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A research article published in the american mathematical society, which discusses the properties of expansive homeomorphisms and their relationship to tree-like continua. The article shows that tree-like continua do not admit expansive homeomorphisms, and provides proof for this result. Expansive homeomorphisms are a type of continuous function that stretches subcontinua, and are known to exhibit chaotic behavior. The document also mentions examples of continua that do admit expansive homeomorphisms, such as plykin's attractors and the dyadic solenoid.
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PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 130, Number 11, Pages 3409– S 0002-9939(02)06447-X Article electronically published on May 14, 2002
CHRISTOPHER MOURON
(Communicated by Alan Dow)
Abstract. A homeomorphism h : X → X is called expansive provided that for some fixed c > 0 and every x, y ∈ X there exists an integer n, dependent only on x and y, such that d(hn(x), hn(y)) > c. It is shown that if X is a tree-like continuum, then h cannot be expansive.
Received by the editors August 16, 2000 and, in revised form, June 21, 2001. 2000 Mathematics Subject Classification. Primary 54H20, 54F50; Secondary 54E40. Key words and phrases. Expansive homeomorphism, tree-like continua. The author is pleased to acknowledge the many useful comments and suggestions made by Charles Hagopian.
© c2002 American Mathematical Society 3409
3410 C. MOURON
of U. A planar continuum X is a non-separating plane continuum provided that R^2 − X is connected. It is important to note that all 1-dimensional non-separating plane continua are tree-like. However, not all tree-like continua can be embedded in the plane. In order for a homeomorphism h to be expansive, h must stretch subcontinua. Since compactness must be preserved, these subcontinua must either be stretched and wrapped or stretched and folded. If a continuum is tree-like, some folding must occur. In this paper, we will see that stretching and folding is not enough to produce an expansive homeomorphism since folding also pushes points closer together. However, stretching and folding is enough to produce continuum-wise expansive homeomorphisms, as there are several examples of arc-like and tree-like continua that admit continuum-wise expansive homeomorphisms [2]. In [3], Kato has shown that arc-like continua do not admit expansive home- omorphisms by first showing that the pseudo-arc does not admit an expansive homeomorphism and then by lifting a homeomorphism of an arc-like continuum to a homeomorphism of the pseudo-arc. Unfortunately, these techniques cannot be extended to tree-like continua. In F.W. Worth’s Dissertation, it was shown that shift homeomorphisms from the inverse limit of tree graphs cannot be expansive homeomorphisms [7], and Kato has also shown that no hereditarily decomposable tree-like continuum can admit an expansive homeomorphism. In the sequel, these results are generalized and it is shown that no tree-like continuum can admit an expansive homeomorphism.
such that d(xkα, xkβ ) < γk, d(hi(xkα), hi(xkβ )) < for all i ≤ 0, and there exists an integer n, 0 ≤ n ≤ k, such that / 3 ≤ d(hn(xkα), hn(xkβ )) < . Let yk = hn(xkα) and
zk = hn(xkβ ).
Proposition 1. Suppose X is connected and a, b ∈ X. For every > 0 there exists a finite sequence {xi}ni=1 ⊂ X such that x 1 = a, xn = b, and d(xi, xi+1) < .
The previous sequence is called a simple chain sequence from a to b with mesh less than . The next theorem is due to Kato [2].
3412 C. MOURON
Case 3. ≤ Qn(xα 1 +1, xβ 1 − 1 ) and xα 1 +1 and xβ 1 − 1 are in same element of T , say T 2. Then W (T 2 , {xi}β α^11 − +1^1 ) = {xα 2 , xβ 2 }, where α 2 < β 2. Suppose T 2 ,... , Tj and xα 2 , xβ 2 ,... , xαj , xβj have been found, again we have 3 cases to consider: Case 1-j. Qn(xαj +1, xβj − 1 ) < and xαj +1, xβj − 1 are in the same element of T. As in Case 1, this implies that we are done. Case 2-j. xαj +1 and xβj − 1 are not contained in the same element of T. As in Case 2, this implies that we are done. Case 3-j. ≤ Qn(xαj +1, xβj − 1 ) and xαj +1 and xβj − 1 are in the same element of T , say Tj+1.
Then W (Tj+1, {xi}β i=j^ −α^1 j +1) = {xαj+1 , xβj+1 }, where αj+1 < βj+1, and the induc- tion continues. Eventually, the induction must stop at some j 1. Otherwise, since αj+1 > αj and βj+1 < βj , there would be a j 2 such that |βj 2 − αj 2 | ≤ 1, which would in turn imply Qn(xαj 2 , xβj 2 ) < /6 which is impossible. So if the induction stops at j 1 , then Case 3-j 1 cannot be satisfied. Hence either Case 1-j 1 or Case 2-j 1 must be satisfied and the lemma is satisfied.
