Topologically mixing maps and the pseudoarc
It is known that the pseudoarc can be constructed as the inverse limit of the copies of [0, 1] with one bonding map f which is topologically exact. On the other hand, the shift homeomorphism σ f is topologically mixing in this case. Thus, it is natural to ask whether f can be only mixing or must be...
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irk-123456789-1651322020-02-13T01:27:18Z Topologically mixing maps and the pseudoarc Drwiega, T. Oprocha, P. Статті It is known that the pseudoarc can be constructed as the inverse limit of the copies of [0, 1] with one bonding map f which is topologically exact. On the other hand, the shift homeomorphism σ f is topologically mixing in this case. Thus, it is natural to ask whether f can be only mixing or must be exact. It has been recently observed that, in the case of some hereditarily indecomposable continua (e.g., pseudocircles) the property of mixing of a bonding map implies its exactness. The main aim of the present article is to show that the indicated kind of forcing of recurrence is not the case for the bonding map defining the pseudoarc. Відомо, що псевдодугу можна отримати як обернену границю копій відрізка [0,1] з єдиним зв'язуючим відображенням f, що є топологічно точним. З іншого боку, в цьому випадку зсувний гомеоморфізм σf є топологічно перемішуючим. Таким чином, природно запитати чи може f бути тільки перемішуючим, чи воно обов'язково повинно бути точним. Нещодавно було встановлено, що для деяких спадково нерозкладних континуумів (наприклад, псевдокіл) точність зв'язуючого відображення є наслідком властивості перемішування. В даній статті показано, що розглянутий тип примусового повернення не реалізується для зв'язуючого відображення, що визначає псевдодугу. 2014 Article Topologically mixing maps and the pseudoarc / T. Drwiega, P. Oprocha // Український математичний журнал. — 2014. — Т. 66, № 2. — С. 176–186. — Бібліогр.: 16 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/165132 517.9 en Український математичний журнал Інститут математики НАН України |
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Статті Статті Drwiega, T. Oprocha, P. Topologically mixing maps and the pseudoarc Український математичний журнал |
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It is known that the pseudoarc can be constructed as the inverse limit of the copies of [0, 1] with one bonding map f which is topologically exact. On the other hand, the shift homeomorphism σ f is topologically mixing in this case. Thus, it is natural to ask whether f can be only mixing or must be exact. It has been recently observed that, in the case of some hereditarily indecomposable continua (e.g., pseudocircles) the property of mixing of a bonding map implies its exactness. The main aim of the present article is to show that the indicated kind of forcing of recurrence is not the case for the bonding map defining the pseudoarc. |
| format |
Article |
| author |
Drwiega, T. Oprocha, P. |
| author_facet |
Drwiega, T. Oprocha, P. |
| author_sort |
Drwiega, T. |
| title |
Topologically mixing maps and the pseudoarc |
| title_short |
Topologically mixing maps and the pseudoarc |
| title_full |
Topologically mixing maps and the pseudoarc |
| title_fullStr |
Topologically mixing maps and the pseudoarc |
| title_full_unstemmed |
Topologically mixing maps and the pseudoarc |
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topologically mixing maps and the pseudoarc |
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Інститут математики НАН України |
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2014 |
| topic_facet |
Статті |
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http://dspace.nbuv.gov.ua/handle/123456789/165132 |
| citation_txt |
Topologically mixing maps and the pseudoarc / T. Drwiega, P. Oprocha // Український математичний журнал. — 2014. — Т. 66, № 2. — С. 176–186. — Бібліогр.: 16 назв. — англ. |
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Український математичний журнал |
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AT drwiegat topologicallymixingmapsandthepseudoarc AT oprochap topologicallymixingmapsandthepseudoarc |
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2025-07-14T17:57:15Z |
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2025-07-14T17:57:15Z |
| _version_ |
1837646045545234432 |
| fulltext |
UDC 517.9
T. Drwięga (AGH Univ. Sci. and Technology, Kraków, Poland),
P. Oprocha (AGH Univ. Sci. and Technology, Kraków; Inst. Math. Polish Acad. Sci., Warszawa, Poland)
TOPOLOGICALLY MIXING MAPS AND THE PSEUDOARC*
ТОПОЛОГIЧНО ПЕРЕМIШУЮЧI ВIДОБРАЖЕННЯ ТА ПСЕВДОДУГА
It is known that the pseudoarc can be constructed as the inverse limit of the copies of [0, 1] with one bonding map f
which is topologically exact. On the other hand, the shift homeomorphism σf is topologically mixing in this case. Thus, it
is natural to ask whether f can be only mixing or must be exact. It has been recently observed that, in the case of some
hereditarily indecomposable continua (e.g., pseudocircles) the property of mixing of a bonding map implies its exactness.
The main purpose of this article is to show that the indicated kind of forcing of recurrence is not the case for the bonding
map defining the pseudoarc.
Вiдомо, що псевдодугу можна отримати як обернену границю копiй вiдрiзка [0, 1] з єдиним зв’язуючим вiдобра-
женням f, що є топологiчно точним. З iншого боку, в цьому випадку зсувний гомеоморфiзм σf є топологiчно
перемiшуючим. Таким чином, природно запитати чи може f бути тiльки перемiшуючим, чи воно обов’язково по-
винно бути точним. Нещодавно було встановлено, що для деяких спадково нерозкладних континуумiв (наприклад,
псевдокiл) точнiсть зв’язуючого вiдображення є наслiдком властивостi перемiшування. В данiй статтi показано, що
розглянутий тип примусового повернення не реалiзується для зв’язуючого вiдображення, що визначає псевдодугу.
