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AN ARC-LIKE CONTINUUM THAT ADMITS A HOMEOMORPHISM WITH
ENTROPY FOR ANY GIVEN VALUE
CHRISTOPHER MOURON
Abstract. An arc-like continuum X is constructed with the following properties: (1) for every ∈ [0, ∞] there exists a homeomorphism g : X −→ X such that the entropy of g is and (2) X does not contain a pseudo-arc. This answers a question by W. Lewis.
- Introduction Let f : X −→ X be a continuous function. The topological entropy of f is a number in [0, ∞] that measures the rate of expansion and mixing of f. For more on entropy see [1], [8] and [9]. The purpose of this work is to construct an arc-like continuum X such that for every ∈ [0, ∞] there exists a homeomorphism g : X −→ X such that the entropy of g is and such that X contains no pseudo-arc. This answers a question by Wayne Lewis in the negative (see [4]). A continuum is a compact, connected metric space. A continuum is arc-like if it is the inverse limit of arcs. Arc-like continua are also called chainable. A continuum X is decomposable if there exist proper subcontinua A and B such that X = A ∪ B. A continuum is indecomposable if it is not decomposable. A continuum is hereditarily indecomposable if every subcontinuum is indecom- posable. The pseudo-arc is an arc-like hereditarily indecomposable continuum. For more on the pseudo-arc see [4]. It has been shown in [6] that every arc-like continuum that admits an homeomor- phism with positive entropy must contain an indecomposable subcontinuum. Also, the pseudo-arc admits an continuum-wise expansive homeomorphism which implies that it admits an infinite en- tropy homeomorphism (see [3] and [7]). However, it is unknown to whether the pseudo-arc admits a positive but finite entropy homeomorphism.
- Preliminaries on Entropy and Inverse Limits U is a cover of X if X ⊂
U ∈U U^.^ U^ is an^ open cover^ if every element of^ U^ is open. Let^ X^ be a compact metric space, f : X −→ X be a map, and U be a finite cover of X. Define N (U, X) be the number of sets in a subcover of U on X with smallest cardinality. If U and V are two open covers of X, let U ∨ V = {U ∩ V |U ∈ U, V ∈ V} and f −^1 (U) = {f −^1 (U )|U ∈ U}. Also, define ∨n− 1 i=0 f^
−i(U) = U ∨ f − (^1) (U) ∨ ... ∨ f −n+1(U), where f 0 = id
and
Ent(f, U) = limn→∞(1/n) log N (
∨n− 1 i=0 f^ −i(U), X).
Then the topological entropy of f is defined as
Ent(f ) = sup{Ent(f, U, X) : U is an open cover of X}.
2000 Mathematics Subject Classification. Primary: 54H20, 54F50, Secondary: 54E40. Key words and phrases. entropy, inverse limit, indecomposable continuum. 1
2 C. MOURON
If U and V are finite covers of X such that for every V ∈ V there exists a U ∈ U such that V ⊂ U , then V refines U. If for every V ∈ V there exists a U ∈ U such that V ⊂ U , then V closure refines U. The following propositions are well known and can be found in several texts such as [8].
Proposition 1. If V, U are covers of X such that V refines U, then Ent(f, V) ≥ Ent(f, U).
Proposition 2. For each positive integer k, Ent(f k) = kEnt(f ).
Let T : [0, 1] −→ [0, 1] be defined by
T (x) =
2 x if x ∈ [0, 1 /2] 2 − 2 x if x ∈ (1/ 2 , 1],
T (x) is often called the tent map. The following theorem is well-known. For further details see [1]:
Theorem 3. Ent(T ) = log(2).
Theorem 4. Suppose that f , g, and h are continuous functions such that the following diagram commutes:
X f −−−−→ X yh
yh
Y g −−−−→ Y then Ent(g) ≤ Ent(f ).
Proof. Follows from the fact that if U is a finite open cover of Y then Ent(f, h−^1 (U)) = Ent(g, U).
Corollary 5. Suppose that f and g are continuous functions and h is a homeomorphism such that the following diagram commutes:
X f −−−−→ X yh
yh
Y g −−−−→ Y then Ent(g) = Ent(f ).
Let I be the unit interval and fi : I −→ I be a continuous function called a bonding map. The collection {I, fi}∞ i=1 is called an inverse system. Each interval I is called a factor space of the inverse system. If each bonding map is the same map f , then the inverse system can be written as {I, f }∞ i=1. Every inverse system {I, fi}∞ i=1 determines a topological space X called the inverse limit of the system and is written X = lim ←−{I, fi}∞ i=1. The space X is the subspace of the Cartesian product
i=1 I^ given by X = lim ←−{I, fi}∞ i=1 = {〈xi〉∞ i=1 ∈
i=1 I|fi(xi+1) =^ xi}. X has the subspace topology induced on it by
i=1 I.^ If^ x^ = (xi) ∞ i=1 and^ y^ = (yi) ∞ i=1 are two points of the inverse limit, we define distance to be
d(x, y) =
i=
|xi−yi| 2 i^.
