• Source: Arens square

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
,
τ
)
,


{\displaystyle (X,\tau ),}

where




X
=
(
(
0
,
1

)

2


∩


Q


2


)
∪
{
(
0
,
0
)
}
∪
{
(
1
,
0
)
}
∪
{
(
1

/

2
,
r


2


)

|


r
∈

Q

,

0
<
r


2


<
1
}


{\displaystyle 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

The topology



τ


{\displaystyle \tau }

is defined from the following basis. Every point of



(
0
,
1

)

2


∩


Q


2




{\displaystyle (0,1)^{2}\cap \mathbb {Q} ^{2}}

is given the local basis of relatively open sets inherited from the Euclidean topology on



(
0
,
1

)

2




{\displaystyle (0,1)^{2}}

. The remaining points of



X


{\displaystyle X}

are given the local bases





U

n


(
0
,
0
)
=
{
(
0
,
0
)
}
∪
{
(
x
,
y
)

|


0
<
x
<
1

/

4
,

0
<
y
<
1

/

n
}


{\displaystyle U_{n}(0,0)=\{(0,0)\}\cup \{(x,y)|\ 0





U

n


(
1
,
0
)
=
{
(
1
,
0
)
}
∪
{
(
x
,
y
)

|


3

/

4
<
x
<
1
,

0
<
y
<
1

/

n
}


{\displaystyle U_{n}(1,0)=\{(1,0)\}\cup \{(x,y)|\ 3/4





U

n


(
1

/

2
,
r


2


)
=
{
(
x
,
y
)

|

1

/

4
<
x
<
3

/

4
,


|

y
−
r


2



|

<
1

/

n
}


{\displaystyle U_{n}(1/2,r{\sqrt {2}})=\{(x,y)|1/4




Properties


The space



(
X
,
τ
)


{\displaystyle (X,\tau )}

is:

T2½, since neither points of



(
0
,
1

)

2


∩


Q


2




{\displaystyle (0,1)^{2}\cap \mathbb {Q} ^{2}}

, nor



(
0
,
0
)


{\displaystyle (0,0)}

, nor



(
0
,
1
)


{\displaystyle (0,1)}

can have the same second coordinate as a point of the form



(
1

/

2
,
r


2


)


{\displaystyle (1/2,r{\sqrt {2}})}

, for



r
∈

Q



{\displaystyle r\in \mathbb {Q} }

.
not T3 or T3½, since for



(
0
,
0
)
∈

U

n


(
0
,
0
)


{\displaystyle (0,0)\in U_{n}(0,0)}

there is no open set



U


{\displaystyle U}

such that



(
0
,
0
)
∈
U
⊂


U
¯


⊂

U

n


(
0
,
0
)


{\displaystyle (0,0)\in U\subset {\overline {U}}\subset U_{n}(0,0)}

since





U
¯




{\displaystyle {\overline {U}}}

must include a point whose first coordinate is



1

/

4


{\displaystyle 1/4}

, but no such point exists in




U

n


(
0
,
0
)


{\displaystyle U_{n}(0,0)}

for any



n
∈

N



{\displaystyle n\in \mathbb {N} }

.
not Urysohn, since the existence of a continuous function



f
:
X
→
[
0
,
1
]


{\displaystyle f:X\to [0,1]}

such that



f
(
0
,
0
)
=
0


{\displaystyle f(0,0)=0}

and



f
(
1
,
0
)
=
1


{\displaystyle f(1,0)=1}

implies that the inverse images of the open sets



[
0
,
1

/

4
)


{\displaystyle [0,1/4)}

and



(
3

/

4
,
1
]


{\displaystyle (3/4,1]}

of



[
0
,
1
]


{\displaystyle [0,1]}

with the Euclidean topology, would have to be open. Hence, those inverse images would have to contain




U

n


(
0
,
0
)


{\displaystyle U_{n}(0,0)}

and




U

m


(
1
,
0
)


{\displaystyle U_{m}(1,0)}

for some



m
,
n
∈

N



{\displaystyle m,n\in \mathbb {N} }

. Then if



r


2


<
min
{
1

/

n
,
1

/

m
}


{\displaystyle r{\sqrt {2}}<\min\{1/n,1/m\}}

, it would occur that



f
(
1

/

2
,
r


2


)


{\displaystyle f(1/2,r{\sqrt {2}})}

is not in



[
0
,
1

/

4
)
∩
(
3

/

4
,
1
]
=
∅


{\displaystyle [0,1/4)\cap (3/4,1]=\emptyset }

. Assuming that



f
(
1

/

2
,
r


2


)
∉
[
0
,
1

/

4
)


{\displaystyle f(1/2,r{\sqrt {2}})\notin [0,1/4)}

, then there exists an open interval



U
∋
f
(
1

/

2
,
r


2


)


{\displaystyle U\ni f(1/2,r{\sqrt {2}})}

such that





U
¯


∩
[
0
,
1

/

4
)
=
∅


{\displaystyle {\overline {U}}\cap [0,1/4)=\emptyset }

. But then the inverse images of





U
¯




{\displaystyle {\overline {U}}}

and






[
0
,
1

/

4
)

¯




{\displaystyle {\overline {[0,1/4)}}}

under



f


{\displaystyle f}

would be disjoint closed sets containing open sets which contain



(
1

/

2
,
r


2


)


{\displaystyle (1/2,r{\sqrt {2}})}

and



(
0
,
0
)


{\displaystyle (0,0)}

, respectively. Since



r


2


<
min
{
1

/

n
,
1

/

m
}


{\displaystyle r{\sqrt {2}}<\min\{1/n,1/m\}}

, these closed sets containing




U

n


(
0
,
0
)


{\displaystyle U_{n}(0,0)}

and




U

k


(
1

/

2
,
r


2


)


{\displaystyle U_{k}(1/2,r{\sqrt {2}})}

for some



k
∈

N



{\displaystyle k\in \mathbb {N} }

cannot be disjoint. Similar contradiction arises when assuming



f
(
1

/

2
,
r


2


)
∉
(
3

/

4
,
1
]


{\displaystyle f(1/2,r{\sqrt {2}})\notin (3/4,1]}

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



X


{\displaystyle X}

is countable and each point has a countable local basis. On the other hand



(
X
,
τ
)


{\displaystyle (X,\tau )}

is neither weakly countably compact, nor locally compact.
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
)
}


{\displaystyle \{(0,0),(1,0)\}}

which is a two-point quasi-component.
not scattered (every nonempty subset



A


{\displaystyle A}

of



X


{\displaystyle X}

contains a point isolated in



A


{\displaystyle A}

), since each basis set is dense-in-itself.
not zero-dimensional, since



(
0
,
0
)


{\displaystyle (0,0)}

doesn't have a local basis consisting of open and closed sets. This is because for



x
∈
[
0
,
1
]


{\displaystyle x\in [0,1]}

small enough, the points



(
x
,
1

/

4
)


{\displaystyle (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. ISBN 0-486-68735-X (Dover edition).

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