1. Introduction. Inverse limits provide an excellent tool in theory of continua, allowing construction
of continua of many different types by a careful selection of a sequence of bonding maps. In theory,
it is enough to specify some sufficient conditions that a continuous map f should satisfy (e.g., some
level of crookedness), and if succeeded, the space obtained as an inverse limit with f as the bonding
map should have desired properties. In practice it is not that simple, since one property can interact
with other, making the final construction hard or even impossible. Additionally, small modifications
in the construction can have large impact on the resulting inverse limit. For example, in 1990s Tom
Ingram asked a question if it is true that all the inverse limits induced by maps in the family of tent
maps are nonhomeomorphic. This question, after around 20 years of research and numerous partial
steps towards it, finally received complete affirmative answer in [1]. Note that while the graphs of all
the maps in this family seem to be very similar, nevertheless the corresponding inverse limits are not
homeomorphic.
Another question of this type was related with the pseudoarc, a classical example of hereditarily
indecomposable continuum. The first construction leading to the pseudoarc was obtained by Knaster
in [11], and many years later extended by Moise [13] who gave it the name pseudoarc (since the
continuum he obtained was similar to the arc, in the sense that it was homeomorphic to each of
its subcontinua). Finally, Bing proved [5] that observation of Moise was not accidental, since all
pseudoarcs are homeomorphic. So from topological point of view the pseudoarc is unique. But it
was not the end of the story. For a long time it was unclear, if pseudoarc can be obtained as an
inverse limit of the unit interval with one bonding map. In [9], Henderson provided such example.
While topologically significant, this example was quite trivial from dynamical point of view. Its
nonwandering set was consisting of two fixed points (attractor-repeller pair). In fact, Henderson’s
* The research of P. Oprocha was supported by the Polish Ministry of Science and Higher Education from sources for
science in the years 2013-2014, grant No. IP2012 004272.
c© T. DRWIĘGA, P. OPROCHA, 2014
176 ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 2
TOPOLOGICALLY MIXING MAPS AND THE PSEUDOARC 177
approach is probably the only known technique of construction of a map on the unit interval which
gives pseudoarc in its inverse limit and has zero topological entropy. It is also easy to check that
most of topological properties of dynamics (e.g., value of topological entropy, transitivity, topological
mixing etc.) are shared by bonding map and induced shift homeomorphism σf on the inverse limit.
There were made numerous attempts to make dynamics on the pseudoarc more rich. One of the
most accessible techniques (and important form the point of view of present work) was developed by
Minc and Transue in [12]. They provided a method of perturbation (in fact a sequence of perturba-
tions), such that topologically exact map is transformed to a transitive map f such that inverse limit
with f as bonding map gives the pseudoarc (it is worth noting, that a few years earlier a map on the
pseudoarc with positive topological entropy was constructed by J. Kennedy in [10]). In fact, it can
be proved that if f is a transitive map on a topological graph, and inverse limit of this graph with
f as a bonding map is hereditarily indecomposable, then f must be mixing [15]. So in fact, all the
maps constructed in [12] are topologically mixing. In [14] it was shown, that technique of Minc and
Transue can be extended in such a way that map f is not only transitive but has shadowing property.
The authors have also shown that if we start with properly chosen map f then it will lead to a topo-
logically exact map. However, recently it was proved that on some topological graphs (e.g., circle), if
f is transitive and its inverse limit is hereditarily indecomposable then map f is topologically exact.
Then a natural question arises:
Can method of Minc and Transue lead to a topologically mixing but not exact map?
To convince the reader even more, that the answer to the above question is not that easy, let us
mention another question related to Minc and Transue technique. In their paper [6], Block, Keesling
and Uspenskij asked if there exists a map of the unit interval, which has a periodic point of odd
period but does not have a point of period 3. This question remains unanswered so far, despite the
fact that there are known numerous techniques of construction of maps of type 5, 7, etc. in the
Sharkovsky’s ordering. In fact there are many more open questions related to dynamical properties of
maps admitted on pseudoarc. Of particular interest is the question on admissible values of topological
entropy. It was proved by Mouron [7] that entropy of shift homeomorphism on pseudoarc (induced
as inverse limit with one bonding map) is either zero or infinity. It is an open question if other values
of entropy for homeomorphisms of pseudoarc can be attained. It is worth mentioning that entropy of
transitive interval map is always positive, and entropy of shift homeomorphism on the inverse limit is
the same as the entropy of bonding map [16], hence if a transitive interval map f gives pseudoarc in
its inverse limit then by the above mentioned result of Mouron [7] implies that f has infinite entropy.
While maps of the unit interval which are topologically mixing but not exact are fairly special
(e.g., they have infinitely many fixed points), it is not sufficient obstruction for inducing pseudoarc as
an inverse limit. The main purpose of this article is to propose an update of Minc – Transue technique,
so that a positive answer to the above question is obtained.
2. Preliminaries. In this article we consider continuous maps f : [0, 1] → [0, 1], where [0, 1]
is endowed with the standard Euclidean metric (and this metric is induced on all subintervals). The
space of all continuous f : [a, b]→ [a, b], where a < b, is denoted by C([a, b]). We endow the space
C([a, b]) with the standard supremum metric
ρ(f, g) = sup
t∈[a,b]
|f(t)− g(t)|.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 2
178 T. DRWIĘGA, P. OPROCHA
Note that the space C([a, b]) is a complete metric space.
A map f ∈ C([0, 1]) is:
topologically mixing if for all nonempty open sets U, V there is N such that fn(U) ∩ V 6= ∅
for every n > N,
topologically exact if for every nonempty open set U there is n > 0 such that fn(U) = [0, 1].
The inverse limit of f ∈ C([0, 1]) is the following subset of the Hilbert cube I (the Cartesian product
of countably many copies of [0, 1] with metric induced by the Tikhonov product topology):
Xf = lim
←−
(f, [0, 1]) = {(x0, x1, . . .) : f(xn+1) = xn} ⊂ I.
