Английская Википедия:Arens square

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Шаблон:Short description Шаблон:Multiple issues In mathematics, the Arens square is a topological space, named for Richard Friederich Arens. Its role is mainly to serve as a counterexample.

Definition

The Arens square is the topological space <math>(X,\tau),</math> where

<math>X=((0,1)^2\cap\mathbb{Q}^2)\cup\{(0,0)\}\cup\{(1,0)\}\cup\{(1/2,r\sqrt{2})|\ r\in\mathbb{Q},\ 0<r\sqrt{2}<1\}</math>

The topology <math>\tau</math> is defined from the following basis. Every point of <math>(0,1)^2\cap\mathbb{Q}^2</math> is given the local basis of relatively open sets inherited from the Euclidean topology on <math>(0,1)^2</math>. The remaining points of <math>X</math> are given the local bases

• <math>U_n(0,0)=\{(0,0)\}\cup\{(x,y)|\ 0<x<1/4,\ 0<y<1/n\}</math>
• <math>U_n(1,0)=\{(1,0)\}\cup\{(x,y)|\ 3/4<x<1,\ 0<y<1/n\}</math>
• <math>U_n(1/2,r\sqrt{2})=\{(x,y)|1/4<x<3/4,\ |y-r\sqrt{2}|<1/n\}</math>

Properties

The space <math>(X,\tau)</math> is:

1. T_2½, since neither points of <math>(0,1)^2\cap\mathbb{Q}^2</math>, nor <math>(0,0)</math>, nor <math>(0,1)</math> can have the same second coordinate as a point of the form <math>(1/2,r\sqrt{2})</math>, for <math>r\in\mathbb{Q}</math>.
2. not T₃ or T_3½, since for <math>(0,0)\in U_n(0,0)</math> there is no open set <math>U</math> such that <math>(0,0)\in U\subset \overline{U}\subset U_n(0,0)</math> since <math>\overline{U}</math> must include a point whose first coordinate is <math>1/4</math>, but no such point exists in <math>U_n(0,0)</math> for any <math>n\in\mathbb{N}</math>.
3. not Urysohn, since the existence of a continuous function <math>f:X\to [0,1]</math> such that <math>f(0,0)=0</math> and <math>f(1,0)=1</math> implies that the inverse images of the open sets <math>[0,1/4)</math> and <math>(3/4,1]</math> of <math>[0,1]</math> with the Euclidean topology, would have to be open. Hence, those inverse images would have to contain <math>U_n(0,0)</math> and <math>U_m(1,0)</math> for some <math>m,n\in\mathbb{N}</math>. Then if <math>r\sqrt{2}<\min\{1/n,1/m\}</math>, it would occur that <math>f(1/2,r\sqrt{2})</math> is not in <math>[0,1/4)\cap(3/4,1]=\emptyset</math>. Assuming that <math>f(1/2,r\sqrt{2})\notin[0,1/4)</math>, then there exists an open interval <math>U\ni f(1/2,r\sqrt{2})</math> such that <math>\overline{U}\cap[0,1/4)=\emptyset</math>. But then the inverse images of <math>\overline{U}</math> and <math>\overline{[0,1/4)}</math> under <math>f</math> would be disjoint closed sets containing open sets which contain <math>(1/2,r\sqrt{2})</math> and <math>(0,0)</math>, respectively. Since <math>r\sqrt{2}<\min\{1/n,1/m\}</math>, these closed sets containing <math>U_n(0,0)</math> and <math>U_k(1/2,r\sqrt{2})</math> for some <math>k\in\mathbb{N}</math> cannot be disjoint. Similar contradiction arises when assuming <math>f(1/2,r\sqrt{2})\notin(3/4,1]</math>.
4. semiregular, since the basis of neighbourhood that defined the topology consists of regular open sets.
5. second countable, since <math>X</math> is countable and each point has a countable local basis. On the other hand <math>(X,\tau)</math> is neither weakly countably compact, nor locally compact.
6. totally disconnected but not totally separated, since each of its connected components, and its quasi-components are all single points, except for the set <math>\{(0,0),(1,0)\}</math> which is a two-point quasi-component.
7. not scattered (every nonempty subset <math>A</math> of <math>X</math> contains a point isolated in <math>A</math>), since each basis set is dense-in-itself.
8. not zero-dimensional, since <math>(0,0)</math> doesn't have a local basis consisting of open and closed sets. This is because for <math>x\in[0,1]</math> small enough, the points <math>(x,1/4)</math> would be limit points but not interior points of each basis set.

References

• Lynn Arthur Steen and J. Arthur Seebach, Jr., Counterexamples in Topology. Springer-Verlag, New York, 1978. Reprinted by Dover Publications, New York, 1995. Шаблон:ISBN (Dover edition).