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On the classification of one dimensional continua that admit expansive homeomorphisms. Christopher G. Mouron Rhodes College July 28, 2016 Christopher G. Mouron On the classification of one dimensional continua that admit expansiv


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On the classification of one dimensional continua that admit expansive homeomorphisms.

Christopher G. Mouron

Rhodes College

July 28, 2016

Christopher G. Mouron On the classification of one dimensional continua that admit expansiv

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Preliminaries

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A continuum is a compact connected metric space. A continuum is 1-dimensional if for every ǫ > 0 there exists a finite

  • pen cover U of X with mesh less than ǫ such that every x ∈ X is

contained in at most 2 elements of U.

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A continuum is a compact connected metric space. A continuum is 1-dimensional if for every ǫ > 0 there exists a finite

  • pen cover U of X with mesh less than ǫ such that every x ∈ X is

contained in at most 2 elements of U.

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Christopher G. Mouron On the classification of one dimensional continua that admit expansive

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A continuum is

1 chainable (also known as arc-like) 2 tree-like 3 G-like 4 k-cyclic

if it is the inverse limit of (or if for every ǫ > 0 there exist an open cover whose nerve is a(n))

1 arc(s) 2 tree(s) 3 topological graph(s) homeomorphic to the same graph G 4 topological graph(s) each having at most k distinct simple

closed curves respectively.

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A continuum is

1 chainable (also known as arc-like) 2 tree-like 3 G-like 4 k-cyclic

if it is the inverse limit of (or if for every ǫ > 0 there exist an open cover whose nerve is a(n))

1 arc(s) 2 tree(s) 3 topological graph(s) homeomorphic to the same graph G 4 topological graph(s) each having at most k distinct simple

closed curves respectively.

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3

f

2

f

1

f

Figure: Arc-like

X = {(x1, x2, x3, ...) | fi(xi+1) = xi}.

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f

2 3

f

1

f

Figure: Tree-like

X = {(x1, x2, x3, ...) | fi(xi+1) = xi}.

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f f

2 3

f

1

Figure: G-like

X = {(x1, x2, x3, ...) | fi(xi+1) = xi}.

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f f

2 3

f

1

Figure: k-cyclic

X = {(x1, x2, x3, ...) | fi(xi+1) = xi}.

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f

2 3

f

1

f

Figure: Not k-cyclic

X = {(x1, x2, x3, ...) | fi(xi+1) = xi}.

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Arc−like Tree−Like continua All 1−dimensional −cyclic −like k G

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Figure: Arc-like and G-like (circle-like).

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Figure: arc-like

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Figure: tree-like

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Figure: arc-like

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Figure: not k-cyclic

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A continuum X is decomposable if it is the union of 2 of its proper subcontinua. A continuum is indecomposable if it is not decomposable. Equivalently, X is indecomposable if every proper subcontinuum is nowhere dense.

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Figure: Decomposable

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Figure: Decomposable

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Figure: Decomposable

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Figure: Decomposable

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Figure: Decomposable

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Figure: Decomposable

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Figure: Indecomposable

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Figure: Indecomposable

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Figure: Indecomposable

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Figure: Indecomposable

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Expansive Homemorphisms

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A homeomorphism h : X → X is called expansive provided that there exists a constant c > 0 such that for every x, y ∈ X there exists an integer n such that d(hn(x), hn(y)) > c. Here, c is called the expansive constant. Expansive homeomorphisms exhibit sensitive dependence on initial conditions in the strongest sense in that no matter how close any two points are, either their images or pre-images will at some point be at least a certain distance apart.

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A homeomorphism h : X → X is called expansive provided that there exists a constant c > 0 such that for every x, y ∈ X there exists an integer n such that d(hn(x), hn(y)) > c. Here, c is called the expansive constant. Expansive homeomorphisms exhibit sensitive dependence on initial conditions in the strongest sense in that no matter how close any two points are, either their images or pre-images will at some point be at least a certain distance apart.

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A homeomorphism h : X → X is called expansive provided that there exists a constant c > 0 such that for every x, y ∈ X there exists an integer n such that d(hn(x), hn(y)) > c. Here, c is called the expansive constant. Expansive homeomorphisms exhibit sensitive dependence on initial conditions in the strongest sense in that no matter how close any two points are, either their images or pre-images will at some point be at least a certain distance apart.

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Example.

Let f : R − → R be defined by f (x) = 2x and let the expansive constant be 1. Suppose that x and y are distinct real numbers. Then there exists an integer n such that 2n|x − y| = |2nx − 2ny| = |f n(x) − f n(y)| > 1. However, R is not compact.

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Example.

Let f : R − → R be defined by f (x) = 2x and let the expansive constant be 1. Suppose that x and y are distinct real numbers. Then there exists an integer n such that 2n|x − y| = |2nx − 2ny| = |f n(x) − f n(y)| > 1. However, R is not compact.

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Example.

Let f : R − → R be defined by f (x) = 2x and let the expansive constant be 1. Suppose that x and y are distinct real numbers. Then there exists an integer n such that 2n|x − y| = |2nx − 2ny| = |f n(x) − f n(y)| > 1. However, R is not compact.

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Example: The shift homeomorphism of the dyadic solenoid, Σ2, is expansive. Define f : S − → S by f (x) = 2x mod 1. Let Σ2 = lim ← −(S, f ) Define the shift homeomorphism f : Σ2 − → Σ2 by

  • f (x) =

f (x1, x2, x3...) = f (x1), f (x2), f (x3), ... = f (x1), x1, x2, .... Also, notice that

  • f −1(x1, x2, x3...) = x2, x3, x4, ....

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Example: The shift homeomorphism of the dyadic solenoid, Σ2, is expansive. Define f : S − → S by f (x) = 2x mod 1. Let Σ2 = lim ← −(S, f ) Define the shift homeomorphism f : Σ2 − → Σ2 by

  • f (x) =

f (x1, x2, x3...) = f (x1), f (x2), f (x3), ... = f (x1), x1, x2, .... Also, notice that

  • f −1(x1, x2, x3...) = x2, x3, x4, ....