In 1985 Barge and Martin (see [2]) have shown that many maps on the interval [0, 1] give rise
to the inverse limit space Xf which is indecomposable (i.e., is not the union of two proper sub-
continua) or at least which contains an indecomposable subcontinuum. In fact many of the inverse
limits are hereditarily indecomposable, that is every subcontinuum of Xf is indecomposable. Any
two nondegenerate hereditarily indecomposable continua are homeomorphic, so are in fact a topo-
logically unique continuum known as the pseudoarc (see Knaster [11], Bing [3, 4] and Moise [13]).
Actually, it is the motivation behind the name pseudoarc, since every nondegenerate subcontinuum
of hereditarily indecomposable continuum C is itself hereditarily indecomposable, therefore homeo-
morphic with C by the above mentioned results. It is the situation similar to the interval [0, 1] where
any nondegenerate subcontiunuum is a closed interval, in particular is homeomorphic to the original
continuum [0, 1].
The main goal of this article is a construction of a continuous map f : [0, 1] → [0, 1] such that
f is topologically mixing but not topologically exact and such that the inverse limit of copies [0, 1]
with f as the bonding map is the pseudoarc.
Definition 2.1. Let f ∈ C([0, 1]), let a, b ∈ [0, 1] and let δ > 0. We say that f is δ-crooked
between a and b if, for every two points c < d such that f(c) = a and f(d) = b, there are points
c < c′ < d′ < d such that |b − f(c′)| < δ and |a − f(d′)| < δ. We say that f is δ-crooked if it is
δ-crooked between every pair of points.
Assume that f ∈ C([0, 1]) is such that f(x) = a and f(y) = b. We say that f is linear on [x, y]
if f(z) = a + (z − x) b− a
y − x
and f is piecewise linear if there is a finite partition of [0, 1] such that
f restricted to each element of the partition is a linear function. We say that x ∈ [0, 1] is a critical
point of f if f ′(x) vanishes or is undefined.
Definition 2.2. Fix any 0 ≤ c < d ≤ 1. We say that a map f ∈ C([c, d]) is admissible on [c, d]
if it is piecewise linear, |f ′(t)| ≥ 4 for every noncritical t ∈ (c, d) , and for every c ≤ a < b ≤ d
there is positive integer m such that fm ([a, b]) = [c, d]. If [c, d] = [0, 1], then we simply say that f
is admissible.
Now, we recall some useful facts from [12], which provides a nice practical method for a con-
struction of the pseudoarc:
Lemma 2.1 ([12], Proposition 2). Let f, F ∈ C([0, 1]) be two maps such that ρ(f, F ) < ε. If f
is δ-crooked, then F is (δ + 2ε)-crooked.
Lemma 2.2 ([12], Proposition 3). Let {fi}∞i=1 be a sequence of δ-crooked continuous maps from
[0, 1] into [0, 1]. If the sequence converges uniformly, then the limit is also δ-crooked.
Lemma 2.3 ([12], Proposition 4). Let f ∈ C([0, 1]) be a map with the property that, for every
δ > 0 there is an integer n > 0 such that fn is δ-crooked. Then the inverse limit of copies of [0, 1]
with f as the bonding map is the pseudoarc.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 2
TOPOLOGICALLY MIXING MAPS AND THE PSEUDOARC 179
3. Mixing but not exact maps and pseudoarc. The following result is known and not hard to
prove (see, e.g., [14], Theorem 10).
Lemma 3.1. Let f ∈ C([0, 1]) be a topologically exact map such that f(0) = 0 and f(1) = 1.
For every ε > 0 there exists an admissible map F ∈ C([0, 1]) such that F (0) = 0, F (1) = 1 and
ρ(f, F ) < ε.
Lemma 3.2. Let f ∈ C([0, 1]) be a topologically exact map such that f(0) = 0 and f(1) = 1
and let ε > 0. There exists τ ∈ (0, ε) and g : [0, 1]→ [0, 1] such that g|[τ,1−τ ] is admissible, g(t) = t
for every t ∈ [0, τ ] ∪ [1− τ, 1] and ρ(f, g) < ε.
Proof. By Lemma 3.1 there exists an admissible map F ∈ C([0, 1]) with F (0) = 0, F (1) = 1
and ρ(F, f) < ε/2. Let m > 0 be an integer such that |F ′(t)| < m for every noncritical t ∈ [0, 1].
Denote τ = ε/20m. We define g : [0, 1]→ [0, 1] by putting:
g(x) =
(1− 2τ)F
(
x− τ
1− 2τ
)
+ τ, if x ∈ [τ, 1− τ ],
x, otherwise.
Note that g(τ) = (1−2τ)F (0)+τ = τ and g(1−τ) = (1−2τ)F (1)+τ = 1−τ so g is continuous.
Easy calculation yields that g′(t) = F ′
(
t− τ
1− 2τ
)
for every noncritical t ∈ [τ, 1− τ ]. It is also easy
to note that ĝ = g|[τ,1−τ ] is topologically exact. Namely ĝ = π ◦F ◦π−1, where π : [0, 1]→ [τ, 1−τ ]
is the linear homeomorphism π(x) = (1 − 2τ)x + τ, and hence for any open set U ⊂ [τ, 1 − τ ]
and any s ≥ 0 we have ĝs(U) = π(F s(π−1(U))). In particular, if s > 0 is sufficiently large then
ĝs(U) = π([0, 1]) = [τ, 1− τ ], so indeed ĝ is topologically exact.
