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2. A WAY TO CONSTRUCT DIRECTED TOPOLOGICAL SPACES

24

3

. ∆

>

(

a

) =

d

F

n

]

a

;

y

[

y

A

,x<a

y>a

o

;

4

. ∆

(

a

) =

d

F

n

]

x

;

a

]

x

A

,x<a

o

;

5

. ∆

<

(

a

) =

d

F

n

]

x

;

a

[

x

A

,x<a

o

;

6

. ∆

6

=

(

a

) = ∆(

a

)

\ {

a

}

.

Obvious

2159

.

1

. ∆

(

a

) = ∆(

a

)

u

F

@

J

(

a

);

2

. ∆

>

(

a

) = ∆(

a

)

u

F

@

J

>

(

a

);

3

. ∆

(

a

) = ∆(

a

)

u

F

@

J

(

a

);

4

. ∆

<

(

a

) = ∆(

a

)

u

F

@

J

<

(

a

);

5

. ∆

6

=

(

a

) = ∆(

a

)

u

F

@

J

6

=

(

a

).

Definition

2160

.

Given a partial order

A

and

x

A

, the following defines

complete funcoids:

1

.

h|

A

|i

{

x

}

= ∆(

x

);

2

.

h|

A

|

i

{

x

}

= ∆

(

x

);

3

.

h|

A

|

>

i

{

x

}

= ∆

>

(

x

);

4

.

h|

A

|

i

{

x

}

= ∆

(

x

);

5

.

h|

A

|

<

i

{

x

}

= ∆

<

(

x

);

6

.

h|

A

|

6

=

i

{

x

}

= ∆

6

=

(

x

).

Proposition

2161

.

The complete funcoid corresponding to the order topol-

ogy

1

is equal to

|

A

|

.

Proof.

Because every open set is a finite union of open intervals, the com-

plete funcoid

f

corresponding to the order topology is described by the formula:

h

f

i

{

x

}

=

d

F

n

]

a

;

b

[

a,b

A

,a<x

b>x

o

= ∆(

x

) =

h|

A

|i

{

x

}

. Thus

f

=

|

A

|

.

Exercise

2162

.

Show that

|

A

|

(in general) is not the same as “right order

topology”

2

.

Proposition

2163

.

1

.

D

|

A

|

1

E

@

X

= @

n

a

A

y

A

:(

y>a

X

[

a

;

y

[

6

=

)

o

;

2

.

|

A

|

1

>

@

X

= @

n

a

A

y

A

:(

y>a

X

]

a

;

y

[

6

=

)

o

;

3

.

D

|

A

|

1

E

@

X

= @

n

a

A

x

A

:(

x<a

X

]

x

;

a

]

6

=

)

o

;

4

.

|

A

|

1

<

@

X

= @

n

a

A

x

A

:(

x<a

X

]

x

;

a

[

6

=

)

o

.

Proof.

a

D

|

A

|

1

E

@

X

@

{

a

} 6

D

|

A

|

1

E

@

X

⇔ h|

A

|

i

@

{

a

} 6

@

X

(

a

)

6

@

X

⇔ ∀

y

A

: (

y > a

X

[

a

;

y

[

6

=

).

a

|

A

|

1

>

@

X

@

{

a

} 6

|

A

|

1

>

@

X

⇔ h|

A

|

>

i

@

{

a

} 6

@

X

>

(

a

)

6

@

X

⇔ ∀

y

A

: (

y > a

X

]

a

;

y

[

6

=

).

The rest follows from duality.

Remark

2164

.

On trivial ultrafilters these obviously agree:

1

.

h|

R

|

i

{

x

}

=

h|

R

|u ≥i

{

x

}

;

2

.

h|

R

|

>

i

{

x

}

=

h|

R

|u

>

i

{

x

}

;

3

.

h|

R

|

i

{

x

}

=

h|

R

|u ≤i

{

x

}

;

4

.

h|

R

|

<

i

{

x

}

=

h|

R

|u

<

i

{

x

}

.

1

See Wikipedia for a definition of “Order topology”.

2

See Wikipedia