suppose that there exists a subset Mj of M such that each A ∈ Mj is j -cyclic, but not (j − 1 )-cyclic, M =
A∈M A^ =^
A∈Mj A, and that^ M^ is tree-like. Pick any A 1 ∈ Mj. There exists a sequence of finite open covers, {Un} ∞ n= 1 , of A 1 such that mesh(Un ) → 0 as n → ∞, Un+ 1 refines Un for each n, the nerve of Un has exactly j simple closed curves, and the nerve of Un (M) is a tree. Since M =
A∈Mj A, there exist an^ A^2 ∈^ Mj^ which is a proper subcontinuum of^ A^1 such that the nerve of U 1 (A 2 ) is a tree. Now, since A 2 is not (j − 1 )-cyclic, there exists an integer N 2 such that for every n N 2 , the nerve of Un (A 2 ) contains exactly j simply closed curves. Let
V 1 =
UN 2 − UN 2 (A 2 )
U ⊂V
U | U ∈ UN 2 (A 2 ) and V ∈ UN 1 (A 2 )
V 1 is a finite open cover of A 1 , and the nerve of V 1 (A 2 ) is a tree. Since the nerve of UN 2 contains exactly j simple closed curves, and the circular chains of UN 2 (A 2 ) have been collapsed, the nerve of V 1 must contain no more than j − 1 simple closed curves. Notice that mesh(V 1 ) mesh(U 1 ).
Suppose that UNi has been found. Since M =
A∈Mj A, there exists an^ A^ i+^1 ∈^ Mj which is a proper subcontinuum of A 1 such that the nerve of UNi (A (^) i+ 1 ) is a tree. Now since A (^) i+ 1 is not (j − 1 )-cyclic, there exists an integer Ni+ 1 such that for every n Ni+ 1 , the nerve of Un (A (^) i+ 1 ) has exactly j simple closed curves. Let
Vi =
UNi+ 1 − UNi+ 1 (A (^) i+ 1 )
U ⊂V
U | U ∈ UNi+ 1 (A (^) i+ 1 ) and V ∈ UNi (A (^) i+ 1 )
Vi is a finite open cover of A 1 , and the nerve of Vi (A 2 ) is a tree. Since the nerve of UNi+ 1 contains exactly j simple closed curves, and the circular chains of UNi+ 1 (A 2 ) have been collapsed, the nerve of Vi must contain no more than j − 1 simple closed curves. Notice that mesh(Vi ) mesh(UNi ). Thus, mesh(Vi ) → 0 as i → ∞, and hence, A 1 must be (j − 1 )-cyclic. This is a contradiction. Thus, M is not tree-like, and the claim is proved.
If B is a proper subcontinuum of M, then B is tree-like. Also, the same properties that hold for M must hold for hj^ (M) for every j. Consider the collection { M, h(M),... , hk^ (M), hk+^1 (M)
and suppose there exist i, j ∈ { 0 , 1 ,... , k + 1 } such that i = j and hi^ (M) = hj^ (M). Then M has period k + 1 which is a contradiction. Thus, either M = X or hi^ (M) = hj^ (M) for every i = j. Suppose hi^ (M) ∩ hj^ (M) has a component C that is not tree-like. Since C is a continuum, and hi^ (M) and hj^ (M) are minimal, hi^ (M) = C = hj^ (M). This is a contradiction. Thus, every component of hi^ (M) ∩ hj^ (M) is tree-like. Also, since X is k-cyclic, the number of components of hi^ (M) ∩ hj^ (M) must be finite by Corollary 9. Let C = {C | C is a component of hi^ (M) ∩ hj^ (M) for some i = j }. We may suppose that X is k-cyclic but not (k − 1 )-cyclic. Since each hi^ (M) is not tree-like, there exists an ε > 0 such that if U is a finite open cover of X such that mesh(U) < ε, then the nerve of U(hi^ (M)) is not a tree for each i. Let {Un} be a sequence of finite open covers of X such that mesh(Un ) < ε for each n, the nerve of each Un has exactly k simple closed curves, limn→∞(mesh(Un )) = 0, each Un+ 1 refines Un, and the nerve of Un (C) is a tree for each C ∈ C.
Claim. There exists an integer N such that for every n N and each i = j , the nerves of Un (hi^ (M)) and Un (hj^ (M)) contain distinct simple closed curves.