Finally for every x ∈ [τ, 1− τ ], by mean value theorem we have (with some point cx ∈ [0, 1])
|F (x)− g(x)| ≤
∣∣∣∣(1− 2τ)F
(
x− τ
1− 2τ
)
+ τ − F (x)
∣∣∣∣ ≤
≤
∣∣∣∣(1− 2τ)F
(
x− τ
1− 2τ
)
− (1− 2τ)F (x)
∣∣∣∣+ |(1− 2τ)F (x)− F (x)|+ τ ≤
≤ (1− 2τ)|F ′(cx)|
∣∣∣∣ x− τ1− 2τ
− x
∣∣∣∣+ 2τ |F (x)|+ τ ≤
≤ (1− 2τ)
mτ
1− 2τ
+ 2τ + τ ≤
≤ τ(m+ 3) ≤ 4mτ ≤ ε
5
.
If x ∈ [0, τ ] then we see that (with some cx ∈ [0, 1])
|F (x)− g(x)| = |F (x)− x| ≤ |F (x)− F (0)| − |F (0)− x| ≤
≤ |F ′(cx)| · τ + τ ≤ τ(m+ 1) ≤ ε
5
.
Similar calculations yield that |F (x) − g(x)| ≤ ε
5
for x ∈ [1 − τ, 1]. Combining together the above
three cases we see that ρ(F, g) ≤ ε
5
and hence ρ(f, g) < ε.
Lemma 3.2 is proved.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 2
180 T. DRWIĘGA, P. OPROCHA
Lemma 3.3. Let γ > 0 and let 0 ≤ a < b ≤ 1. There is a piecewise linear map h ∈ C([a, b])
such that h is admissible on [a, b], h (a) = a, h (b) = b and supt∈[a,b] |h(t)− t| < γ.
Proof. Decrease γ if necessary, so that γ < 1. Let q =
⌊
1
3γ
⌋
+1, where btc is the integer part of
t. Let µ =
b− a
q
and ξ =
µ
5
for each i = 0, 1, . . . , q− 1, let ui = a+ iµ. Let h be a piecewise linear
map on [a, b] which is defined over each interval [ui, ui+1] as the connect-the-dots map by points
(ui, ui), (ui + ξ, ui+1), (ui + 2ξ,max {a, ui − ξ}), (ui+1 − 2ξ,min {b, ui+1 + ξ}), (ui+1 − ξ, ui),
(ui+1, ui+1).
Observe that by the construction |h′(t)| ≥ 4 for every noncritical t and
sup
t∈[0,1]
|h(t)− t| ≤ max
0≤i<q
|ui+1 + ξ − (ui − ξ)| ≤ 3µ < γ.
Fix a nondegenerate closed interval J ⊂ [a, b]. We claim that there is an integer m > 0 and
0 ≤ i < q such that [ui, ui+1] ⊂ hm(J). Observe that if J contains two critical points, then there is i
such that two consecutive points in the sequence ui, ui+ξ, ui+2ξ, ui+1−2ξ, ui+1−ξ, ui+1 belong
to J. But then [ui, ui+1] ⊂ h(J). So the remaining possibility is that J contains at most one critical
point. But then there is an interval [p, q] ⊂ J such that h is linear on [p, q] and q − p ≥ diam(J)/2.
Then
diam(h(J)) ≤ 4|q − p| ≤ 2 diam(J)
and so for some k > 0 the interval hk(J) must contain two critical points, in particular [ui, ui+1] ⊂
⊂ hk+1(J) for some i. The proof of the claim is finished.
Finally, observe that if 0 ≤ i < j ≤ q are such that [ui, uj ] ⊂ J then by the definition of h we
obtain that [ui′ , uj′ ] ⊂ h2(J) where i′ = max {0, i− 1} and j′ = min {1, j + 1} .
Combining all the previous observations we see that for every non-degenerate closed interval
J ⊂ [0, 1] there are integers s, l ≥ 0 and 0 ≤ i < q such that [ui, ui+1] ⊂ hs(J) and hl([ui, ui+1]) =
= [0, 1]. In particular, if we put m = s+ l then hm(J) = [0, 1] which shows that h is admissible.
Lemma 3.3 is proved.
To prove Lemma 3.5, let us first present the following fact which is [12] (Lemma on p. 1167):
Lemma 3.4. Let f ∈ C([0, 1]) be an admissible map. Let η and δ be two positive numbers.
Then there is an admissible map F ∈ C([0, 1]), and there is a positive integer n such that Fn is
δ-crooked and |F (t)− f(t)| < η for every t ∈ [0, 1]. Moreover, if 0 ≤ a < b ≤ 1 and b − a ≥ η,
then f([a, b]) ⊂ F ([a, b]) and Fn([a, b]) = [0, 1].
Lemma 3.5. Let 0 < θ <
1
2
and let η and δ be two positive numbers. Let f ∈ C([θ, 1− θ]) be
an admissible map on [θ, 1 − θ] with the property that, for every ε > 0, there is an integer n > 0
such that fn([a, b]) = [θ, 1− θ] provided that b− a ≥ ε and θ ≤ a < b ≤ 1− θ.
Then there is an admissible map F : [θ, 1− θ]→ [θ, 1− θ] and an integer m > 0 such that Fm
is δ-crooked and |F (t)− f(t)| < η for every t ∈ [θ, 1− θ].
Moreover, if θ ≤ a < b ≤ 1 − θ and b − a ≥ η, then f([a, b]) ⊂ F ([a, b]) and Fm([a, b]) =
= [θ, 1− θ].
Proof. Let us define a function f∗ = φ∗◦f◦φ from [0, 1] into [0, 1], where φ and φ∗ are two linear
homeomorphisms defined by: φ : [0, 1]→ [θ, 1−θ], φ(t) = (1− 2θ) t+θ and φ∗ : [θ, 1− θ]→ [0, 1],
φ∗(t) = φ−1 (t) =
1
1− 2θ
t − θ
1− 2θ
. Then f∗ is a continuous and piecewise linear function and
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 2
TOPOLOGICALLY MIXING MAPS AND THE PSEUDOARC 181
additionally, for each noncritical t we have∣∣(f∗)′ (t)∣∣ = |φ∗ (f (φ(t)))| · ∣∣f ′ (φ(t))∣∣ · ∣∣φ′(t)∣∣ ≥ 1
1− 2θ
· 4 · (1− 2θ) = 4.
