# Are all Scott-continuous functions computable?

A chain-complete partial order (equivalently, a pointed dcpo) is a set $$D$$ with a partial order $$\leq$$ such that all chains of $$D$$ have a supremum. The least upper bound ($$\bigsqcup$$) of the empty chain is the least element $$\bot$$ of the CCPO.

A function $$f\colon M \to N$$ is monotone if for all $$a, b \in M$$, the following holds: $$a \leq b \implies f(a) \leq f(b)$$

A function $$f\colon M \to N$$ between two CCPOs is Scott-continuous if it is monotone and for every chain $$C$$ of $$M$$, we have

$$f(\bigsqcup_{c \in C} c) = \bigsqcup_{m \in C} f(c)\,.$$

Scott-continuous functions play an important part in defining denotational semantics of programs, and as is well-known in computing science, every Turing-computable function is Scott-continuous$$^0$$. Is the converse true? Is every Scott-continuous function computable?

1. See this question, from which I took and edited some definitions.

$$\newcommand{\TM}{T\!M}$$
Let $$\TM$$ be the set of Turing machines and $$2 = \{0,1\}$$.
The powerset of any set with $$\leq$$ equal to $$\subseteq$$ is a CCPO, its least upper bounds being the unions of the chains.
Take the function $$f\colon P(\TM) \to P(\TM \times 2)$$ that maps a set of Turing machines $$S$$ to $$\{ (t, h) \mid t \in S \text{ and h equals whether t halts on empty input } \}$$
The function $$f$$ is Scott-continuous $$-$$ it is defined pointwise, so it's monotonous and $$f(\bigcup_{c \in C} c) = \bigcup_{c \in C} f(c)\,,$$ and therefore $$f(\bigsqcup_{c \in C} c) = \bigsqcup_{c \in C} f(c)\,.$$