4 C. MOURON
Proof. Let U be any finite open cover of [0, 1]. Then there exists finite open covers V 0 and V 1 of [0, a] and [a, 1] respectively such that V 0 ∪ V 1 refines U. Then
N (
n∨− 1
i=
(h)−i(V 0 ∪ V 1 ), [0, 1]) = N (
n∨− 1
i=
(f )−i(V 0 ), [0, a]) + N (
n∨− 1
i=
(g)−i(V 1 )[a, 1])
≤ 2 max{N (
n∨− 1
i=
(f )−i(V 0 ), [0, a]), N (
n∨− 1
i=
(g)−i(V 1 ), [a, 1])}.
So
Ent(h, U) ≤ Ent(h, V 0 ∪ V 1 )
= lim sup n−→∞
log N (
∨n− 1 i=0 (h)
−i(V 0 ∪ V 1 ), [0, 1]) n
≤ lim sup n−→∞
log(2 max{N (
∨n− 1 i=0 (f^ )
−i(V 0 ),^ [0, a]), N^ (
∨n− 1 i=0 (g)
−i(V 1 ),^ [a,^ 1])}) n
= lim sup n−→∞
log(max{N (
∨n− 1 i=0 (f^ ) −i(V 0 ), [0, a]), N (∨n−^1 i=0 (g) −i(V 1 ), [a, 1])}) n = max{Ent(f, V 0 ), Ent(g, V 1 )}.
Thus Ent(h) ≤ max{Ent(f ), Ent(g)}. It follows from Lemma 9 that Ent(h) = max{Ent(f ), Ent(g)}.
Theorem 11. Suppose that {an}∞ n=0 is a strictly increasing sequence in [0, 1) that converges at 1 where a 0 = 0. Let fn : [an, an+1] −→ [an, an+1],
such that fn(an+1) = fn+1(an+1) for every n ≥ 0. Define f : [0, 1] −→ [0, 1] by
f (x) =
fn(x) if x ∈ [an, an+1] 1 if x = 1,
Then Ent(f ) = sup{Ent(fn)}.
Proof. By Lemma 10, it follows that
Ent(f ) = max{{Ent(f |[0,an])}n i=0−^1 , Ent(f |[an,1]} = max{{Ent(fi)}n i=0−^1 , Ent(f |[an,1]} = sup n
{Ent(fn)}.
Let f be a continuous function on then interval I. Define the extension of f by
E(f )(x) =
1 / 2 f (2x) if x ∈ [0, 1 /2] (2 − f (1))x + f (1) − 1 if x ∈ (1/ 2 , 1],
and the two extension of f by
E 2 (f )(x) =
(f (0) + 1)x if x ∈ [0, 1 /3) 1 / 3 f (3(x − 1 /3)) + 1/ 3 if x ∈ [1/ 3 , 2 /3] (2 − f (1))x + f (1) − 1 if x ∈ (2/ 3 , 1].
ENTROPY HOMEOMORPHISMS 5
Figure 1. E(T )(x) and E 2 (T )(x)
(See Figure 1.) Let {gi}∞ i=0 be such that the domain of gi is the range of gi+1. Then define gi i+1 = gi and gii+ k= gi ◦ gi+1 ◦ ... ◦ gi+k− 1.
Theorem 12. Ent(E 2 (f )) = Ent(E(f )) = Ent(f ).
Proof. Proof will be for Ent(E(f )) = Ent(f ). Proof that Ent(E 2 (f )) = Ent(f ) is similar. Let a = E(f )(1/2) There are 2 Cases: Case 1: a < 1 /2. First let a 0 = 1/2, a 1 = (E(f ))−^1 (a 0 )∩(1/ 2 , 1] and then inductively define an = (E(f ))−^1 (an− 1 ). (Note that E(f ) is 1-1 on [a 1 , 1]). Also notice that an → 1 as n increases. Next let
g : [0, 1 /2] −→ [0, 1 /2], g 0 : [1/ 2 , a 1 ] −→ [a, 1 /2], g 1 : [a 1 , a 2 ] −→ [1/ 2 , a 1 ], . . .
gk : [ak, ak+1] −→ [ak− 1 , ak],
where g(x) = E(f )(x) and gi(x) = E(f )|ai,ai+1. Notice that for each i ≥ 0, gi is 1-1 and onto. Let V 0 be an open cover of [0, 1 /2] then define V 1 = {g− 0 1 (V ∩ [a, 1 /2])|V ∈ V 0 } = {(E(f ))−^1 (V ) ∩ [1/ 2 , a 1 ]|V ∈ V 0 }.
Continuing inductively define
Vk = {g k−−^11 (V )|V ∈ Vk− 1 } = {(E(f ))−k(V ) ∩ [ak− 1 , ak]|V ∈ V 0 }.
Clearly, Vk is an open cover of [ak− 1 , ak]. Next define Wn(V 0 ) = (
⋃^ n
k=
Vk) ∪ {(an, 1]}.