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Example: The shift homeomorphism of the dyadic solenoid, Σ2, is expansive. Define f : S − → S by f (x) = 2x mod 1. Let Σ2 = lim ← −(S, f ) Define the shift homeomorphism f : Σ2 − → Σ2 by

  • f (x) =

f (x1, x2, x3...) = f (x1), f (x2), f (x3), ... = f (x1), x1, x2, .... Also, notice that

  • f −1(x1, x2, x3...) = x2, x3, x4, ....

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Example: The shift homeomorphism of the dyadic solenoid, Σ2, is expansive. Define f : S − → S by f (x) = 2x mod 1. Let Σ2 = lim ← −(S, f ) Define the shift homeomorphism f : Σ2 − → Σ2 by

  • f (x) =

f (x1, x2, x3...) = f (x1), f (x2), f (x3), ... = f (x1), x1, x2, .... Also, notice that

  • f −1(x1, x2, x3...) = x2, x3, x4, ....

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Figure: Doubling map f (x) = 2x mod 1.

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Figure: Doubling map f (x) = 2x mod 1.

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Figure: Doubling map f (x) = 2x mod 1.

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Figure: Inverse limit of f (z) is the solenoid Σ2.

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Theorem The shift homeomorphism on Σ2 is expansive. Proof. Let the expansive constant be 1

4.Notice that if x, y ∈ S and

dS(x, y) < 1

4 then dS(f (x), f (y)) = 2dS(x, y).Let x, y be distinct

elements in Σ2. Then there exists i such that xi = yi.Furthermore, there exists a nonnegative natural number n such that

1 4 < 2ndS(xi, yi) ≤ 1/2. Hence,

d( f n−i+1(x), f n−i+1(y)) = d( f n( f −i+1(x)), f n( f −i+1(y))) = d( f n(xi, xi+1, ...), f (yi, yi+1, ...)) > 2ndS(xi, yi) > 1 4.

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Theorem The shift homeomorphism on Σ2 is expansive. Proof. Let the expansive constant be 1

4.Notice that if x, y ∈ S and

dS(x, y) < 1

4 then dS(f (x), f (y)) = 2dS(x, y).Let x, y be distinct

elements in Σ2. Then there exists i such that xi = yi.Furthermore, there exists a nonnegative natural number n such that

1 4 < 2ndS(xi, yi) ≤ 1/2. Hence,

d( f n−i+1(x), f n−i+1(y)) = d( f n( f −i+1(x)), f n( f −i+1(y))) = d( f n(xi, xi+1, ...), f (yi, yi+1, ...)) > 2ndS(xi, yi) > 1 4.

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Theorem The shift homeomorphism on Σ2 is expansive. Proof. Let the expansive constant be 1

4.Notice that if x, y ∈ S and

dS(x, y) < 1

4 then dS(f (x), f (y)) = 2dS(x, y).Let x, y be distinct

elements in Σ2. Then there exists i such that xi = yi.Furthermore, there exists a nonnegative natural number n such that

1 4 < 2ndS(xi, yi) ≤ 1/2. Hence,

d( f n−i+1(x), f n−i+1(y)) = d( f n( f −i+1(x)), f n( f −i+1(y))) = d( f n(xi, xi+1, ...), f (yi, yi+1, ...)) > 2ndS(xi, yi) > 1 4.

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Theorem The shift homeomorphism on Σ2 is expansive. Proof. Let the expansive constant be 1

4.Notice that if x, y ∈ S and

dS(x, y) < 1

4 then dS(f (x), f (y)) = 2dS(x, y).Let x, y be distinct

elements in Σ2. Then there exists i such that xi = yi.Furthermore, there exists a nonnegative natural number n such that

1 4 < 2ndS(xi, yi) ≤ 1/2. Hence,

d( f n−i+1(x), f n−i+1(y)) = d( f n( f −i+1(x)), f n( f −i+1(y))) = d( f n(xi, xi+1, ...), f (yi, yi+1, ...)) > 2ndS(xi, yi) > 1 4.

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Theorem The shift homeomorphism on Σ2 is expansive. Proof. Let the expansive constant be 1

4.Notice that if x, y ∈ S and

dS(x, y) < 1

4 then dS(f (x), f (y)) = 2dS(x, y).Let x, y be distinct

elements in Σ2. Then there exists i such that xi = yi.Furthermore, there exists a nonnegative natural number n such that

1 4 < 2ndS(xi, yi) ≤ 1/2. Hence,

d( f n−i+1(x), f n−i+1(y)) = d( f n( f −i+1(x)), f n( f −i+1(y))) = d( f n(xi, xi+1, ...), f (yi, yi+1, ...)) > 2ndS(xi, yi) > 1 4.

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Theorem The shift homeomorphism on Σ2 is expansive. Proof. Let the expansive constant be 1

4.Notice that if x, y ∈ S and

dS(x, y) < 1

4 then dS(f (x), f (y)) = 2dS(x, y).Let x, y be distinct

elements in Σ2. Then there exists i such that xi = yi.Furthermore, there exists a nonnegative natural number n such that

1 4 < 2ndS(xi, yi) ≤ 1/2. Hence,

d( f n−i+1(x), f n−i+1(y)) = d( f n( f −i+1(x)), f n( f −i+1(y))) = d( f n(xi, xi+1, ...), f (yi, yi+1, ...)) > 2ndS(xi, yi) > 1 4.

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A homeomorphism is continuum-wise expansive if there exists a constant c > 0 such that for any subcontinuum Y of X there exists an integer n such that diam(hn(Y )) > c. Every expansive homeomorphism is continuum-wise expansive, but the converse is not true.

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A homeomorphism is continuum-wise expansive if there exists a constant c > 0 such that for any subcontinuum Y of X there exists an integer n such that diam(hn(Y )) > c. Every expansive homeomorphism is continuum-wise expansive, but the converse is not true.

