Discontinuous Birkhoff theorem
A topological space X is called totally recurrent if every mapping f : X → X has a recurrent point. We prove that a Hausdorff space X is totally recurrent if and only if X is either finite or a one-point compactification of an infinite discrete space.
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| Дата: | 2007 |
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Інститут прикладної математики і механіки НАН України
2007
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| Назва видання: | Український математичний вісник |
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| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Цитувати: | Discontinuous Birkhoff theorem / O. Petrenko, I.V. Protasov // Український математичний вісник. — 2007. — Т. 4, № 3. — С. 434-436. — Бібліогр.: 2 назв. — англ. |
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irk-123456789-1245272017-09-30T03:03:29Z Discontinuous Birkhoff theorem Petrenko, O. Protasov, I.V. A topological space X is called totally recurrent if every mapping f : X → X has a recurrent point. We prove that a Hausdorff space X is totally recurrent if and only if X is either finite or a one-point compactification of an infinite discrete space. 2007 Article Discontinuous Birkhoff theorem / O. Petrenko, I.V. Protasov // Український математичний вісник. — 2007. — Т. 4, № 3. — С. 434-436. — Бібліогр.: 2 назв. — англ. 1810-3200 2000 MSC. 37B20, 54C10, 58K15. http://dspace.nbuv.gov.ua/handle/123456789/124527 en Український математичний вісник Інститут прикладної математики і механіки НАН України |
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English |
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A topological space X is called totally recurrent if every mapping f : X → X has a recurrent point. We prove that a Hausdorff space X is totally recurrent if and only if X is either finite or a one-point compactification of an infinite discrete space. |
| format |
Article |
| author |
Petrenko, O. Protasov, I.V. |
| spellingShingle |
Petrenko, O. Protasov, I.V. Discontinuous Birkhoff theorem Український математичний вісник |
| author_facet |
Petrenko, O. Protasov, I.V. |
| author_sort |
Petrenko, O. |
| title |
Discontinuous Birkhoff theorem |
| title_short |
Discontinuous Birkhoff theorem |
| title_full |
Discontinuous Birkhoff theorem |
| title_fullStr |
Discontinuous Birkhoff theorem |
| title_full_unstemmed |
Discontinuous Birkhoff theorem |
| title_sort |
discontinuous birkhoff theorem |
| publisher |
Інститут прикладної математики і механіки НАН України |
| publishDate |
2007 |
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http://dspace.nbuv.gov.ua/handle/123456789/124527 |
| citation_txt |
Discontinuous Birkhoff theorem / O. Petrenko, I.V. Protasov // Український математичний вісник. — 2007. — Т. 4, № 3. — С. 434-436. — Бібліогр.: 2 назв. — англ. |
| series |
Український математичний вісник |
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AT petrenkoo discontinuousbirkhofftheorem AT protasoviv discontinuousbirkhofftheorem |
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2025-07-09T01:34:20Z |
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2025-07-09T01:34:20Z |
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1837131241402400768 |
| fulltext |
Український математичний вiсник
Том 4 (2007), № 3, 434 – 436
Discontinuous Birkhoff theorem
O. Petrenko, Igor V. Protasov
Abstract. A topological space X is called totally recurrent if every
mapping f : X → X has a recurrent point. We prove that a Hausdorff
space X is totally recurrent if and only if X is either finite or a one-point
compactification of an infinite discrete space.
2000 MSC. 37B20, 54C10, 58K15.
Key words and phrases. Recurrent point, totally recurrent space.
Let X be a topological space, f : X → X. A point x ∈ X is said
to be recurrent if, x is a limit point of the orbit {fm(x) : m ∈ ω}. By
Birkhoff Theorem ([1, 2]), every continuous mapping of a compact space
has a recurrent point.
We say that a topological space X is totally recurrent if every map-
ping f : X → X has a recurrent point.
Example 1. Every finite space is totally recurrent.
Example 2. Let X be an infinite discrete space, Ẋ = X ∪ {∞} be a
one-point compactification of X, f : Ẋ → Ẋ. If ∞ is not a recurrent
point of f then there exist a neighbourhood U of x and n ∈ ω such that
fm(x) /∈ U for every m > n. Since Ẋ \ U is finite, at least one point of
Ẋ \ U is recurrent, so Ẋ is totally recurrent.
Example 3. Let X be an infinite set endowed with a topology in which
a subset U is open if and only if X \ U is finite. Every point of X is
a limit point of any infinite subset of X. It follows that X is a totally
recurrent T1-space.
To prove our main result we need some auxiliary lemmas.
Lemma 1. Every closed subspace F of a totally recurrent space X is
totally recurrent.