ENTROPY HOMEOMORPHISMS 7
Figure 2. D(T )(x) and H(T )(x)
Thus Ent(E(f )) ≤ Ent(f ). Since both 2x and (1/2)x are homeomorphisms, it follows from Corollary 5 that Ent(f ) = Ent(1/ 2 f (2x)). Also from Lemma 9 it follows that Ent(1/ 2 f (2x)) ≤ Ent(E(f )). Hence, Ent(E(f )) = Ent(f ). Case 2 a = 1/2. Then E(f ) is the identity on [1/ 2 , 1]. Thus it also follows from Lemma 9 and Corollary 5 that Ent(E(f )) = Ent(f ).
Define the halving of f by
H(f )(x) =
1 − 1 / 2 E(f )(2x) if x ∈ [0, 1 /2] 1 − x if x ∈ [1/ 2 , 1],
and the doubling f by
D(f )(x) =
1 / 2 E(f )(2x) if x ∈ [0, 1 /2] 1 − 1 / 2 E(f )(2 − 2 x) if x ∈ [1/ 2 , 1].
(See Figure 2.) Define H^2 (f ) = H(H(f )) and continuing inductively define Hn(f ) = H(Hn−^1 (f )). On the other hand (H(f ))^2 = H(f ) ◦ H(f ). Notice that (H(f ))^2 = D(f ).
Theorem 13. Ent(D(f )) = Ent(f ).
Proof. Notice that h 1 (x) = (1/2)x and h 2 (x) = 1−(1/2)x are homeomorphisms. Thus by Corollary 5, Ent(E(f )) = Ent(h 1 ◦ E(f ) ◦ h− 1 1 ) = Ent(1/ 2 E(f )(2x))
and Ent(E(f )) = Ent(h 2 ◦ E(f ) ◦ h− 2 1 ) = Ent(1 − 1 / 2 E(f )(2 − 2 x)).
So by Lemma 10 and Theorem 12,
Ent(D(f )) = max{Ent(1/ 2 E(f )(2x)), Ent(1 − 1 / 2 E(f )(2 − 2 x))} = Ent(E(f )) = Ent(f ).
8 C. MOURON
Theorem 14. Ent(Hn(f )) = (^21) n Ent(f ).
Proof. First, since (H(f ))^2 = D(f ), it follows by Proposition 2 and Theorem 13 that
Ent(H(f )) =
Ent(D(f )) =
Ent(f ).
Continuing inductively suppose that Ent(Hn−^1 (f )) = (^2) n^1 − 1 Ent(f ). Then it follows that
Ent(Hn(f )) = Ent(H(Hn−^1 (f )))
=
Ent(Hn−^1 (f ))
=
2 n^
Ent(f ).
The next theorem is the main result of this paper:
Theorem 15. There exists an arc-like continuum X such that for every ∈ [0, ∞] there exists a homomorphism g : X −→ X such that Ent(g) = . Furthermore, X does not contain a pseudo-arc.
Proof. Now for the construction of the arc-like continuum. For each integer n ≥ 1 let
sn : [
2 n^ − 1 2 n^
2 n+1^ − 1 2 n+^
] −→ [
2 n^ − 1 2 n^
2 n+1^ − 1 2 n+^
]
be defined by
sn(x) =
2 n^ − 1 2 n^
2 n^
E 2 (Hn(T ))(2n(x −
2 n^ − 1 2 n^
Then by Theorems 3, 12 and 14, Ent(sn) = (^21) n log(2). Next define
g(x) =
sn(x) if x ∈ [ 2
n− 1 2 n^ ,^
2 n+1− 1 2 n+1^ ] for^ n^ = 0,^1 ,^2 , ... 1 if x = 1.
Let X = lim ←−{[0, 1], g}∞ i=1 and Xn = lim ←−{[ 2
n− 1 2 n^ ,^
2 n+1− 1 2 n+1^ ], sn}
∞ i=1 for^ n^ = 0,^1 ,^2 , ...^ Then since^ sn is composed of a finite number of monotone pieces, each Xn does not contain a pseudo-arc by Corollary 8. Thus, X = {〈 1 〉∞ i=1} ∪
n=0 Xn^ does not contain a pseudo-arc (see Figure 3). Given > 0, there exist positive integers {cn}∞ n=0 such that c 2 nn log(2) ≤ and limn−→ c 2 nn log(2) = . For = 0 let cn = 0 and for = ∞ let cn = 2^2 n. Then take
g(x) =
sc nn (x) if x ∈ [ 2
n− 1 2 n^ ,^
2 n+1− 1 2 n+1^ ] for^ n^ = 0,^1 ,^2 , ... 1 if x = 1.
So by Proposition 2, Ent(sc nn (x)) = c 2 nn log(2). Thus, it follows from Theorem 11 that Ent(g) = . Also since sn ◦sc nn = sc nn ◦sn, it follows that g ◦g = g ◦g. Let ̂g : X −→ X be defined by ̂g(〈xi〉) =
〈g(xi)〉. Then by Theorem 6, Ent(̂g) = Ent(g) = . Also, notice that ĝ|[ 2 n− 1 2 n^ ,^2
n+1− 1 2 n+1^ ]^
= ŝc nn which
is a shift homeomorphism on Xn. Thus, it follows that ̂g is a homeomorphism.
