domain
A poset $P$ is said to be a directed complete partially ordered set, or dcpo for short, if every directed set^{} $D\subseteq P$ has a supremum^{}. Since the empty set^{} is directed, a dcpo must have a bottom element.
A domain $P$ is a continuous dcpo. Here, continuous means that $P$ is a continuous poset.
Example. Let $A,B$ be sets. Consider the set $P$ of all partial functions^{} from $A$ to $B$. This means that any $f\in P$ is a function $C\to B$, for some subset $C$ of $A$. We show that $P$ is a domain.

1.
$P$ is a poset: Define a binary relation^{} on $P$ as follows: $f\le g$ iff $g$ is an extension^{} of $f$. In other words, if $f:C\to B$ and $g:D\to B$, then $C\subseteq D$ and $f(x)=g(x)$ for all $x\in C$. Clearly, $\le $ is reflexive^{}, antisymmetric, and transitive^{}. So $\le $ turns $P$ into a poset.

2.
$P$ is a dcpo: Suppose that $D$ is a directed subset of $P$. Set $E=\bigcup \{\mathrm{dom}(f)\mid f\in D\}$. Define $g:E\to B$ as follows: for any $x\in E$, $g(x)=f(x)$ where $x\in \mathrm{dom}(f)$ for some $f\in D$. Is this welldefined? Suppose $x\in \mathrm{dom}({f}_{1})\cap \mathrm{dom}({f}_{2})$. Since $D$ is directed, there is an $f\in D$ extending both ${f}_{1}$ and ${f}_{2}$. This means that ${f}_{1}(x)=f(x)={f}_{2}(x)$. Therefore, $g:=\bigvee D$ is a welldefined function (on $E$). Hence $P$ is a dcpo.

3.
If $f,g\ll h$, then $f\vee g\ll h$: Since $h$ extends both $f$ and $g$, $a:=f\vee g:\mathrm{dom}(f)\cup \mathrm{dom}(g)\to B$ is welldefined (the construction is the same as above). To show that $a\ll h$, suppose $D\subseteq P$ is directed and $h\le \bigvee D$ (note that $\bigvee D$ exists by 2 above). Since $f\ll h$, there is $r\in D$ such that $f\le r$. Similarly, $g\ll h$ implies an $s\in D$ with $g\le s$. Since $D$ is directed, there is $t\in D$ with $r,s\le t$. This means $f\le t$ and $g\le t$, or $a=f\vee g\le t$.

4.
$P$ is continuous: Let $\mathrm{wb}(h)=\{f\in P\mid f\ll h\}$. Then by 3 above, $\mathrm{wb}(h)$ is a directed set. By 2, $b:=\bigvee \mathrm{wb}(h)$ exists, and $b\le h$. Suppose $x\in \mathrm{dom}(h)$. Then the function ${c}_{x}:\{x\}\to B$ defined by ${c}_{x}(x)=h(x)$ is way below $h$, for if $h\le \bigvee D$, then $x\in \mathrm{dom}(f)$ for some $f\in D$, or $\mathrm{dom}({c}_{x})=\{x\}\subseteq \mathrm{dom}(f)$, which means ${c}_{x}\le f$. Therefore, ${c}_{x}\le b$. This implies that $\mathrm{dom}(h)=\bigvee \{\mathrm{dom}({c}_{x})\mid x\in \mathrm{dom}(h)\}\subseteq \mathrm{dom}(b)$. As a result, $h\le b$.
Remark. Domain theory is a branch of order theory that is used extensively in theoretical computer science. As in the example, one can think of a domain as a collection^{} of partial pictures or pieces of partial information. Being continuous is the same as saying that any picture or piece of information can be pieced together by partial ones by way of “approximations”.
Title  domain 
Canonical name  Domain12 
Date of creation  20130322 16:49:25 
Last modified on  20130322 16:49:25 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  8 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06B35 
Synonym  directed complete 
Synonym  directed complete poset 
Synonym  directed complete partially ordered set 
Related topic  CompleteLattice 
Defines  domain 
Defines  dcpo 