Scott continuous

Scott continuous

Let

P

1

,

P

2

be two dcpos (

http://planetmath.org/Dcpo

). A

function

f

:

P

1

→

P

2

is said to be

Scott continuous

if for any

directed set

D

⊆

P

1

,

f

⁢

(

⋁

D

)

=

⋁

f

⁢

(

D

)

.

First, observe that

f

is

monotone

. If

a

≤

b

, then

f

⁢

(

b

)

=

f

⁢

(

⋁

{

a

,

b

}

)

=

⋁

{

f

⁢

(

a

)

,

f

⁢

(

b

)

}

, so that

f

⁢

(

a

)

≤

f

⁢

(

b

)

. As a result, if

D

is directed, so is

f

⁢

(

D

)

.

Proposition 1

.

f

:

P

1

→

P

2

is Scott continuous iff it is

continuous

when

P

1

and

P

2

are equipped with the

Scott topologies

.

Before proving this, let’s make one additional observation:

Lemma 1

.

If

f

is continuous (under Scott topologies), then

f

is monotone.

Proof.

Suppose

a

≤

b

∈

P

1

. We wish to show that

f

⁢

(

a

)

≤

f

⁢

(

b

)

, or

f

(

a

)

∈

↓

f

(

b

)

. Assume the contrary. Consider

U

=

P

2

-

↓

f

⁢

(

b

)

. Then

f

⁢

(

a

)

∈

U

and

U

is Scott open, hence

a

∈

f

-

1

⁢

(

U

)

is Scott open also. Since

a

≤

b

and

f

-

1

⁢

(

U

)

is upper,

b

∈

f

-

1

⁢

(

U

)

, which implies

f

⁢

(

b

)

∈

U

=

P

2

-

↓

f

⁢

(

b

)

, a

contradiction

. Therefore,

f

⁢

(

a

)

≤

f

⁢

(

b

)

.
∎

Now the proof of the

proposition

.

Proof.

Suppose first that

f

is Scott continuous. Take an open set

U

∈

P

2

. We want to show that

V

:=

f

-

1

⁢

(

U

)

is open in

P

1

. In other words,

V

is upper and that

V

has non-empty

intersection

with any directed set

D

∈

P

1

whenever its

supremum

⋁

D

lies in

V

. If

a

∈

↑

V

, then some

b

∈

V

with

b

≤

a

, which implies

f

⁢

(

b

)

≤

f

⁢

(

a

)

. Since

f

⁢

(

b

)

∈

U

,

f

(

a

)

∈

↑

U

=

U

, so

a

∈

f

-

1

⁢

(

U

)

=

V

,

V

is upper. Now, suppose

⋁

D

∈

V

. So

⋁

f

⁢

(

D

)

=

f

⁢

(

⋁

D

)

∈

U

. Since

f

⁢

(

D

)

is directed, there is

y

∈

f

⁢

(

D

)

∩

U

, which means there is

x

∈

P

1

such that

f

⁢

(

x

)

=

y

and

x

∈

D

∩

V

. This shows that

V

is Scott open.

Conversely

, suppose

f

is continuous (

inverse

of a Scott open set is Scott open). Let

D

be a directed subset of

P

1

and let

d

=

⋁

D

. We want to show that

f

⁢

(

d

)

=

⋁

f

⁢

(

D

)

. First, for any

e

∈

D

, we have that

e

≤

d

so that

f

⁢

(

e

)

≤

f

⁢

(

d

)

since

f

is monotone. This shows

⋁

f

⁢

(

D

)

≤

f

⁢

(

d

)

. Now suppose

r

is any

upper bound

of

f

⁢

(

D

)

. We want to show that

f

⁢

(

d

)

≤

r

, or

f

(

d

)

∈

↓

r

. Assume not. Then

f

⁢

(

d

)

lies in

U

:=

P

2

-

↓

r

, a Scott open set. So

⋁

D

=

d

∈

f

-

1

⁢

(

U

)

, also Scott open, which implies some

e

∈

D

with

e

∈

f

-

1

⁢

(

U

)

, or

f

⁢

(

e

)

∈

U

. This means

f

⁢

(

e

)

≰

r

, a contradiction. Thus

f

⁢

(

d

)

≤

r

, and the proof is

complete

.
∎

Remark

. This notion of

continuity

is attributed to Dana Scott when he was trying to come up with a model for the

formal system

of untyped

lambda calculus

.

References

1

G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove, D. S. Scott,

Continuous Lattices and Domains

, Cambridge University Press, Cambridge (2003).

Title

Scott continuous

Canonical

name

ScottContinuous

Date of creation

2013-03-22 16:49:53

Last modified on

2013-03-22 16:49:53

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06B35

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