In order to prove that f∗ is an admissible we need to show that for every 0 ≤ a < b ≤ 1 there is a
positive integer m such that (f∗)m ([a, b]) = [0, 1]. Let us notice that
(f∗)m = φ∗ ◦ fm ◦ φ
and from the assumption that f is an admissible map on [θ, 1− θ] we get that f∗ is an admissible
map. Take δ∗ = δ/(1−2θ) and η∗ = η/(1−2θ). Let F ∗ : [0, 1]→ [0, 1] be a map provided for f∗ by
application of Lemma 3.4, that is (F ∗)n is δ∗-crooked for some integer n > 0, |F ∗(t)− f∗(t)| < η∗
for every t ∈ [0, 1] and if 0 ≤ a < b ≤ 1 and b − a ≥ η, then f∗([a, b]) ⊂ F ∗ ([a, b]) and
(F ∗)n([a, b]) = [0, 1].
Put F = φ ◦ F ∗ ◦ φ∗. Note that if |a− b| < η then |φ∗(a)− φ∗(b)| < η
1− 2θ
≤ η∗. Similarly, if
|a− b| < η∗ then |φ(a)− φ(b)| < (1− 2θ)η∗ ≤ η. Therefore
sup
t∈[θ,1−θ]
|F (t)− f(t)| ≤ sup
s∈[0,1]
|φ(F ∗(s))− φ(f∗(s))| ≤ η.
Similar calculations show that all the other required conditions on F are satisfied.
Lemma 3.5 is proved.
Lemma 3.6. Let 0 < θ <
1
2
, let δ > 0 and let f ∈ C([θ, 1 − θ]) satisfy f(θ) = θ, f(1 − θ) =
= 1− θ. Let F ∈ C([0, 1]) be a continuous map defined by
F (t) =
f(t) for t ∈ [θ, 1− θ],
t for t /∈ [θ, 1− θ].
If there is an integer k > 0 such that fk is δ-crooked on [θ, 1 − θ] then F k is (θ + δ)-crooked
on [0, 1].
Proof. Fix any nondegenerate interval J = [a, b] ⊂ [0, 1]. We need to find points a < c < d < b
such that |F k(a)−F k(d)| < θ+δ and |F k(b)−F k(c)| < θ+δ. If J ⊂ [θ, 1−θ] then there is nothing
to prove, since fk is δ-crooked. Similarly, if J ⊂ [0, θ] or J ⊂ [1− θ, 1] we are done by taking any
a < c < d < b since then |F k(a)− F k(d)| = |a− d| ≤ θ and |F k(b)− F k(c)| = |b− c| ≤ θ. In the
remaining case, let p = max {θ, a} and q = min {p, 1− θ}. Then [p, q] ⊂ [θ, 1−θ] is a nondegenerate
interval, and so there are p < c < d < q such that |fk(p) − fk(d)| = |F k(p) − F k(d)| < δ and
|fk(q)−fk(c)| = |F k(q)−F k(c)| < δ. Finally, note that if a < p then |F k(a)−F k(p)| = |a−p| < θ
and so
|F k(a)− F k(d)| ≤ |F k(a)− F k(p)|+ |F k(p)− F k(d)| ≤ θ + δ
and similarly |F k(c)− F k(b)| ≤ θ + δ.
Lemma 3.6 is proved.
Definition 3.1. A piecewise linear map f : [0, 1] → [0, 1] is Markov if there is a sequence
0 = a0 < a1 < . . . < an = 1 such that f is linear on each interval [ai, ai+1] and f([ai, ai+1]) =
= [ak, am] for some k < m.
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182 T. DRWIĘGA, P. OPROCHA
The main application of Markov maps is that it is relatively easy to verify if such a map is
transitive or topologically exact. It will be a useful tool in many constructions presented later.
Lemma 3.7. For every admissible map f ∈ C([a, b]) and every ε > 0 there exists an admissible
Markov map F ∈ C([a, b]) such that supt∈[a,b] |f(t)−F (t)| < ε and F (a) ≤ f(a) and F (b) ≥ f(b).
Proof. Let ε > 0 and a = u0 < u1 < . . . < un = b be a partition of [a, b] into n intervals of
equal length, that is |ui+1 − ui| = |uj+1 − uj | for any 0 ≤ i < j < n. Assume additionally that n is
large enough, so that the following condition is satisfied:
max
i=0,1,...,n−1
{|ui − ui+1|, diam f([ui, ui+1])} <
ε
6
.
We are going to define a piecewise linear map F : [a, b]→ [a, b]. First, we define F at the points
of the partition by the formula
F (ui) = max{uj : uj ≤ min f([ui, ui+1])} for i < n,
F (un) = min{uj : uj ≥ max f([un−1, un])}.
Now, we are going to define F on each of the intervals [ui, ui+1] for i = 0, 1, . . . , n − 1. Fix
0 ≤ i < n and denote s = s(i) = max{j : uj ≤ min f([ui, ui+1])} and t = t(i) = min{j : uj ≥
≥ max f([ui, ui+1])}. Note f is not constant on any interval, hence we have s < t and
f([ui, ui+1]) ⊂ [us, ut].
Denote αi = F (ui+1)− us. We are going to define F on [ui, ui+1] depending on the value of αi. If
αi ≤ 0 then we divide interval [ui, ui+1] into four intervals of equal length, that is we introduce 3
middle points pj = ui + j · (ui+1 − ui)/4, for j = 1, 2, 3. Next, we define F as the connect-the-dots
map by points (ui, F (ui)), (p1, ut), (p2, us), (p3, ut), (ui+1, F (ui+1)). Note that each interval in the
partition ui < p1 < p2 < p3 < ui+1 has length |ui+1 − ui|/4 and |ut − F (ui+1)| ≥ |ut − us|, so F
has slope 4 in each of the intervals of the partition.