Lemma 5. Let h : X −→ X be a homeomorphism of a compact space onto itself. Suppose that there exist sequences {yi}∞ i=1, {zi}∞ i=1 such that d(hk(yn), hk(zn)) < for all k ≤ n. Then there exists a limit point y of {yi}∞ i=1 and a limit point z of {zi}∞ i=1 such that d(hk(y), hk(z)) < 2 for all k.
Proof. Let Y be the set of limit points of {yi}∞ i=1. Pick y in Y and let {yαi }∞ i=1 be a subseqence that converges to y. Let Zα be the set of limit points of {zαi }∞ i=1. Pick z ∈ Zα and let {zβi }∞ i=1 be a subsequence of {zαi }∞ i=1 that converges to z. Then {yβi }∞ i=1 is a subsequence of {yαi }∞ i=1 and hence also converges to y. For each positive integer n, there exists mn ≥ n such that d(yβmn , y) < L(h, n, /2) and d(zβmn , z) < L(h, n, /2). Thus,
d(hk(y), hk(z)) < d(hk^ (y), hk(yβmn )) + d(hk(yβmn ), hk(zβmn ))
for all −n ≤ k ≤ n. Since n is arbitrary, the lemma holds.
Theorem 6. Tree-like continua do not admit expansive homeomorphisms.
Proof. Suppose that h : X −→ X is an expansive homeomorphism of tree-like continuum X with expansive constant c. Let be chosen such that 0 < < c/3. By Theorem 2, there exists a nondegenerate subcontinuum M such that either limn→∞ diam hn(M ) = 0 or limn→−∞ diam hn(M ) = 0. Without loss of generality, we may assume that diam(hi(M )) < for all i ≤ 0. Let {δk}∞ k=1 be a sequence of positive numbers such that each δk < L(h, k, /6). Let Tk be a tree-cover of X with mesh < δk. Let Ak be any |Tk | + 1 elements of M. By the pigeon-hole principle, for each N , there must be at least 2 elements akN , bkN ∈ Ak such that hN^ (akN ), hN^ (bkN ) are in a common element of Tk. Since Ak is finite, we may conclude that there are two elements ak, bk ∈ Ak and a sequence of increasing integers {Nj }∞ j=1 such that hNj^ (ak) and hNj^ (bk) are in a common element of Tk for each j. Also, since h is expansive, there exists an integer nk such that d(hnk^ (ak), hnk^ (bk)) ≥ c > .
TREE-LIKE CONTINUA 3413
Pick Njk ≥ nk. By Lemma 4, there exists xkα, xkβ ∈ hNjk^ (M ) such that / 3 ≤
QNjk (xkα, xkβ ) < and d(xkα, xkβ ) < δk. Hence, d(hi(xkα), hi(xkβ )) < for all i ≤ k. Now, let mk be the positive integer such that d(h−mk^ (xkα), h−mk^ (xkβ )) ≥ /3. Let
yk = h−mk^ (xkα) and zk = h−mk^ (xkβ ). Then d(hi(yk), hi(zk)) < for all i < k + mk. By Lemma 5, there exist limit points y of {yk}∞ k=1 and z of {zk}∞ k=1 such that d(hi(y), hi(z)) ≤ 2 < c for all i. However, since d(yk, zk) ≥ /3, y and z must be distinct. Therefore, h is not expansive.
A continuum is decomposable if it is the union of two of its proper subcontinuum and indecomposable otherwise. A continuum is hereditarily indecomposable if every subcontinuum is indecomposable.
Question 1 ([3]). Does there exist a hereditarily indecomposable continuum that admits an expansive homeomorphism?
Question 2. Does there exist a non-separating plane continuum that admits an expansive homeomorphism?
If so, then it cannot be 1-dimensional.
References [1] H. Kato, Expansive homeomorphisms in continuum theory, Topology Appl., Proceedings of General Topology and Geometric Topology Symposium, (eds. Y. Kodama and T. Hoshina), 45 (1992), no.3, 223-243. MR 93j: 54023 [2] H. Kato, Continuum-wise expansive homeomorphisms, Can. J. Math. 45 (1993), no. 3, 576-
Department of Mathematical Sciences, University of Delaware, Newark, Delaware 19716 E-mail address: mouron@math.udel.edu