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A homeomorphism is positively continuum-wise fully expansive if for every pair ǫ, δ > 0 there is a N(ǫ, δ) > 0 such that if Y is a subcontinuum of X with diam(Y ) ≥ δ, then dH(hn(Y ), X) < ǫ for all n ≥ N(ǫ, δ).

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Question What properties must a continuum have in order for a continuum to admit and expansive (continuum-wise expansive) homeomorphism? Well, in order for a continuum to admit an expansive (continuum-wise) homeomorphism, all of the proper subcontinuum must be continuous stretched and there must be room for this to

  • happen. This is how indecomposable continua are created.

Theorem (Kato) If a G-like continuum admits an expansive homeomorphism, then it must contain an indecomposable subcontinuum.

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Question What properties must a continuum have in order for a continuum to admit and expansive (continuum-wise expansive) homeomorphism? Well, in order for a continuum to admit an expansive (continuum-wise) homeomorphism, all of the proper subcontinuum must be continuous stretched and there must be room for this to

  • happen. This is how indecomposable continua are created.

Theorem (Kato) If a G-like continuum admits an expansive homeomorphism, then it must contain an indecomposable subcontinuum.

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Question What properties must a continuum have in order for a continuum to admit and expansive (continuum-wise expansive) homeomorphism? Well, in order for a continuum to admit an expansive (continuum-wise) homeomorphism, all of the proper subcontinuum must be continuous stretched and there must be room for this to

  • happen. This is how indecomposable continua are created.

Theorem (Kato) If a G-like continuum admits an expansive homeomorphism, then it must contain an indecomposable subcontinuum.

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Theorem (M.) If a k-cyclic continuum admits an expansive homeomorphism, then it must contain an indecomposable subcontinuum. Theorem (Kato) If a continuum admits a positively continuum-wise expansive homeomorphism, then the continuum must be indecomposable. Question Suppose that X is a one-dimensional continuum that admits an expansive homeomorphism, must X be indecomposable? No!

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Theorem (M.) If a k-cyclic continuum admits an expansive homeomorphism, then it must contain an indecomposable subcontinuum. Theorem (Kato) If a continuum admits a positively continuum-wise expansive homeomorphism, then the continuum must be indecomposable. Question Suppose that X is a one-dimensional continuum that admits an expansive homeomorphism, must X be indecomposable? No!

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Theorem (M.) If a k-cyclic continuum admits an expansive homeomorphism, then it must contain an indecomposable subcontinuum. Theorem (Kato) If a continuum admits a positively continuum-wise expansive homeomorphism, then the continuum must be indecomposable. Question Suppose that X is a one-dimensional continuum that admits an expansive homeomorphism, must X be indecomposable? No!

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Theorem (M.) If a k-cyclic continuum admits an expansive homeomorphism, then it must contain an indecomposable subcontinuum. Theorem (Kato) If a continuum admits a positively continuum-wise expansive homeomorphism, then the continuum must be indecomposable. Question Suppose that X is a one-dimensional continuum that admits an expansive homeomorphism, must X be indecomposable? No!

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Define f : S − → S by f (x) = 2x mod 1. Let Σ2 = lim ← −(S, f ) Then Σ2 is the dyadic solenoid and as we said before, the shift homeomorphism is expansive. Let S be the unit circle with 1 sticker in the complex plane and let

  • f :

S − → S in the following way:

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Figure: Doubling and stretch map f (x).

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Figure: Doubling and stretch map f (x).

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Figure: Doubling and stretch map f (x).

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Figure: Inverse limit of f (x) is a ray limiting to the soleniod

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Expansive Homeomorphisms of Plane Continua

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Question Does there exists an expansive homeomorphism of a plane continuum? Yes, the Plykin Attractor is a one dimensional plane continuum that admits an expansive homeomorphism.

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Question Does there exists an expansive homeomorphism of a plane continuum? Yes, the Plykin Attractor is a one dimensional plane continuum that admits an expansive homeomorphism.

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P

Figure: Plykin attractor admits an expansive homeomorphism

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However, the Plykin Attractor is a 1-dimensional 4-separating plane continuum that admits an expansive homeomorphism. Question Does there exist an 1-dimensional plane separating continuum that admits an expansive homeomorphism? No!

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However, the Plykin Attractor is a 1-dimensional 4-separating plane continuum that admits an expansive homeomorphism. Question Does there exist an 1-dimensional plane separating continuum that admits an expansive homeomorphism? No!

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All 1-dimensional non-separating plane continua are tree-like. (Converse is not true.) Theorem (M.) Tree-like continua do not admit expansive homeomorphisms. The proof of this result contains many important ideas and techniques, so it will be valuable to examine it.

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All 1-dimensional non-separating plane continua are tree-like. (Converse is not true.) Theorem (M.) Tree-like continua do not admit expansive homeomorphisms. The proof of this result contains many important ideas and techniques, so it will be valuable to examine it.

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Let h : X − → X be a homeomorphism. M is an unstable subcontinuum of h if diam(hn(M)) → 0 as n → −∞. M is an stable subcontinuum of h if diam(hn(M)) → 0 as n → ∞.

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Let h : X − → X be a homeomorphism. M is an unstable subcontinuum of h if diam(hn(M)) → 0 as n → −∞. M is an stable subcontinuum of h if diam(hn(M)) → 0 as n → ∞.

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Let h : X − → X be a homeomorphism. M is an unstable subcontinuum of h if diam(hn(M)) → 0 as n → −∞. M is an stable subcontinuum of h if diam(hn(M)) → 0 as n → ∞.

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Theorem (Kato) If h : X − → X is an continuum-wise expansive homeomorphism of a continuum, then there exists a stable or unstable subcontinuum. Since h is expansive if and only if h−1 is expansive, we will always assume the existence of an unstable subcontinuum.

