Example. Monotonic but not continuous [dt-001M]

Liam O'Connor

Consider this function from $\mathbb {N} \cup \{\infty \} \rightarrow \{\top ,\bot \}$ , defined by: $f(x) = \begin {cases} \bot & \text {if}\ x \in \mathbb {N}\\ \top & \text {otherwise} \end {cases}$ This function is monotonic, but, taking $X = \mathbb {N}$ (which is a directed subset of the cpo $\mathbb {N} \cup \{\infty \}$ ), then $f(\bigsqcup \mathbb {N}) = f(\infty ) = \top$ , but $\bigsqcup \{ f(n) \mid n \in \mathbb {N} \} = \bigsqcup \{ \bot \} = \bot$ . Thus, this function is not continuous.

Example. Monotonic but not continuous [dt-001M]

Problem. Monotonicity is insufficient [dt-001K]

Definition. Continuity [dt-001J]

Theorem. Continuity implies monotonicity [dt-001L]

Proof.

Example. Monotonic but not continuous [dt-001M]

Exercise. [dt-001N]

Solution.

Theorem. Alternative characterisation of continuity [dt-001O]

Proof.

Thesis.

Exercise. [dt-001W]

Solution.

Theorem. The Kleene fixed point theorem [dt-001P]

Proof.

Definition. The $\textbf {fix}$ operator [dt-001Q]

Exercise. [dt-001S]

Solution.

Solution.

Solution.