Proper subsets

Proper subsets

Idea

A subset $A$ of a set $S$ is proper if it is not the improper subset of $S$ ($S$ as a subset of itself). We may also say (in the context of concrete sets) that $S$ is a proper superset of $A$ or that $S$ properly contains? $A$.

Definition

In material set theory

In material set theory, we may state that $A$ is proper in any of these equivalent ways:

1. $A$ is not equal (as a subset) to $S$;
2. There exists an element $x$ of $S$ such that $x \notin A$;
3. Given any way of expressing $A$ as the intersection of a family of subsets of $S$, this family is inhabited.

Actually, these three definitions are equivalent only if we accept the principle of excluded middle; in constructive mathematics, we usually prefer (3). (For example, consider the notion of proper filter on a set $X$, thought of as a subset of the power set of $X$.) However, (3) is not predicative; see positive element for discussion of this in the dual context. Also, (2) may be strengthened using an inequality relation other than the denial inequality.

In structural set theory

In structural set theory, subsets are represented by injections, and so we may state that the injection $m:A \hookrightarrow S$ is proper in any of these equivalent ways:

1. The injection $m:A \hookrightarrow S$ is not an bijection;
2. There exists an element $x$ of $S$ such that for all elements $y \in A$, $m(x) \neq y$;
3. Given any way of expressing $m:A \hookrightarrow S$ as the intersection of a family of injections into $S$, this family is inhabited.

Similarly as the case in material set theory, these three definitions are equivalent only if we accept the principle of excluded middle.

More generally

In any category $C$, subsets become subobjects and are thus represented by the monomorphisms in $C$. Then a proper subobject is defined in any of the following ways:

1. The monomorphism $m:A \hookrightarrow S$ is not an isomorphism;
2. If $C$ is a well-pointed category with terminal object $1$, then there exists a global element $x:1 \to S$ such that for all global elements $y:1 \to A$, $m \circ x \neq y$;
3. If the subobject preorder of every object in $C$ is a pre-inflattice, then given any way of expressing $m:A \hookrightarrow S$ as the intersection of a family of monomorphisms into $S$, this family is inhabited.