In the second case αi > 0, we divide interval [ui, ui+1] into 5 pieces of equal length by introduc-
ing additional partition points qj = ui + j · (ui+1 − ui)/5, for j = 1, 2, 3, 4. Then we define func-
tion F as the connect-the-dots map given by points (ui, F (ui)), (q1, ut), (q2, us), (q3, ut), (q4, us),
(ui+1, F (ui+1)). Note that |F (ui+1) − us| ≥ |ui+1 − ui| so again F has slope at least 4 on each
interval of linearity in [ui, ui+1].
Therefore the function F defined above is a piecewise linear continuous map on [a, b]. Addition-
ally, for every 0 ≤ i < n there are k ≤ s(i) < t(i) ≤ m such that F ([ui, ui+1]) = [uk, um], and
|F ′(t)| ≥ 4 for every noncritical t ∈ (a, b). To prove that F is admissible, it remains to show that F
is topologically exact.
Denote J(i) = {m : [um, um+1] ⊂ F ([ui, ui+1])} and observe that by the definition of F we
have
f([ui, ui+1]) ⊂ [us, ut] ⊂ F ([ui, ui+1]) =
⋃
j∈J(i)
[uj , uj+1].
Then, using mathematical induction, we obtain that
fk+1([ui, ui+1]) ⊂ fk(F ([ui, ui+1])) ⊂ fk
⋃
j∈J(i)
[uj , uj+1]
=
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TOPOLOGICALLY MIXING MAPS AND THE PSEUDOARC 183
=
⋃
j∈J(i)
fk([uj , uj+1]) ⊂
⋃
j∈J(i)
F k([uj , uj+1]) = F k
⋃
j∈J(i)
[uj , uj+1]
=
= F k+1([ui, ui+1]).
In particular, since f is an exact map, for every 0 ≤ i < n there is k > 0 such that F k([ui, ui+1]) =
= [a, b].
Now let J be an open interval. Since F ′(t) ≥ 4 for every noncritical value t ∈ (a, b) there is
j > 0 such that F j(J) contains three consecutive critical points. But then, by the definition of F
there is 0 ≤ i < n such that [ui, ui+1] ⊂ F j+1(J). Indeed, F is an exact map.
Observe that for every 0 ≤ i < n we have
sup
t∈[ui,ui+1]
|F (t)− f(t)| ≤ diamF ([ui, ui+1]).
Denote r(i) = i+ 2 if i < n− 1 and put r(i) = n otherwise. Then by the definition of F we obtain
that
diamF ([ui, ui+1]) ≤ 2 · ε
6
+ diam f([ui, ur(i)]) ≤
ε
3
+ 2 · ε
6
< ε.
This yields that supt∈[a,b] |f(t)− F (t)| < ε and so the proof is finished.
Lemma 3.7 is proved.
Lemma 3.8. Let ε > 0. Let f ∈ C([a, b]) and g ∈ C([b, c]), be two admissible maps on [a, b] and
[b, c] , respectively, such that f(b) = g(b) = b. Then there is an admissible map F ∈ C([a, c]) such
that for every t ∈ [a, b] we have |F (t)− f(t)| < ε and for every t ∈ [b, c] we have |F (t)− g(t)| < ε.
Proof. Note that if we use Lemma 3.7 to perturb f, g to Markov maps, then the condition
f(b) = g(b) remains satisfied. Hence, by Lemma 3.7, without loss of generality we may assume
that both maps f, g are Markov. Let {li}ti=1 be an increasing sequence giving Markov partition for
functions f and g, that is f is Markov with respect to partition a = l0 < l1 < . . . < ls = b and
g is Markov with respect to partition b = ls < ls+1 < . . . < lt = c. Since periodic points are
dense in [a, b] and in [c, d], including periodic orbits in the partition if necessary, we may assume that
|li − li+1| < ε/3, diam f([li, li+1]) < ε/3 for 0 ≤ i < s and diam g([li, li+1])) < ε/3 for s ≤ i < t.
Fix points p < q in [ls−1, b] dividing it into three intervals of equal length (i.e., p = ls−1 +
+ (b − ls−1)/3, q = ls−1 + 2(b − ls−1)/3) and let p′ < q′ divide [b, ls+1] into three intervals
of equal length. We define F (t) = f(t) for every t ∈ [a, ls−1] and F (t) = g(t) for every t ∈
∈ [ls+1, c]. We also put F (p) = F (p′) = g(ls+1) > ls+1 and F (q) = F (q′) = f(ls−1) > ls−1 (these
inequalities are obtained by the fact that both f, g have slope 4 on intervals of linearity). Finally, we
put F (b) = b and define F as the connect-the-dots map, so it becomes piecewise linear also on the
interval [ls−1, ls+1]. Note that F ′(t) ≥ 4 in every noncritical point and F is Markov with respect
to partition {li}ti=0 ∪ {p, q, p′, q′}. Note that F ([ls−1, p]) ∩ F ([p, q]) ∩ F ([q, b]) ⊃ f([ls−1, b]) and
F ([b, p′]) ∩ F ([p′, q′]) ∩ F ([q′, ls+1]) ⊃ g([b, ls+1]).