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Theorem (Kato) If h : X − → X is an continuum-wise expansive homeomorphism of a continuum, then there exists a stable or unstable subcontinuum. Since h is expansive if and only if h−1 is expansive, we will always assume the existence of an unstable subcontinuum.

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Let h : X − → X be a homeomorphism of a continuum X. Define dn

k(x, y) = max{d(hi(x), hi(y)) : k ≤ i ≤ n}.

And define dn

−∞(x, y) = sup{d(hi(x), hi(y)) : −∞ < i ≤ n}.

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Let h : X − → X be a homeomorphism of a continuum X. Define dn

k(x, y) = max{d(hi(x), hi(y)) : k ≤ i ≤ n}.

And define dn

−∞(x, y) = sup{d(hi(x), hi(y)) : −∞ < i ≤ n}.

Christopher G. Mouron On the classification of one dimensional continua that admit expansive

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Lemma Let h : X − → X be a homeomorphism of a compact space X. Suppose that 0 < ǫ < c and for each n ∈ N there exists points xn, yn ∈ X such that ǫ/3 ≤ d(xn, yn) and dn

−n(xn, yn) < ǫ.

Then c cannot be an expansive constant.

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Proof. There exist converging subsequences {xn(i)}∞

i=1 → x and

{yn(i)}∞

i=1 → y.

Since d(xn(i), yn(i)) ≥ ǫ/3, x and y must be distinct. Since {n(i)}∞

i=1 is strictly increasing, it follows that given k ∈ Z,

then −n(i) ≤ k ≤ n(i) for all i ≥ |k|. So d(hk(xn(i)), hk(yn(i))) < ǫ for all i ≥ |k|. Thus, d(hk(x), hk(y)) ≤ ǫ < c for all k ∈ Z.

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Proof. There exist converging subsequences {xn(i)}∞

i=1 → x and

{yn(i)}∞

i=1 → y.

Since d(xn(i), yn(i)) ≥ ǫ/3, x and y must be distinct. Since {n(i)}∞

i=1 is strictly increasing, it follows that given k ∈ Z,

then −n(i) ≤ k ≤ n(i) for all i ≥ |k|. So d(hk(xn(i)), hk(yn(i))) < ǫ for all i ≥ |k|. Thus, d(hk(x), hk(y)) ≤ ǫ < c for all k ∈ Z.

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Proof. There exist converging subsequences {xn(i)}∞

i=1 → x and

{yn(i)}∞

i=1 → y.

Since d(xn(i), yn(i)) ≥ ǫ/3, x and y must be distinct. Since {n(i)}∞

i=1 is strictly increasing, it follows that given k ∈ Z,

then −n(i) ≤ k ≤ n(i) for all i ≥ |k|. So d(hk(xn(i)), hk(yn(i))) < ǫ for all i ≥ |k|. Thus, d(hk(x), hk(y)) ≤ ǫ < c for all k ∈ Z.

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Proof. There exist converging subsequences {xn(i)}∞

i=1 → x and

{yn(i)}∞

i=1 → y.

Since d(xn(i), yn(i)) ≥ ǫ/3, x and y must be distinct. Since {n(i)}∞

i=1 is strictly increasing, it follows that given k ∈ Z,

then −n(i) ≤ k ≤ n(i) for all i ≥ |k|. So d(hk(xn(i)), hk(yn(i))) < ǫ for all i ≥ |k|. Thus, d(hk(x), hk(y)) ≤ ǫ < c for all k ∈ Z.

Christopher G. Mouron On the classification of one dimensional continua that admit expansive

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Proof. There exist converging subsequences {xn(i)}∞

i=1 → x and

{yn(i)}∞

i=1 → y.

Since d(xn(i), yn(i)) ≥ ǫ/3, x and y must be distinct. Since {n(i)}∞

i=1 is strictly increasing, it follows that given k ∈ Z,

then −n(i) ≤ k ≤ n(i) for all i ≥ |k|. So d(hk(xn(i)), hk(yn(i))) < ǫ for all i ≥ |k|. Thus, d(hk(x), hk(y)) ≤ ǫ < c for all k ∈ Z.

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Lemma Suppose

1 T is a tree-cover of continuum X 2 a and b are elements of X that are in the same element of T

such that dn

k(a, b) ≥ ǫ

Then there exists xα, xβ ∈ X such that ǫ/3 ≤ dn

k(xα, xβ) < ǫ and

xα, xβ are in the same element of T .

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Figure: Tree cover of X and unstable subcontinuum M.

Christopher G. Mouron On the classification of one dimensional continua that admit expansive

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Figure: Simple chain from a to b such that the distance between consecutive points is less that ǫ/3 under dn

k.

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Figure: We only need to consider the simple chain and not the subcontinuum.

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b

a

6

/

ε

distance is less than

Figure: Simple chain from a to b such that the distance between consecutive points is less that ǫ/3 under dn

k.

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα, xβ) ≥ ǫ or ǫ > dn k(xα, xβ) ≥ ǫ/3. If it is the

latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−1, xβ+1) ≥ ǫ or ǫ > dn k(xα−1, xβ+1) ≥ ǫ/3. If

it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−2, xβ+2) ≥ ǫ or ǫ > dn k(xα−2, xβ+2) ≥ ǫ/3. If

it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−3, xβ+3) ≥ ǫ or ǫ > dn k(xα−3, xβ+3) ≥ ǫ/3. If

it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−4, xβ+4) ≥ ǫ or ǫ > dn k(xα−4, xβ+4) ≥ ǫ/3. If

it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−5, xβ+5) ≥ ǫ or ǫ > dn k(xα−5, xβ+5) ≥ ǫ/3. If

it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: If this the case we can use the triangle inequality!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−6, xβ+γ) ≥ ǫ or ǫ > dn k(xα−6, xβ+γ) ≥ ǫ/3. If

it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−7, xβ+γ+1) ≥ ǫ or

ǫ > dn

k(xα−7, xβ+γ+1) ≥ ǫ/3. If it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−8, xβ+γ+2) ≥ ǫ or

ǫ > dn

k(xα−8, xβ+γ+2) ≥ ǫ/3. If it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−9, xβ+γ+3) ≥ ǫ or

ǫ > dn

k(xα−9, xβ+γ+3) ≥ ǫ/3. If it is the latter, we are done!