Received 11.05.2007
ISSN 1810 – 3200. c© Iнститут математики НАН України
O. Petrenko, I. V. Protasov 435
Proof. We take an arbitrary mapping f : F → F , fix some point y ∈ F
and define a mapping h : X → X by the rule
h|F = f |F , h|X\F ≡ y.
Since X is totally recurrent, there exists a recurrent point x of h. By the
definition of h, we have x ∈ F , so x is a recurrent point of f and F is
totally recurrent.
Lemma 2. Let X be a topological space, γ be a limit ordinal. Assume
that there exists a family {Fα : α < γ} of non-empty closed subspaces of
X such that F0 = X and
(i) Fα ⊃ Fβ for all α < β < γ;
(ii) Fβ =
⋂
α<β Fα for every limit ordinal β < γ;
(iii)
⋂
α<γ Fα = ∅.
Then X is not totally recurrent.
Proof. For every α < γ, we fix some point xα ∈ Fα \ Fα+1 and define a
mapping f : X → X by the rule f |Fα\Fα+1
≡ xα+1. Given an arbitrary
point x ∈ X, we choose the minimal ordinal β such that x /∈ Fβ . By (ii),
β is not a limit ordinal, so β = α+ 1 for some α < γ and x ∈ Fα \Fα+1.
By definition of f , we have fn(x) ∈ Fα+1 for every natural number n, so
fn(x) /∈ X \Fα+1. Since X \Fα+1 is a neighbourhood of X, we conclude
that x is not a recurrent point of f .
Lemma 3. Let X be a topological space. Assume that there exist two
families {Fn : n ∈ ω}, {Hn : n ∈ ω} of closed subspaces of X such that
F0 ∩H0 = ∅ and Fn ⊃ Fn+1, Hn ⊃ Hn+1 for every n ∈ ω. Then X is
not totally recurrent.
Proof. For every n ∈ ω, we fix some points xn ∈ Fn\Fn+1, yn ∈ Hn\Hn+1
and define a mapping f : X → X by the rule
f |Fn\Fn+1
≡ xn+1, f |Hn\Hn+1
≡ yn+1,
f |∩n∈ωFn ≡ y0, f |∩n∈ωHn ≡ x0, f |X\(F0∪H0) ≡ x0.
It is a routine verification that f has no recurrent points.
Lemma 4. Let X be an infinite totally recurrent space such that every
infinite closed subspace of X has an infinite proper closed subspace. Then
the following statements hold
436 Discontinuous Birkhoff theorem
(i) F ∩H 6= ∅ for any two closed infinite subspaces of X;
(ii) there exists a non-empty finite subset A of X such that X \ U is
finite for every open subset U of X containing A.
Proof. (i) follows directly from Lemma 3.
(ii) Using the assumption of lemma, we can construct inductively,
a family {Fα : α < γ} of infinite closed subspaces of X satisfying (i),
(ii) of Lemma 2 and such that
⋂
α<γ Fα is finite. Put A =
⋂
α<γ Fα.
By Lemma 2, A is non-empty. Let U be an open subset of X such
that A ⊆ U . Assume that X \ U is infinite and, for every α < γ, put
Hα = (X \ U) ∩ Fα. By (i) of Lemma 4, Hα 6= ∅ for every α < γ. Then
the family {Hα : α < γ} of closed subsets of X \U satisfies Lemma 2, so
X \ U is not totally recurrent contradicting Lemma 1.
Theorem 1. Let X be an infinite Hausdorff totally recurrent space.
Then X is a one-point compactification of a discrete space.
Proof. Let A = {a1, . . . , an} be a subset ofX given by Lemma 4. SinceX
is Hausdorff, every point from X \A is isolated, so it suffices to show that
A has only one non-isolated point in X. We assume the contrary that
a1, a2 are non-isolated, and choose pairwise disjoint open sets U1, . . . , Un
containing a1, . . . , an. Since every point from X \ (U1, . . . , Un) is isolated
then U1, . . . , Un are closed, so U1, U2 are infinite disjoint closed subsets
of X and we get a contradiction to Lemma 4.
Example 3 shows that this theorem does not hold for T1-spaces.
References
[1] G. D. Birkhoff, Dynamical systems, Amer. Math. Soc. Coll. Publ., 9, 1927.
[2] R. Ellis, Lectures on topological dynamics, Benjamin, New York, 1969.
Contact information
O. Petrenko,
Igor V. Protasov
Department of Cybernetics,
National T. Shevchenko University of Kyiv,
Volodimirska 64,
Kyiv 01033,
Ukraine
E-Mail: protasov@unicyb.kiev.ua
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