Arens square explained

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

(X,\tau), whereX = (( 0 , 1 ) ^{2\capQ} ^{2)\cup\{(0,0)\}\cup\{(1,0)\}\cup\{(1/2,r\sqrt{2})| r\inQ, 0<r\sqrt{2}<1\}}

The topology

\tau is defined from the following basis. Every point of ( 0 , 1 ) ^{2\capQ} ^{2} is given the local basis of relatively open sets inherited from the Euclidean topology on ( 0 , 1 ) ^{2} . The remaining points of X are given the local basesU _{n(0,0)=\{(0,0)\}\cup\{(x,y)| 0<x<1/4, 0<y<1/n\}} U _{n(1,0)=\{(1,0)\}\cup\{(x,y)| 3/4<x<1, 0<y<1/n\}} U _{n(1/2,r\sqrt{2})=\{(x,y)|1/4<x<3/4, |y-r\sqrt{2}|<1/n\}}

Properties

The space

(X,\tau) is:

1. T_2½, since neither points of

( 0 , 1 ) ^{2\capQ} ^{2} , nor ( 0 , 0 ) , nor ( 0 , 1 ) can have the same second coordinate as a point of the form ( 1 / 2 ,r\sqrt{ 2 }) , for r\in Q .

1. not T₃ or T_3½, since for

( 0 , 0 )\in U _{n(0,0)} there is no open set U such that ( 0 , 0 )\in U\subset \overline{U}\subset U _{n(0,0)} since \overline{U} must include a point whose first coordinate is 1 / 4 , but no such point exists in U _{n(0,0)} for any n\in N .

1. not Urysohn, since the existence of a continuous function

f:X\to [ 0 , 1 ] such that f( 0 , 0 ) =0 and f( 1 , 0 ) =1 implies that the inverse images of the open sets [ 0 , 1 / 4 ) and ( 3 / 4 , 1 ] of [ 0 , 1 ] with the Euclidean topology, would have to be open. Hence, those inverse images would have to contain U _{n(0,0)} and U _{m(1,0)} for some m,n\in N . Then if r\sqrt{ 2 }< min \{ 1 /n, 1 /m\} , it would occur that f( 1 / 2 ,r\sqrt{ 2 }) is not in [ 0 , 1 / 4 )\cap( 3 / 4 , 1 ] = \emptyset . Assuming that f( 1 / 2 ,r\sqrt{ 2 })\notin[ 0 , 1 / 4 ) , then there exists an open interval U\ni f( 1 / 2 ,r\sqrt{ 2 }) such that \overline{U}\cap[ 0 , 1 / 4 ) = \emptyset . But then the inverse images of \overline{U} and \overline{[ 0 , 1 / 4 )} under f would be disjoint closed sets containing open sets which contain ( 1 / 2 ,r\sqrt{ 2 }) and ( 0 , 0 ) , respectively. Since r\sqrt{ 2 }< min \{ 1 /n, 1 /m\} , these closed sets containing U _{n(0,0)} and U _{k(1/2,r\sqrt{2})} for some k\in N cannot be disjoint. Similar contradiction arises when assuming f( 1 / 2 ,r\sqrt{ 2 })\notin( 3 / 4 , 1 ] .

1. semiregular, since the basis of neighbourhood that defined the topology consists of regular open sets.
2. second countable, since

X is countable and each point has a countable local basis. On the other hand (X,\tau) is neither weakly countably compact, nor locally compact.

1. totally disconnected but not totally separated, since each of its connected components, and its quasi-components are all single points, except for the set

\{( 0 , 0 ),( 1 , 0 )\} which is a two-point quasi-component.

1. not scattered (every nonempty subset

A of X contains a point isolated in A ), since each basis set is dense-in-itself.

1. not zero-dimensional, since

( 0 , 0 ) doesn't have a local basis consisting of open and closed sets. This is because for x\in[ 0 , 1 ] small enough, the points (x, 1 / 4 ) 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. (Dover edition).