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b

a

6

/

ε

distance is less than

Figure: Hence either dn

k(xα−10, xβ+γ+4) ≥ ǫ or

ǫ > dn

k(xα−10, xβ+γ+4) ≥ ǫ/3. Oops! Contradiction!

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Lemma Suppose

1 T is a tree-cover of continuum X 2 a and b are elements of X that are in the same element of T

such that dn

k(a, b) ≥ ǫ

Then there exists xα, xβ ∈ X such that ǫ/3 ≤ dn

k(xα, xβ) < ǫ and

xα, xβ are in the same element of T .

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Theorem Let h : X − → X be a homeomorphism and 0 < ǫ < c. Suppose that M is an unstable subcontinuum of h such that for every δ > 0 there exist an integer k = k(δ) and a tree-cover Tk of hk(M) with the following properties:

1 mesh(Tk) < δ 2 there exist points xk, yk ∈ hk(M) that are in the same

element of Tk

3 d0

−∞(xk, yk) > ǫ.

Then c cannot be an expansive constant for h.

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Proof. Since subcontinua of unstable subcontinua are unstable, we may choose M such that diam(hi(M)) < ǫ/2 for all i ≤ 0. Choose δk such that if d(x, y) < δk then d(hi(x), hi(y)) < ǫ for all 0 ≤ i ≤ k. By (3) and a previous Lemma, there exists ˆ xk, ˆ yk ∈ hk(M) such that ˆ xk, ˆ yk are in the same element of Tk and ǫ/3 ≤ d0

−k(ˆ

xk, ˆ yk) < ǫ.

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Proof. Since subcontinua of unstable subcontinua are unstable, we may choose M such that diam(hi(M)) < ǫ/2 for all i ≤ 0. Choose δk such that if d(x, y) < δk then d(hi(x), hi(y)) < ǫ for all 0 ≤ i ≤ k. By (3) and a previous Lemma, there exists ˆ xk, ˆ yk ∈ hk(M) such that ˆ xk, ˆ yk are in the same element of Tk and ǫ/3 ≤ d0

−k(ˆ

xk, ˆ yk) < ǫ.

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Proof. Since subcontinua of unstable subcontinua are unstable, we may choose M such that diam(hi(M)) < ǫ/2 for all i ≤ 0. Choose δk such that if d(x, y) < δk then d(hi(x), hi(y)) < ǫ for all 0 ≤ i ≤ k. By (3) and a previous Lemma, there exists ˆ xk, ˆ yk ∈ hk(M) such that ˆ xk, ˆ yk are in the same element of Tk and ǫ/3 ≤ d0

−k(ˆ

xk, ˆ yk) < ǫ.

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Proof. Since d(ˆ xk, ˆ yk) < δk, it follows that ǫ/2 < dk

−∞(ˆ

xk, ˆ yk) < ǫ. Let i ≤ 0 be such that d(hi(ˆ xk), hi(ˆ yk)) ≥ ǫ/3 and define wk = hi(ˆ xk) and zk = hi(ˆ yk). Then ǫ/3 ≤ d(wk, zk) and dk

−k(wk, zk) < ǫ.

Hence, c cannot be an expansive constant by the previous Lemma.

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Proof. Since d(ˆ xk, ˆ yk) < δk, it follows that ǫ/2 < dk

−∞(ˆ

xk, ˆ yk) < ǫ. Let i ≤ 0 be such that d(hi(ˆ xk), hi(ˆ yk)) ≥ ǫ/3 and define wk = hi(ˆ xk) and zk = hi(ˆ yk). Then ǫ/3 ≤ d(wk, zk) and dk

−k(wk, zk) < ǫ.

Hence, c cannot be an expansive constant by the previous Lemma.

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Proof. Since d(ˆ xk, ˆ yk) < δk, it follows that ǫ/2 < dk

−∞(ˆ

xk, ˆ yk) < ǫ. Let i ≤ 0 be such that d(hi(ˆ xk), hi(ˆ yk)) ≥ ǫ/3 and define wk = hi(ˆ xk) and zk = hi(ˆ yk). Then ǫ/3 ≤ d(wk, zk) and dk

−k(wk, zk) < ǫ.

Hence, c cannot be an expansive constant by the previous Lemma.

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Proof. Since d(ˆ xk, ˆ yk) < δk, it follows that ǫ/2 < dk

−∞(ˆ

xk, ˆ yk) < ǫ. Let i ≤ 0 be such that d(hi(ˆ xk), hi(ˆ yk)) ≥ ǫ/3 and define wk = hi(ˆ xk) and zk = hi(ˆ yk). Then ǫ/3 ≤ d(wk, zk) and dk

−k(wk, zk) < ǫ.

Hence, c cannot be an expansive constant by the previous Lemma.

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Corollary Tree-like continua do not admit expansive homeomorphisms.

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Proof. Suppose on the contrary that h : X − → X is expansive with expansive constant c. Let 0 < ǫ < c. We may assume that M is an unstable subcontinuum of h. Let δ > 0 and Tδ be a tree-cover of X with mesh less than δ. Let Aδ be |Tδ| + 1 points of M. Since h is expansive, there exists a k such that dk

∞(x, y) > c for all

distinct x, y ∈ Aδ. By the pigeon-hole principal there exists distinct x′, y′ ∈ Aδ such that w = hk(x′) and z = hk(y′) are in the same element of Tδ (Notice dk

∞(w, z) > c > ǫ).

Thus by the previous Theorem, c cannot be an expansive constant.