Also, since F ′(t) ≥ 4, for every nondegenerate interval J there is an integer k > 0 such that
F k(J) contains three consecutive points of the partition and hence there is j such that [lj , lj+1] ⊂
⊂ F k+1(J) for some 0 ≤ j < t. Additionally, since for every 0 ≤ i < s we get f([li, li+1]) ⊂
⊂ F ([li, li+1]), and for s ≤ i < t we have g([li, li+1]) ⊂ F ([li, li+1]) then there exists m > 0 such
that Fm([li, li+1]) ⊃ [a, b] for every 0 ≤ i < s and Fm([li, li+1]) ⊃ [b, c] for every s ≤ i < t. But
[ls−1, ls+1] ⊂ F ([a, b]) ∩ F ([b, c]). By the above observations we obtain that there is 0 ≤ j < t such
that
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184 T. DRWIĘGA, P. OPROCHA
F k+2m+2(J) ⊃ F 2m+1([lj , lj+1]) ⊃ Fm([ls−1, ls+1]) ⊃
⊃ Fm([ls−1, b]) ∪ Fm([b, ls+1]) ⊃ [a, b] ∪ [b, c] = [a, c].
Indeed F is topologically exact.
Lemma 3.8 is proved.
Theorem 3.1. For every ε > 0 and every topologically exact map g ∈ C([0, 1]) such that
g(0) = 0 and g(1) = 1 there exists a mixing but not topologically exact map f ∈ C([0, 1]) such that
ρ(f, g) < ε and the inverse limit of copies of [0, 1] with f as the bonding map is the pseudoarc.
Proof. Fix ε > 0. First, use Lemma 3.2 to obtain 0 < τ < ε/16 and a map F ∈ C([0, 1]) such
that ρ(F, g) < ε/2, F restricted to [τ, 1− τ ] is admissible and it is an identity mapping outside this
interval.
We claim that there is a sequence of maps f1, f2, . . . ∈ C([0, 1]), an increasing sequence of
positive integers n(1) < n(2) < . . . , and three sequences of positive numbers ε1, ε2, . . . , δ1, δ2, . . . ,
τ1, τ2, . . . such that ρ(f1, F ) < ε/4, τ1 = τ, δ1 =
τ1
4
and additionally the following conditions are
satisfied:
(i) |fi+1(t)− fi(t)| < εi < 2−i for all i = 1, 2, . . . and for each t ∈ [0, 1],
(ii) fi restricted to [τi, 1−τi] is an admissible mapping, fi([τi, 1−τi]) = [τi, 1−τi] and fi(t) = t
for each t ∈ [0, τi] ∪ [1− τi, 1] and for i = 1, 2, . . . ,
(iii) for every i = 1, 2, . . . and every 1 ≤ k ≤ i the map fn(k)i is (2−k − 2−k−i)-crooked,
(iv) for every i = 1, 2, . . . and every 1 ≤ k ≤ i, if b− a ≥ 2−k, then fn(k)i ([a, b]) ⊃ [τk + ξ, 1−
− ξ − τk], where ξ =
∑i
j=k
δj ,
(v) δi+1 <
δi
2
, εi+1 <
εi
2
, τi+1 <
τi
2
, εi <
τi
4
.
We prove the above claim by the induction on i. For i = 1 apply Lemma 3.5 with θ = τ1, η =
= ε1 = min{τ1/8, 2−2}, δ = 2−1 − 2−1−1 = 1/4 and map F |[τ1,1−τ1] obtaining an admissible map
f1 : [τ1, 1 − τ1] → [τ1, 1 − τ1] and an integer n(1) > 0 such that (3) and (3) are satisfied. Next we
extend f1 onto whole [0, 1] putting f1(t) = t for all t ∈ [0, τ1]∪ [1− τ1, 1], hence by our construction
also (3) holds. This completes the first step of induction.
Next assume that the claim holds for all i = 1, . . . , s for some s ≥ 1. We will prove that it also
holds for i = s+1. Put τs+1 = τs/4 and fix any δs+1 < min
{
δs/2, 2
−2s−4}. By uniform continuity
of fs we can find εs+1 < min
{
2−s−1, εs/2, τs+1/4
}
such that if ρ(h, fs) < εs+1 then ρ(hj , f js ) <
< δs+1 for j = 0, 1, . . . , n(s). In particular, for any 1 ≤ k ≤ s we have (2−k − 2−k−s + 2δs+1) <
< (2−k − 2−k−s−1). Hence, by Lemma 2.1, for every 1 ≤ k ≤ s the map hn(k) is (2−k − 2−k−s−1)-
crooked. Similarly hn(k)([a, b]) ⊃ [τk + ξ, 1− ξ − τk] where ξ =
∑s+1
j=k
δj .
Note that fs is an admissible map on [τs, 1 − τs], endpoints of this interval are fixed points of
fs and by Lemma 3.3 there are also admissible maps on [τs+1, τs] and [1 − τs, 1 − τs+1] which
are arbitrarily small perturbations of identity on these intervals (and endpoints of these intervals
are fixed points for them). Therefore, we can apply twice Lemma 3.8 obtaining an admissible map
G : [τs+1, 1 − τs+1] such that supt∈[τs+1,1−τs+1] |G(t) − fs(t)| < εs+1/2. We can extend G to the
whole [0, 1] putting G(t) = t for all t ∈ [0, τs+1] ∪ [1 − τs+1, 1] and then G ∈ C([0, 1]) and
additionally ρ(G, fs) < εs+1/2. Next we apply Lemma 3.5 with θ = τs+1, η = min{εs+1/2, 2
−s−2},
δ = 2−s−1 − 2−2s−2 and map G|[τs+1,1−τs+1] obtaining an admissible map fs+1 : [τs+1, 1− τs+1]→
→ [τs+1, 1 − τs+1] and an integer n(s + 1) > 0 such that (3) and (3) are satisfied with k = s + 1.
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TOPOLOGICALLY MIXING MAPS AND THE PSEUDOARC 185
As before, we extend fs+1 to [0, 1] putting identity on the points outside [τs+1, 1 − τs+1]. Since
ρ(fs+1, fs) < εs+1, conditions (3) and (3) are satisfied for every 1 ≤ k ≤ s + 1 and also (3) holds.
This completes the induction.
