# Scott continuous

Let ${P}_{1},{P}_{2}$ be two dcpos (http://planetmath.org/Dcpo). A function $f:{P}_{1}\to {P}_{2}$ is said to be *Scott continuous* if for any directed set^{} $D\subseteq {P}_{1}$, $f(\bigvee D)=\bigvee f(D)$.

First, observe that $f$ is monotone^{}. If $a\le b$, then $f(b)=f(\bigvee \{a,b\})=\bigvee \{f(a),f(b)\}$, so that $f(a)\le f(b)$. As a result, if $D$ is directed, so is $f(D)$.

###### Proposition 1.

$f:{P}_{1}\to {P}_{2}$ is Scott continuous iff it is continuous^{} when ${P}_{\mathrm{1}}$ and ${P}_{\mathrm{2}}$ 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 $a\le b\in {P}_{1}$. We wish to show that $f(a)\le f(b)$, or $f(a)\in \downarrow f(b)$. Assume the contrary. Consider $U={P}_{2}-\downarrow f(b)$. Then $f(a)\in U$ and $U$ is Scott open, hence $a\in {f}^{-1}(U)$ is Scott open also. Since $a\le b$ and ${f}^{-1}(U)$ is upper, $b\in {f}^{-1}(U)$, which implies $f(b)\in U={P}_{2}-\downarrow f(b)$, a contradiction^{}. Therefore, $f(a)\le f(b)$.
∎

Now the proof of the proposition^{}.

###### Proof.

Suppose first that $f$ is Scott continuous. Take an open set $U\in {P}_{2}$. 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 $D\in {P}_{1}$ whenever its supremum^{} $\bigvee D$ lies in $V$. If $a\in \uparrow V$, then some $b\in V$ with $b\le a$, which implies $f(b)\le f(a)$. Since $f(b)\in U$, $f(a)\in \uparrow U=U$, so $a\in {f}^{-1}(U)=V$, $V$ is upper. Now, suppose $\bigvee D\in V$. So $\bigvee f(D)=f(\bigvee D)\in U$. Since $f(D)$ is directed, there is $y\in f(D)\cap U$, which means there is $x\in {P}_{1}$ such that $f(x)=y$ and $x\in D\cap V$. 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 $d=\bigvee D$. We want to show that $f(d)=\bigvee f(D)$. First, for any $e\in D$, we have that $e\le d$ so that $f(e)\le f(d)$ since $f$ is monotone. This shows $\bigvee f(D)\le f(d)$. Now suppose $r$ is any upper bound^{} of $f(D)$. We want to show that $f(d)\le r$, or $f(d)\in \downarrow r$. Assume not. Then $f(d)$ lies in $U:={P}_{2}-\downarrow r$, a Scott open set. So $\bigvee D=d\in {f}^{-1}(U)$, also Scott open, which implies some $e\in D$ with $e\in {f}^{-1}(U)$, or $f(e)\in U$. This means $f(e)\nleqq r$, a contradiction. Thus $f(d)\le r$, 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 |