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Proof. Suppose on the contrary that h : X − → X is expansive with expansive constant c. Let 0 < ǫ < c. We may assume that M is an unstable subcontinuum of h. Let δ > 0 and Tδ be a tree-cover of X with mesh less than δ. Let Aδ be |Tδ| + 1 points of M. Since h is expansive, there exists a k such that dk

∞(x, y) > c for all

distinct x, y ∈ Aδ. By the pigeon-hole principal there exists distinct x′, y′ ∈ Aδ such that w = hk(x′) and z = hk(y′) are in the same element of Tδ (Notice dk

∞(w, z) > c > ǫ).

Thus by the previous Theorem, c cannot be an expansive constant.

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Proof. Suppose on the contrary that h : X − → X is expansive with expansive constant c. Let 0 < ǫ < c. We may assume that M is an unstable subcontinuum of h. Let δ > 0 and Tδ be a tree-cover of X with mesh less than δ. Let Aδ be |Tδ| + 1 points of M. Since h is expansive, there exists a k such that dk

∞(x, y) > c for all

distinct x, y ∈ Aδ. By the pigeon-hole principal there exists distinct x′, y′ ∈ Aδ such that w = hk(x′) and z = hk(y′) are in the same element of Tδ (Notice dk

∞(w, z) > c > ǫ).

Thus by the previous Theorem, c cannot be an expansive constant.

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Proof. Suppose on the contrary that h : X − → X is expansive with expansive constant c. Let 0 < ǫ < c. We may assume that M is an unstable subcontinuum of h. Let δ > 0 and Tδ be a tree-cover of X with mesh less than δ. Let Aδ be |Tδ| + 1 points of M. Since h is expansive, there exists a k such that dk

∞(x, y) > c for all

distinct x, y ∈ Aδ. By the pigeon-hole principal there exists distinct x′, y′ ∈ Aδ such that w = hk(x′) and z = hk(y′) are in the same element of Tδ (Notice dk

∞(w, z) > c > ǫ).

Thus by the previous Theorem, c cannot be an expansive constant.

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Proof. Suppose on the contrary that h : X − → X is expansive with expansive constant c. Let 0 < ǫ < c. We may assume that M is an unstable subcontinuum of h. Let δ > 0 and Tδ be a tree-cover of X with mesh less than δ. Let Aδ be |Tδ| + 1 points of M. Since h is expansive, there exists a k such that dk

∞(x, y) > c for all

distinct x, y ∈ Aδ. By the pigeon-hole principal there exists distinct x′, y′ ∈ Aδ such that w = hk(x′) and z = hk(y′) are in the same element of Tδ (Notice dk

∞(w, z) > c > ǫ).

Thus by the previous Theorem, c cannot be an expansive constant.

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Proof. Suppose on the contrary that h : X − → X is expansive with expansive constant c. Let 0 < ǫ < c. We may assume that M is an unstable subcontinuum of h. Let δ > 0 and Tδ be a tree-cover of X with mesh less than δ. Let Aδ be |Tδ| + 1 points of M. Since h is expansive, there exists a k such that dk

∞(x, y) > c for all

distinct x, y ∈ Aδ. By the pigeon-hole principal there exists distinct x′, y′ ∈ Aδ such that w = hk(x′) and z = hk(y′) are in the same element of Tδ (Notice dk

∞(w, z) > c > ǫ).

Thus by the previous Theorem, c cannot be an expansive constant.

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Proof. Suppose on the contrary that h : X − → X is expansive with expansive constant c. Let 0 < ǫ < c. We may assume that M is an unstable subcontinuum of h. Let δ > 0 and Tδ be a tree-cover of X with mesh less than δ. Let Aδ be |Tδ| + 1 points of M. Since h is expansive, there exists a k such that dk

∞(x, y) > c for all

distinct x, y ∈ Aδ. By the pigeon-hole principal there exists distinct x′, y′ ∈ Aδ such that w = hk(x′) and z = hk(y′) are in the same element of Tδ (Notice dk

∞(w, z) > c > ǫ).

Thus by the previous Theorem, c cannot be an expansive constant.

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The Plykin Attractor is 4-separating and admits an expansive homeomorphism. One dimensional non-separating plane continua do not admit expansive homeomorphisms. What can be said about 2-separating and 3-separating plane continua? To examine this, I am going to generalize the previous techniques.

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The Plykin Attractor is 4-separating and admits an expansive homeomorphism. One dimensional non-separating plane continua do not admit expansive homeomorphisms. What can be said about 2-separating and 3-separating plane continua? To examine this, I am going to generalize the previous techniques.

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Exponental Wrapping and Fully Expansiveness

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Figure: Let X be a continuum, Y be a tree-like subcontinuum and U be a finite open cover of X. Then define T(Y , U) to be a tree cover of Y of minimal cardinality that refines U.

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Figure: Let X be a continuum, Y be a tree-like subcontinuum and U be a finite open cover of X. Then define T(Y , U) to be a tree cover of Y of minimal cardinality that refines U.

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Figure: Let X be a continuum, Y be a tree-like subcontinuum and U be a finite open cover of X. Then define T(Y , U) to be a tree cover of Y of minimal cardinality that refines U.

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Figure: Let X be a continuum, Y be a tree-like subcontinuum and U be a finite open cover of X. Then define T(Y , U) to be a tree cover of Y of minimal cardinality that refines U.

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Figure: Likewise, if T is a tree cover that refines U, then define T(T , U) to be a tree cover of minimal cardinality that refines U and is refined by T .

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Figure: Likewise, if T is a tree cover that refines U, then define T(T , U) to be a tree cover of minimal cardinality that refines U and is refined by T .

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Lemma Let h : X − → X be a homeomorphism, and M be an unstable subcontinuum for h. Suppose that for every δ > 0 there exists

1 a finite open cover Uδ of X 2 c, k > 0 3 Eδ ⊂ M

such that

1 dk

−∞(x, y) > c for all distinct x, y ∈ Eδ

2 |T(hk(M), Uδ)| < |Eδ|.

