اشترك بالحزمة الذهبية واحصل على وصول غير محدود شمرا أكاديميا
تسجيل مستخدم جديدNonexistence of Countable Extremally Disconnected Groups with Many Open Subgroups
123
0
0.0
(
0
)
تأليف
Olga Sipacheva
اسأل ChatGPT حول البحث
ﻻ يوجد ملخص باللغة العربية
It is proved that the existence of a countable extremally disconnected Boolean topological group containing a family of open subgroups whose intersection has empty interior implies the existence of a rapid ultrafilter.
قيم البحث
اقرأ أيضاً
It is proved that any countable topological group in which the filter of neighborhoods of the identity element is not rapid contains a discrete set with precisely one nonisolated point. This gives a negative answer to Protasovs question on the existe
nce in ZFC of a countable nondiscrete group in which all discrete subsets are closed. It is also proved that the existence of a countable nondiscrete extremally disconnected group implies the existence of a rapid ultrafilter and, hence, a countable nondiscrete extremally disconnected group cannot be constructed in ZFC.
Given an arbitrary measurable cardinal $kappa$, a nondiscrete Hausdorff extremally disconnected topological group of cardinality $kappa$ is constructed.
A group topology is said to be linear if open subgroups form a base of neighborhoods of the identity element. It is proved that the existence of a nondiscrete extremally disconnected group of Ulam nonmeasurable cardinality with linear topology implie
s that of a nondiscrete extremally disconnected group of cardinality at most $2^omega$ with linear topology.
In this paper, we prove that: (1) Let $f:Grightarrow H$ be a continuous $d$-open surjective homomorphism; if $G$ is an $mathbb{R}$-factorizabile paratopological group, then so is $H$. Peng and Zhangs result cite[Theorem 1.7]{PZ} is improved. (2) Let
$G$ be a regular $mathbb{R}$-factorizable paratopological group; then every subgroup $H$ of $G$ is $mathbb{R}$-factorizable if and only if $H$ is $z$-embedded in $G$. This result gives out a positive answer to an question of M.~Sanchis and M.~Tkachenko cite[Problem 5.3]{ST}.
A Hausdorff topological space $X$ is called $textit{superconnected}$ (resp. $textit{coregular}$) if for any nonempty open sets $U_1,dots U_nsubseteq X$, the intersection of their closures $bar U_1capdotscapbar U_n$ is not empty (resp. the complement
$Xsetminus (bar U_1capdotscapbar U_n)$ is a regular topological space). A canonical example of a coregular superconnected space is the projective space $mathbb Qmathsf P^infty$ of the topological vector space $mathbb Q^{<omega}={(x_n)_{ninomega}in mathbb Q^{omega}:|{ninomega:x_n e 0}|<omega}$ over the field of rationals $mathbb Q$. The space $mathbb Qmathsf P^infty$ is the quotient space of $mathbb Q^{<omega}setminus{0}^omega$ by the equivalence relation $xsim y$ iff $mathbb Q{cdot}x=mathbb Q{cdot}y$. We prove that every countable second-countable coregular space is homeomorphic to a subspace of $mathbb Qmathsf P^infty$, and a topological space $X$ is homeomorphic to $mathbb Qmathsf P^infty$ if and only if $X$ is countable, second-countable, and admits a decreasing sequence of closed sets $(X_n)_{ninomega}$ such that (i) $X_0=X$, $bigcap_{ninomega}X_n=emptyset$, (ii) for every $ninomega$ and a nonempty open set $Usubseteq X_n$ the closure $bar U$ contains some set $X_m$, and (iii) for every $ninomega$ the complement $Xsetminus X_n$ is a regular topological space. Using this topological characterization of $mathbb Qmathsf P^infty$ we find topological copies of the space $mathbb Qmathsf P^infty$ among quotient spaces, orbit spaces of group actions, and projective spaces of topological vector spaces over countable topological fields.
سجل دخول لتتمكن من نشر تعليقات
التعليقات
جاري جلب التعليقات
سجل دخول لتتمكن من متابعة معايير البحث التي قمت باختيارها
