Topologies with $T_1$-complements

Steiner, E. F.; Steiner, A. K.

doi:10.4064/fm-61-1-23-28

Cited by 24 publications

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“…The topological notion of T 1 -complementarity of topologies on X corresponds to the algebraic notion of complementarity in the lattice (L(X), ≤). We refer the reader to [10], [9], [1], [5] and [14,Sec. 12] for details and relevant discussions.…”

Section: Introductionmentioning

confidence: 99%

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A compact Hausdorff topology that is a T₁-complement of itself

Shakhmatov

Tkachenko

2002

Fund. Math.

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Abstract. Topologies τ 1 and τ 2 on a set X are called T 1 -complementary if τ 1 ∩ τ 2 = {X \ F : F ⊆ X is finite} ∪ {∅} and τ 1 ∪ τ 2 is a subbase for the discrete topology on X.We provide an example of a compact Hausdorff space of size 2 c which is T 1 -complementary to itself (c denotes the cardinality of the continuum). We prove that the existence of a compact Hausdorff space of size c that is T 1 -complementary to itself is both consistent with and independent of ZFC. On the other hand, we construct in ZFC a countably compact Tikhonov space of size c which is T 1 -complementary to itself and a compact Hausdorff space of size c which is T 1 -complementary to a countably compact Tikhonov space. The last two examples have the smallest possible size: It is consistent with ZFC that c is the smallest cardinality of an infinite set admitting two Hausdorff T 1 -complementary topologies [8]. Our results provide complete solutions to Problems 160 and 161 (both posed by S. Watson [14]) from Open Problems in Topology (North-Holland, 1990). Introduction.Recall that a topology τ on a set X is called a T 1 topology, and the pair (X, τ ) is called a T 1 -space, provided that all singletons {x} of X are τ -closed. The cofinite topology {X \ F : F ⊆ X is finite} ∪ {∅} on X is the smallest (with respect to set inclusion) T 1 topology on X.Two topologies τ 1 and τ 2 on an infinite set X are called:

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Section: Introductionmentioning

confidence: 99%

“…The classical notion of T 1 -complementarity has been studied for a long time (see [10], [9], [1], [5] and [11]). …”

Section: Introductionmentioning

confidence: 99%

A compact Hausdorff topology that is a T₁-complement of itself

Shakhmatov

Tkachenko

2002

Fund. Math.

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“…Theorem 1. There are two Hausdorff topologies σ 0 and σ 1 on a set such that σ 0 ∨ σ 1 is the discrete topology, such that σ 0 ∧ σ 1 is the cofinite topology, and such that σ 0 and σ 1 are homeomorphic topologies.If σ 0 and σ 1 are topologies which satisfy • σ 0 ∨ σ 1 is the discrete topology, • σ 0 ∧ σ 1 is the cofinite topology, then we say σ 0 and σ 1 are T 1 -complements.In 1967, Steiner and Steiner [2] showed that of any pair of T 1 -complements on a countable set, at least one is not Hausdorff. In 1969, Anderson and Stewart [1] showed that of any pair of T 1 -complements, at least one is not first countable Hausdorff.…”

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confidence: 99%

A completely regular space which is the $T_1$-complement of itself

Watson

1996

Proc. Amer. Math. Soc.

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Abstract. Two topologies τ and σ on a fixed set are T 1 -complements if τ ∩ σ is the cofinite topology and τ ∪ σ is a sub-base for the discrete topology. In 1967, Steiner and Steiner showed that of any two T 1 -complements on a countable set, at least one is not Hausdorff. In 1969, Anderson and Stewart asked whether a Hausdorff topology on an uncountable set can have a Hausdorff T 1 -complement. We construct two homeomorphic completely regular T 1 -complementary topologies. Theorem 1. There are two Hausdorff topologies σ 0 and σ 1 on a set such that σ 0 ∨ σ 1 is the discrete topology, such that σ 0 ∧ σ 1 is the cofinite topology, and such that σ 0 and σ 1 are homeomorphic topologies.If σ 0 and σ 1 are topologies which satisfy • σ 0 ∨ σ 1 is the discrete topology, • σ 0 ∧ σ 1 is the cofinite topology, then we say σ 0 and σ 1 are T 1 -complements.In 1967, Steiner and Steiner [2] showed that of any pair of T 1 -complements on a countable set, at least one is not Hausdorff. In 1969, Anderson and Stewart [1] showed that of any pair of T 1 -complements, at least one is not first countable Hausdorff. Anderson and Stewart also asked: Can a Hausdorff topology on an (uncountable) set have a Hausdorff T 1 -complement ? We answer this question affirmatively.First we need a little finite combinatorics:There is a directed graph on a finite set containing the distinguished element p whose edges can be colored with three colors: green, red, and yellow, and two shades: light and dark, so that 1. The subgraph of light edges decomposes into components each of which is one of • the graph formed by identifying the initial vertex of a light green edge with the initial vertex of another light green edge, • the graph formed by identifying the terminal vertices of a light green edge and a light red edge,

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“…The study of complementarity in L 1 (X) was initiated by A. Steiner and E. Steiner in [6,8,7]. Later on, S. Watson used an elaborated combinatorics in [10] to prove that a set X of cardinality c + , where c = 2 ω , admits a Tychonoff self T 1 -complementary topology τ .…”

Section: Introductionmentioning

confidence: 99%