Proof of Claim. Suppose not. Thus, suppose for every p, there exists n p such that i (^) n = j (^) n and the nerves of Un (hin^ (M)) and Un (hjn^ (M)) do not have distinct simple closed curves. Since there is only a finite number of combinations of {i, j }, there exists a subsequence {nm} of {n} such that i (^) m = i (^) l and j (^) m = j (^) l for all positive integers m , l. For ease of notation, let {nm} = {n}, i = i (^) m and j = j (^) m. The nerves of Un (hi^ (M)) and Un (hj^ (M)) must both contain a collection of circle- chains, say Sin and Snj. Since the nerves of Un (hi^ (M)) and Un (hj^ (M)) do not have
distinct simple closed curves, either Sni ⊂ Snj or Snj ⊂ Sni for each n. Thus, without loss of generality, we may assume that there exists an infinite subsequence {nm} of {n} such that Snim ⊂ Snjm for each m. Again, for ease of notation, let {n} represent {nm}.
each g−^1 (y) is a continuum (possibly degenerate). Also, if A is any subcontinuum of g(X), then g−^1 (A) is a subcontinuum of X. A continuum X is irreducible between a and b if a, b ∈ X and if A is a proper subcontinuum of X, then a /∈ A or b /∈ A. The notation, I (a, b), will be used to imply that I (a, b) is a continuum irreducible between a and b. If K is a subcontinuum of X and a, b ∈ K, then let IK (a, b) be a subcontinuum of K irreducible between a and b.
Theorem 15 (Kuratowski [4]). If X is hereditarily decomposable and is irreducible between a and b , then there exists a monotone map ψ : X → [ 0 , 1 ] such that ψ(a) = 0 and ψ(b) = 1_. In fact, there exists a minimal monotone onto map_ φ : X → [ 0 , 1 ] such that if ψ : X → [ 0 , 1 ] is any other monotone onto map and φ−^1 (z) ∩ ψ−^1 (y) = ∅ , then φ−^1 (z) ⊂ ψ−^1 (y).
If φ : X → [ 0 , 1 ] is a minimal monotone onto map, then each φ−^1 (y) is called a tranche of X. Tranches are nowhere dense subcontinua of X. That is, int(φ−^1 (y)) = ∅. Also, if I (a, x) = I (a, y), then x and y are in the same tranche of I (a, b). The following well-known lemma is needed:
Lemma 16. Continuum X is indecomposable if there exists distinct points a, b and c in X such that X is irreducible between each pair of a, b and c_._
Theorem 17. Suppose that X is neither 1 -cyclic nor tree-like and that every proper subcontinuum of X is tree-like, then X is either indecomposable or 2 -indecomposable.
Proof. If X is decomposable, there exist minimal proper subcontinua A, B such that A ∪ B = X. That is, if A′^ is a proper subcontinuum of A then A′^ ∪ B = X and if B′^ is a proper subcontinuum of B then A ∪ B′^ = X. Also, since X is not 1-cyclic, it follows from Theorems 7, 8, and 11 that A ∩ B must have at least 3 components. Let Cx , Cy and Cz be 3 distinct components of A ∩ B that contain the points x, y and z, respectively. Let IA (x, y), IA (x, z) and IA (y, z) be subcontinua of A irreduciblie between x and y, x and z, and y and z, respectively. Suppose that IA (x, y) is a proper subcontinuum of A. Then from Theorem 7, B ∩ IA (x, y) is a proper subcontinuum of X that is not tree-like which is impossible. Hence, IA (x, y) = A. Similarily, it can be shown that IA (x, y) = IA (x, z) = IA (y, z) = A. Thus, it may be concluded that A is indecomposable. Proof is similar to show that B is indecomposable. ✷
Assumption 4. Suppose that X is hereditarily decomposable, 1 -cyclic but not tree-like and that every proper subcontinuum of X is tree-like.
Lemma 18. Under Assumption 4 , there exists a, b ∈ X and distinct minimal subcontinua I (a, b) and I ′(a, b) irreducible about a, b such that X = I (a, b) ∪ I ′(a, b).