Then h is not an expansive homeomorphism. Proof. This follows from a similar “pigeon-hole principal” argument as the tree-like result.

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Lemma Let h : X − → X be a homeomorphism, and M be an unstable subcontinuum for h. Suppose that for every δ > 0 there exists

1 a finite open cover Uδ of X 2 c, k > 0 3 Eδ ⊂ M

such that

1 dk

−∞(x, y) > c for all distinct x, y ∈ Eδ

2 |T(hk(M), Uδ)| < |Eδ|.

Then h is not an expansive homeomorphism. Proof. This follows from a similar “pigeon-hole principal” argument as the tree-like result.

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Combining this with the following theorem: Theorem (Kato) Let h : X − → X be a continuum-wise expansive homeomorphism and M be an unstable subcontinuum. Then there exists p > 1 and a collection of subsets {En}∞

n=1 of M

such that dn

−∞(x, y) > c

for all distinct x, y ∈ En and |En| > pn. We get the following corollary Corollary Let h : X − → X be an expansive homeomorphism, and M be an unstable subcontinuum for h. Then there exists p > 1 such that |T(hk(M), Uδ)| ≥ pk.

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Combining this with the following theorem: Theorem (Kato) Let h : X − → X be a continuum-wise expansive homeomorphism and M be an unstable subcontinuum. Then there exists p > 1 and a collection of subsets {En}∞

n=1 of M

such that dn

−∞(x, y) > c

for all distinct x, y ∈ En and |En| > pn. We get the following corollary Corollary Let h : X − → X be an expansive homeomorphism, and M be an unstable subcontinuum for h. Then there exists p > 1 such that |T(hk(M), Uδ)| ≥ pk.

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Corollary Let h : X − → X be an expansive homeomorphism, and M be an unstable subcontinuum for h. Then there exists p > 1 such that |T(hk(M), Uδ)| ≥ pk. That is unstable tree-like subcontinua must wrap (not fold) in the continuum at an exponential rate!

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Corollary Suppose X is a 1-dimensional continuum that separates the plane into 2 complementary domains. Then X does not admit an expansive homeomorphism. Proof. First, every unstable subcontinuum M must be tree-like. It can be shown (with much work) that if U is a finite open cover of X, then |T(hk(M), U)| has a polynomial growth rate. This is due to the fact that indecomposable 2-separating must be created by more folding than wrapping (as in the solenoid.)

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Corollary Suppose X is a 1-dimensional continuum that separates the plane into 2 complementary domains. Then X does not admit an expansive homeomorphism. Proof. First, every unstable subcontinuum M must be tree-like. It can be shown (with much work) that if U is a finite open cover of X, then |T(hk(M), U)| has a polynomial growth rate. This is due to the fact that indecomposable 2-separating must be created by more folding than wrapping (as in the solenoid.)

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Figure: Although there is some wrapping in a 2-separating plane continuum, it can be shown that there must be “more” bending.

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Conjecture 1-dimensional 3-separating plane continua do not admit expansive homeomorphisms. It appears that there is a “little bit more” bending than wrapping in 1-dimensional 3-separating plane continua. I have some new techniques to “measure ” this bending but currently it is very technical.

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However, there is more we can say: Let X be a continuum. Y is a minimal cyclic subcontinuum of X if Y is not tree-like but every proper subcontinuum is tree-like. Theorem If h : X − → X is a expansive homeomorphism of a minimally cyclic continuum X, then h (or h−1) is positively continuum-wise fully expansive. This follows from the fact that unstable subcontinua must wrap more and more in X. Since X is minimally cyclic, they converge to X in the Hausdorff metric. With a little more work, it can be shown that every subcontinuum has this property.

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However, there is more we can say: Let X be a continuum. Y is a minimal cyclic subcontinuum of X if Y is not tree-like but every proper subcontinuum is tree-like. Theorem If h : X − → X is a expansive homeomorphism of a minimally cyclic continuum X, then h (or h−1) is positively continuum-wise fully expansive. This follows from the fact that unstable subcontinua must wrap more and more in X. Since X is minimally cyclic, they converge to X in the Hausdorff metric. With a little more work, it can be shown that every subcontinuum has this property.

Christopher G. Mouron On the classification of one dimensional continua that admit expansiv

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However, there is more we can say: Let X be a continuum. Y is a minimal cyclic subcontinuum of X if Y is not tree-like but every proper subcontinuum is tree-like. Theorem If h : X − → X is a expansive homeomorphism of a minimally cyclic continuum X, then h (or h−1) is positively continuum-wise fully expansive. This follows from the fact that unstable subcontinua must wrap more and more in X. Since X is minimally cyclic, they converge to X in the Hausdorff metric. With a little more work, it can be shown that every subcontinuum has this property.

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Also, if X is a k-cyclic continuum, then it has a finite number of minimally cyclic subcontinua. Hence if h : X → X is a homeomorphism and Y is a minimally cyclic subcontinuum, then there exists a k such that hk(Y ) = Y . Note: hk is expansive if and only if h is expansive. So, Theorem Suppose that h : X − → X is a expansive homeomorphism of a k-cyclic continuum. Then there exists k ∈ N and a subcontinuum Y such that hk|Y : Y − → Y is fully expansive. Furthermore, Y is indecomposable and minimally cyclic. A fully expansive homeomorphism is one that is expansive and either h or h−1 is positively continuum-wise fully expansive.

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Also, if X is a k-cyclic continuum, then it has a finite number of minimally cyclic subcontinua. Hence if h : X → X is a homeomorphism and Y is a minimally cyclic subcontinuum, then there exists a k such that hk(Y ) = Y . Note: hk is expansive if and only if h is expansive. So, Theorem Suppose that h : X − → X is a expansive homeomorphism of a k-cyclic continuum. Then there exists k ∈ N and a subcontinuum Y such that hk|Y : Y − → Y is fully expansive. Furthermore, Y is indecomposable and minimally cyclic. A fully expansive homeomorphism is one that is expansive and either h or h−1 is positively continuum-wise fully expansive.

