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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)))))))

Vishal I. Sikka
Wed Dec 7 13:23:42 PST 1994