Note that
∑∞
j=1
εj ≤ τ/2, that is fi converges uniformly to a map f(t) = limi→∞ fi(t), in
particular f : [0, 1]→ [0, 1] is continuous. This also proves that ρ(f, g) < ε. Furthermore, by (3) and
Lemma 2.2 for each k the map fn(k) is 2−k-crooked, so the inverse limit of [0, 1] with f as a single
bonding map is the pseudoarc by Lemma 2.3.
Fix an open interval U ⊂ [0, 1] and fix any a < b such that [a, b] ⊂ U. Fix any γ > 0.
Note that ξk =
∑∞
j=k
δj ≤
∑∞
j=k
2−jτ ≤ 2k−1τ so there is k > 0 such that [γ/2, 1 − γ/2] ⊂
⊂ [τk+2ξk, 1−2ξk− τk]. We may also assume that k is sufficiently large so that b−a > 2−k. Then,
by (3) we have that
f
n(k)
i ([a, b]) ⊃ [τk + ξ, 1− ξ − τk] ⊃ [τk + 2ξk, 1− 2ξk − τk] ⊃ [γ/2, 1− γ/2]
for every i ≥ k, and additionally, if i is sufficiently large then ρ(f, fi) < γ/2. This shows that
fn(k)(U) ⊃ fn(k)([a, b]) ⊃ [γ, 1− γ].
Since U and γ were arbitrary, we see that f is mixing.
Finally, observe that
∞∑
j=k
εj ≤
∞∑
j=k
2−j+kεk ≤ 2εk ≤ τk/2.
Since fk([τk, 1− τk]) = [τk, 1− τk] we obtain that f([τk, 1− τk]) ⊂ [τk/2, 1− τk/2] ⊂ (0, 1). This
holds for every k = 1, 2, . . . and hence
f((0, 1)) = f
( ∞⋃
k=1
[τk, 1− τk]
)
⊂ (0, 1).
This shows that f is not exact completing the proof.
Theorem 3.1 is proved.
4. Final comments. In Theorem 3.1 we provided a method of perturbation in the case that we
start with topologically exact map f ∈ C([0, 1]) such that both endpoints are fixed points. But it is
clear that similar technique will work in the case that only one of endpoints is fixed, say f(0) = 0,
however in this case its perturbation F which generates the pseudoarc will have only one inaccessible
fixed point, that is F (0) = 0 and F ((0, 1]) = (0, 1]. It may even happen that F (1) 6= 1.
In fact a similar approach will be successful, if G is a topological graph with an endpoint p and f
is a topologically exact map on G such that f(p) = p. Simply, most of our techniques work locally,
so it is really unimportant that G is not an interval. The only essential ingredient is an endpoint which
is a fixed point [15] (see also [8]).
Finally, it also should be clear that a simple modification of techniques involved in Theorem 3.1
will work in case of topologically exact maps with periodic endpoints, that is f(0) = 1, f(1) = 0.
Simply, instead of attaching identity map at each endpoint, we will attach 2-periodic intervals.
While we present here only rough idea of other cases, a careful reader should be able to make all
the necessary adjustments and prove theorems announced above.
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186 T. DRWIĘGA, P. OPROCHA
1. Barge M., Bruin H., Štimac S. The Ingram conjecture // Geometry & Topology. – 2012. – 16. – P. 2481 – 2516.
2. Barge M., Martin J. Chaos, periodicity and snake-like continua // Trans. Amer. Math. Soc. – 1985. – 289. –
P. 355 – 365.
3. Bing R. H. A homogeneous indecomposable plane continuum // Duke Math. J. – 1948. – 15. – P. 729 – 742.
4. Bing R. H. Snake-like continua // Duke Math. J. – 1951. – 18. – P. 653 – 663.
5. Bing R. H. Concerning hereditarily indecomposable continua // Pacif. J. Math. – 1951. – 1. – P. 43 – 51.
6. Block L., Keesling J., Uspenskij V. V. Inverse limits which are the pseudoarc // Houston J. Math. – 2000. – 26. –
P. 629 – 638.
7. Mouron C. Entropy of shift maps of the pseudo-arc // Topology Appl. – 2012. – 159. – P. 34 – 39.
8. Harańczyk G., Kwietniak D., Oprocha P. Topological entropy and specification property for mixing graph maps //
Ergodic Theory and Dynam. Systems / publ. on-line, DOI:10.1017/etds.2013.6.
9. Henderson G. W. The pseudo-arc as an inverse limit with one binding map // Duke Math. J. – 1964. – 31. –
P. 421 – 425.
10. Kennedy J. Positive entropy homeomorphisms on the pseudoarc // Mich. Math. J. – 1989. – 36, №. 2. – P. 181 – 191.
11. Knaster B. Un continu dont tout sous-continu est indécomposable // Fund. Math. – 1922. – 3. – P. 247 – 286.
12. Minc P., Transue W. R. R. A transitive map on [0, 1] whose inverse limit is the pseudoarc // Proc. Amer. Math. Soc. –
1991. – 111. – P. 1165 – 1170.
13. Moise E. E. An indecomposable plane continuum which is homeomorphic to each of its nondegenerate subcontinua
// Trans. Amer. Math. Soc. – 1948. – 63. – P. 581 – 594.
14. Kościelniak P., Oprocha P. Shadowing, entropy and a homeomorphism of the pseudoarc // Proc. Amer. Math. Soc. –
2010. – 138. – P. 1047 – 1057.
15. Kościelniak P., Oprocha P., Tuncali M. Hereditarily indecomposable inverse limits of graphs: shadowing, mixing and
exactness // Proc. Amer. Math. Soc. – 2014. – 142. – P. 681 – 694.
16. Ye X. Topological entropy of the induced maps of the inverse limits with bonding maps // Topology Appl. – 1995. –
67. – P. 113 – 118.
Received 17.12.12,
after revision — 26.03.13
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