Theorem. Compact basis for algebraic cpos [dt-004B]

Liam O'Connor

If $X$ is a basis for $A$ , then $A$ is algebraic and $\mathsf {K}(A) = X$ .

Proof.

Liam O'Connor

If $a \in \mathsf {K}(A)$ , then $\bigsqcup M = a$ where $M = \{ x \in X \mid x \sqsubseteq a \}$ , as $X$ is a basis. Since $a$ is compact, $a \sqsubseteq b$ for some $b \in M$ . But since $a$ is the lub of $M$ , $b \sqsubseteq a$ as well. By antisymmetry $a = b$ , hence $a \in X$ . Thus $\mathsf {K}(A) \subseteq X$ , so $\mathsf {K}(A) = X$ and $A$ is algebraic.

Theorem. Compact basis for algebraic cpos [dt-004B]

Proof.

Definition. Basis [dt-004A]

Theorem. Compact basis for algebraic cpos [dt-004B]

Proof.

Theorem. Closure of algebraic cpos under product [dt-004C]

Proof.

Problem.

Aside. Scott's original solution: using complete lattices [dt-004D]

Definition. Consistent completeness [dt-004E]

Example. Consistent completeness vs directed completeness [dt-004F]

Definition. Scott domain [dt-004G]

Remark. [dt-004H]

Theorem. Closure of Scott domains [dt-004I]

Thesis.

Proof.