# A closed graph theorem in dcpo’s

Today I will add one closed graph type theorem to the list made by Terence Tao in his post Closed graph theorem in various categories.

I first recall a few basic notions of domain theory. If $X$ is a poset or partially ordered set, a subset $D$ of $X$ is directed if it is nonempty and for all $d, d' \in D$ there is some $d'' \in D$ such that $d \leq d''$ and $d' \leq d''$. The poset $X$ is a dcpo (a directed-complete poset) if every directed subset has a supremum, written $\bigvee D$. A subset $U$ of $X$ is d-open if $\bigvee D \in U \implies D \cap U \neq \emptyset$ for every directed subset $D$ of $X$. The collection of d-open subsets forms a topology called the d-topology.

Then a map $f : X \rightarrow Y$ is called d-continuous if it is continuous when $X$ and $Y$ are equipped with their respective d-topologies. It happens that a map $f : X \rightarrow Y$ is order-preserving and d-continuous if and only if $f(\bigvee D) = \bigvee f(D)$ for every directed subset $D$ of $X$ (which is a characterization of Scott continuity). For a proof of this latter assertion and for more on the d-topology of a poset I refer the reader to the paper D-completions and the d-topology by Klaus Keimel and Jimmie Lawson.

Theorem (Closed graph theorem (dcpo theory)). Let $X, Y$ be dcpo’s. Then an order-preserving map $f : X \rightarrow Y$ is d-continuous if and only if the graph $\Sigma := \{ (x, f(x)) : x \in X \}$ is d-closed (in the poset $X \times Y$ equipped with the product topology).

Before giving a proof to this theorem, it is important to recall that the d-topology on the poset $X \times Y$ does not coincide in general with the product of the d-topologies on $X$ and $Y$.

Proof. Suppose that $\Sigma$ is d-closed, and let $D$ be a directed subset of $X$. We define $\Delta := \{ (d, f(d)) : d \in D \}$, which is a subset of $\Sigma$, and a directed subset of the poset $X \times Y$ since $f$ is order-preserving. It is easily seen that $\Delta$ has a supremum in $X \times Y$ equal to $(\bigvee D, \bigvee f(D))$; since $\Sigma$ is d-closed, this supremum is in $\Sigma$, hence $f(\bigvee D) = \bigvee f(D)$. $\Box$

Question. Do we have a closed graph theorem in (continuous?) dcpo’s equipped with the Lawson topology?