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# Scott continuous

Let $P_{1},P_{2}$ be two dcpos. 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 $a\leq b$, then $f(b)=f(\bigvee\{a,b\})=\bigvee\{f(a),f(b)\}$, so that $f(a)\leq f(b)$. 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 $P_{1}$ 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 $a\leq b\in P_{1}$. We wish to show that $f(a)\leq f(b)$, 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 $a\leq b$ 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, $f(a)\leq f(b)$. ∎

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 $b\leq a$, which implies $f(b)\leq f(a)$. 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 $e\leq d$ so that $f(e)\leq f(d)$ since $f$ is monotone. This shows $\bigvee f(D)\leq f(d)$. Now suppose $r$ is any upper bound of $f(D)$. We want to show that $f(d)\leq r$, 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 $f(e)\not\leq r$, a contradiction. Thus $f(d)\leq r$, 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).

## Mathematics Subject Classification

06B35*no label found*

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