Proof. Since X is decomposable, there exist minimal proper subcontinua A, B such that A ∪ B = X. Since A and B are proper subcontinua, they must be tree-like. Also, it follows from Theorems 7, 8, and 11 that since X is 1-cyclic but not tree-like, A ∩ B must have
exactly 2 components. Let Ca , Cb be the distinct components of A ∪ B where a ∈ Ca and b ∈ Cb. Let IA (a, b) be a subcontinuum of A irreducible about a, b and let IB (a, b) be a subcontinuum of B irreducible about a, b. Since IA (a, b) ∩ IB (a, b) has exactly 2 components, IA (a, b) ∪ IB (a, b) cannot be tree-like. Hence, IA (a, b) ∪ IB (a, b) = X. Also, since A and B are both minimal with respect to the decomposition of X, IA (a, b) = A and IB (a, b) = B. ✷
Theorem 19. Under Assumption 4 , there exists a minimal monotone map Φ : X → S , where S is a simple closed curve.
Proof. Let IA (a, b) and IB (a, b) be described as in Lemma 18. Since IA (a, b) and IB (a, b) are both hereditarily decomposable and irreducible between a and b, there exist minimal monotone maps ΦA : IA (a, b) → [ 0 , 1 ] and ΦB : IB (a, b) → [ 1 , 2 ] such that ΦA (a) = 0, ΦA (b) = 1 = ΦB (b), and ΦB (a) = 2. Now let S be formed by identifying 0 to 2 on the interval [ 0 , 2 ]. Let Φ : X → S be defined in the following manner:
Φ(x) = ΦA (x) if x ∈ IA (a, b), Φ(x) = ΦB (x) if x ∈ IB (a, b). Now, if there exists a y ∈ ( 0 , 1 ) such that Φ A− 1 (y) ∩ IB (a, b) = ∅, then IA (a, b) and IB (a, b) are not both minimal with respect to the decomposition of X. Same is true if there exists a y ∈ ( 1 , 2 ) such that Φ B− 1 (y) ∩ IA (a, b) = ∅. Clearly, if y ∈ ( 0 , 1 ) then
Φ−^1 (y) = Φ A− 1 (y) and is therefore connected. Similarly, if y ∈ ( 1 , 2 ) then Φ−^1 (y) = Φ B− 1 (y) and is also connected. Now, Φ−^1 ( 0 ) = Φ−^1 ( 2 ) = Φ A− 1 ( 0 ) ∪ Φ B− 1 ( 2 ). Since both
Φ A− 1 ( 0 ) and Φ B− 1 ( 2 ) are connected and a ∈ Φ A− 1 ( 0 ) ∩ Φ B− 1 ( 2 ), it follows that Φ−^1 ( 0 )
(equivalently Φ−^1 ( 2 ), Φ− A 1 ( 0 ) ∪ Φ B− 1 ( 2 )) is connected. Proof is similar to show that Φ−^1 ( 1 ) is connected. Next, it must be shown that Φ is minimal. Let Ψ : X → S be another monotone map. If y ∈ ( 0 , 1 ) and Φ−^1 (y) ∩ Ψ −^1 (z) = ∅, then Φ−^1 (y) = Φ− A 1 (y) ⊂ Ψ −^1 (z) since ΦA is
minimal. Likewise, if y ∈ ( 1 , 2 ) and Φ−^1 (y) ∩ Ψ −^1 (z) = ∅, then Φ−^1 (y) = Φ− B 1 (y) ⊂ Ψ −^1 (z) since ΦB is minimal. So, suppose that Φ−^1 ( 0 ) ∩ Ψ −^1 (z) = ∅. Since ΦA and ΦB are both minimal, Φ A− 1 ( 0 ) ⊂ Ψ −^1 (z) and Φ B− 1 ( 2 ) ⊂ Ψ −^1 (z). Hence, Φ−^1 ( 0 ) = Φ−^1 ( 2 ) =
Φ A− 1 ( 0 ) ∪ Φ− B 1 ( 2 ) ⊂ Ψ −^1 (z). Proof is similar to show that if Φ−^1 ( 1 ) ∩ Ψ −^1 (z) = ∅, then Φ−^1 ( 1 ) ⊂ Ψ −^1 (z). Thus, Φ is a minimal, monotone map. ✷
Theorem 20. Under Assumption 1 , if h is any homeomorphism of X onto itself, there exists a homeomorphism f of the circle, S, onto itself such that the following diagram commutes :
X Φ
h (^) X
Φ S
f S
Proof. First, it will be shown that if y, z ∈ S such that h(Φ−^1 (z)) ∩ Φ−^1 (y) = ∅, then h(Φ−^1 (z)) = Φ−^1 (y).
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