Do you want to publish a course? Click here

Oscillator topologies on a paratopological group and related number invariants

338   0   0.0 ( 0 )
 Added by Taras Banakh
 Publication date 2012
  fields
and research's language is English




Ask ChatGPT about the research

We introduce and study oscillator topologies on paratopological groups and define certain related number invariants. As an application we prove that a Hausdorff paratopological group $G$ admits a weaker Hausdorff group topology provided $G$ is 3-oscillating. A paratopological group $G$ is 3-oscillating (resp. 2-oscillating) provided for any neighborhood $U$ of the unity $e$ of $G$ there is a neighborhood $Vsubset G$ of $e$ such that $V^{-1}VV^{-1}subset UU^{-1}U$ (resp. $V^{-1}Vsubset UU^{-1}$). The class of 2-oscillating paratopological groups includes all collapsing, all nilpotent paratopological groups, all paratopological groups satisfying a positive law, all paratopological SIN-group and all saturated paratopological groups (the latter means that for any nonempty open set $Usubset G$ the set $U^{-1}$ has nonempty interior). We prove that each totally bounded paratopological group $G$ is countably cellular; moreover, every cardinal of uncountable cofinality is a precaliber of $G$. Also we give an example of a saturated paratopological group which is not isomorphic to its mirror paratopological group as well as an example of a 2-oscillating paratopological group whose mirror paratopological group is not 2-oscillating.

rate research

Read More

232 - Taras Banakh 2010
We answer several questions of I.Protasov and E.Zelenyuk concerning topologies on groups determined by T-sequences. A special attention is paid to studying the operation of supremum of two group topologies.
(1) Every infinite, Abelian compact (Hausdorff) group K admits 2^|K|-many dense, non-Haar-measurable subgroups of cardinality |K|. When K is nonmetrizable, these may be chosen to be pseudocompact. (2) Every infinite Abelian group G admits a family A of 2^2^|G|-many pairwise nonhomeomorphic totally bounded group topologies such that no nontrivial sequence in G converges in any of the topologies T in A. (For some G one may arrange w(G,T) < 2^|G| for some T in A.) (3) Every infinite Abelian group $G$ admits a family B of 2^2^|G|-many pairwise nonhomeomorphic totally bounded group topologies, with w(G,T) = 2^|G| for all T in B, such that some fixed faithfully indexed sequence in G converges to 0_G in each T in B.
We prove that the existence of a selective ultrafilter implies the existence of a countably compact Hausdorff group topology on the free Abelian group of size continuum. As a consequence, we show that the existence of a selective ultrafilter implies the existence of a Wallace semigroup (i.e., a countably compact both-sided cancellative topological semigroup which is not a topological group).
A topological group $X$ is called $duoseparable$ if there exists a countable set $Ssubseteq X$ such that $SUS=X$ for any neighborhood $Usubseteq X$ of the unit. We construct a functor $F$ assigning to each (abelian) topological group $X$ a duoseparable (abelain-by-cyclic) topological group $FX$, containing an isomorphic copy of $X$. In fact, the functor $F$ is defined on the category of unital topologized magmas. Also we prove that each $sigma$-compact locally compact abelian topological group embeds into a duoseparable locally compact abelian-by-countable topological group.
By the {em Suslinian number} $Sln(X)$ of a continuum $X$ we understand the smallest cardinal number $kappa$ such that $X$ contains no disjoint family $C$ of non-degenerate subcontinua of size $|C|gekappa$. For a compact space $X$, $Sln(X)$ is the smallest Suslinian number of a continuum which contains a homeomorphic copy of $X$. Our principal result asserts that each compact space $X$ has weight $leSln(X)^+$ and is the limit of an inverse well-ordered spectrum of length $le Sln(X)^+$, consisting of compacta with weight $leSln(X)$ and monotone bonding maps. Moreover, $w(X)leSln(X)$ if no $Sln(X)^+$-Suslin tree exists. This implies that under the Suslin Hypothesis all Suslinian continua are metrizable, which answers a question of cite{DNTTT1}. On the other hand, the negation of the Suslin Hypothesis is equivalent to the existence of a hereditarily separable non-metrizable Suslinian continuum. If $X$ is a continuum with $Sln(X)<2^{aleph_0}$, then $X$ is 1-dimensional, has rim-weight $leSln(X)$ and weight $w(X)geSln(X)$. Our main tool is the inequality $w(X)leSln(X)cdot w(f(X))$ holding for any light map $f:Xto Y$.
comments
Fetching comments Fetching comments
Sign in to be able to follow your search criteria
mircosoft-partner

هل ترغب بارسال اشعارات عن اخر التحديثات في شمرا-اكاديميا