Specialization orders and subsumption
Specialization Partial Orders
A specialization partial order is a partial order where moving downward means “more specific” and moving upward means “more general.”
Let $(P, \sqsubseteq)$ be a poset:
- $a \sqsubseteq b$: “$a$ is subsumed by $b$” (or “$a$ specializes $b$”)
- $b \sqsupseteq a$: “$b$ subsumes $a$”
This is still a standard partial order, so it satisfies:
- Reflexivity: $a \sqsubseteq a$
- Antisymmetry: $a \sqsubseteq b$ and $b \sqsubseteq a$ implies $a = b$
- Transitivity: $a \sqsubseteq b$ and $b \sqsubseteq c$ implies $a \sqsubseteq c$
Core Relations
Equivalence
Two elements are equivalent when each subsumes the other:
\[a \equiv b \;\; \text{iff} \;\; a \sqsubseteq b \land b \sqsubseteq a\]In an antisymmetric poset, this collapses to equality.
Comparability
Two elements are comparable if one is below the other:
\[a \text{ comparable to } b \;\; \text{iff} \;\; a \sqsubseteq b \lor b \sqsubseteq a\]If neither holds, they are incomparable.
Top and Bottom
- Top element $\top$: subsumes every element ($x \sqsubseteq \top$ for all $x$)
- Bottom element $\bot$: is subsumed by every element ($\bot \sqsubseteq x$ for all $x$)
Not every poset has both, but many useful specialization hierarchies do.
Why This Matters in Category Theory
Every poset gives a category:
- Objects are elements of the poset
- There is exactly one morphism $a \to b$ when $a \sqsubseteq b$
So specialization hierarchies are a direct source of categories.
See: