As mentioned earlier, the key difference between bounded and unbounded objects is that the former can be members of other sets while the latter cannot. This fact establishes a necessary and sufficient test for boundedness - an object is bounded just in case it is a member of a set. However, this is not very helpful, since we often need to determine whether or not an object is bounded based on other properties, not the sets of which it is a member. In this section, we look at some axioms that help us make this determination for sets.

First of all, we have the *finite set axiom*. Any finite set of bounded sets
is itself a bounded set.

`(bounded (setof @l))`

The *subset axiom* assures that the set of all of subsets of a
bounded set is also a bounded set.

`(=> (bounded ?v) (bounded (setofall ?u (subset ?u ?v))))`

The *union axiom* tells us that the generalized union of any bounded set of
bounded sets is also a bounded set. Since every finite set is bounded, this allows us
to conclude, as a special case, that the union of any finite number of bounded sets
is a bounded set.

`(=> (and (bounded ?u) (forall (?x) (=> (member ?x ?u) (bounded ?x))))`

` (bounded (generalized-union ?u)))`

The *intersection axiom* tells us that the intersection of a bounded set and any
other set is a bounded set. So long as one of the sets defining the intersection is
bounded, the resulting set is bounded.

`(=> (and (bounded ?u) (set ?s))`

` (bounded (intersection ?u ?s)))`

Finally, we have the *axiom of infinity*. There is a bounded set containing a
set, a set that properly contains that set, a third set that properly contains the
second set, and so forth. In short, there is at least one bounded set of infinite
cardinality.

`(exists (?u)`

` (and (bounded ?u)`

` (not (empty ?u))`

` (forall (?x)`

` (=> (member ?x ?u)`

` (exists (?y) (and (member ?y ?u)`

` (proper-subset ?x ?y)))))))`

Wed Dec 7 13:23:42 PST 1994