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Figure: Here, the solenoid is minimally cyclic and the restriction homeomorphism to the solenoid is fully expansive.

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Questions

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Question If h : X − → X is a fully expansive homemorphism of k-cyclic continuum, then is X is homeomorphic to the inverse limit of the bouquet of circles? Note: Theorem If X is a circle-like continuum that admits an expansive homeomorphism, then X is a solenoid formed by the same bonding map.

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Question If h : X − → X is a fully expansive homemorphism of k-cyclic continuum, then is X is homeomorphic to the inverse limit of the bouquet of circles? Note: Theorem If X is a circle-like continuum that admits an expansive homeomorphism, then X is a solenoid formed by the same bonding map.

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This leads to more questions: Question Suppose X is a k-cyclic continuum that admits an expansive

  • homeomorphism. Is X the inverse limit of the same graph G and

same bonding map f : G − → G such that the shift homeomorphism of f is fully expansive? Question Suppose X is a k-cyclic continuum that admits an expansive

  • homeomorphism. Is X the union of bouquet-like continua and a

finite number of rays limiting to these continua? Question If X is 1-dimensional continuum that admits an expansive

  • homeomorphism. Must X contain a k-cyclic continuum that

admits a fully expansive homeomorphism? Need to only consider infinite-cyclic continua.

Christopher G. Mouron On the classification of one dimensional continua that admit expan

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This leads to more questions: Question Suppose X is a k-cyclic continuum that admits an expansive

  • homeomorphism. Is X the inverse limit of the same graph G and

same bonding map f : G − → G such that the shift homeomorphism of f is fully expansive? Question Suppose X is a k-cyclic continuum that admits an expansive

  • homeomorphism. Is X the union of bouquet-like continua and a

finite number of rays limiting to these continua? Question If X is 1-dimensional continuum that admits an expansive

  • homeomorphism. Must X contain a k-cyclic continuum that

admits a fully expansive homeomorphism? Need to only consider infinite-cyclic continua.

Christopher G. Mouron On the classification of one dimensional continua that admit expan

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This leads to more questions: Question Suppose X is a k-cyclic continuum that admits an expansive

  • homeomorphism. Is X the inverse limit of the same graph G and

same bonding map f : G − → G such that the shift homeomorphism of f is fully expansive? Question Suppose X is a k-cyclic continuum that admits an expansive

  • homeomorphism. Is X the union of bouquet-like continua and a

finite number of rays limiting to these continua? Question If X is 1-dimensional continuum that admits an expansive

  • homeomorphism. Must X contain a k-cyclic continuum that

admits a fully expansive homeomorphism? Need to only consider infinite-cyclic continua.

Christopher G. Mouron On the classification of one dimensional continua that admit expan

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Question Suppose that h : X − → X is an expansive homeomorphism of a k-cyclic continuum (or a G-like continuum). Does there exist a graph H and a map f : H − → H such that

1 X is homeomorphic to Y = lim

← −(H, f )

2 the shift homeomorphism

f of lim ← −(H, f ) is expansive

3 there exists a map φ : X −

→ Y such that f ◦ φ = φ ◦ h?

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Figure 5 P_2 P_1 X

Figure: 2-dimensional plane continuum that admits an expansive homeomorphism

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Question Does there exist a 2-dimensional non-separating plane continuum that admits an expansive homeomorphism? Question Does there exists a plane continuum that admits an expansive homeomorphism and separates the plane into an infinite number

  • f complementary domains.

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Question Does there exist a 2-dimensional non-separating plane continuum that admits an expansive homeomorphism? Question Does there exists a plane continuum that admits an expansive homeomorphism and separates the plane into an infinite number

  • f complementary domains.

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Question Does there exist an infinite-cyclic continuum that admits an expansive homeomorphism? In particular, does the Menger curve admit an expansive homeomorphism? Also, must it contain an indecomposable subcontinuum Question If continuum X admits an expansive homeomorphism, must X be the union of arcs? A continuum is hereditarily equivalent if it is homeomorphic to each of its proper nondegenerate subcontinua. (These are arcs and psuedo-arcs.) Question If continuum X admits a continuum-wise expansive homeomorphism, must X be the union of hereditarily equivalent continua?

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Question Does there exist an infinite-cyclic continuum that admits an expansive homeomorphism? In particular, does the Menger curve admit an expansive homeomorphism? Also, must it contain an indecomposable subcontinuum Question If continuum X admits an expansive homeomorphism, must X be the union of arcs? A continuum is hereditarily equivalent if it is homeomorphic to each of its proper nondegenerate subcontinua. (These are arcs and psuedo-arcs.) Question If continuum X admits a continuum-wise expansive homeomorphism, must X be the union of hereditarily equivalent continua?

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Question Does there exist an infinite-cyclic continuum that admits an expansive homeomorphism? In particular, does the Menger curve admit an expansive homeomorphism? Also, must it contain an indecomposable subcontinuum Question If continuum X admits an expansive homeomorphism, must X be the union of arcs? A continuum is hereditarily equivalent if it is homeomorphic to each of its proper nondegenerate subcontinua. (These are arcs and psuedo-arcs.) Question If continuum X admits a continuum-wise expansive homeomorphism, must X be the union of hereditarily equivalent continua?

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Question Does there exist an infinite-cyclic continuum that admits an expansive homeomorphism? In particular, does the Menger curve admit an expansive homeomorphism? Also, must it contain an indecomposable subcontinuum Question If continuum X admits an expansive homeomorphism, must X be the union of arcs? A continuum is hereditarily equivalent if it is homeomorphic to each of its proper nondegenerate subcontinua. (These are arcs and psuedo-arcs.) Question If continuum X admits a continuum-wise expansive homeomorphism, must X be the union of hereditarily equivalent continua?

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Thank You

Christopher G. Mouron On the classification of one dimensional continua that admit expansive