Set Diversity is the variety of members a set has. It overlaps with
Cardinality at some situations, but it diverges from it at others.
As far as finite sets are concerned set diversities parallel their
Cardinalities. But when having infinite sets matters can diverge,
for example the set N of all naturals is to be thought of possessing
more diversity than the set E of all even naturals since it contains
all the variety of members of E but adds to it the variety of the odds
that E lacks. In general
a set has more diversity than any of its proper
subsets. So if a set cannot be of equal Cardinality to any proper
subset of it, then its diversity
would copy its Cardinality.
Now diversity diverges from Cardinality also with comparisons between
overlapping sets, for example if two sets overlap and the non overlapping
portions of them have unequal diversities then those sets would have unequal
diversities even if they had the same Cardinality. If two sets are of unequal
Cardinality then their diversities copy their Cardinalities.
The major problem facing defining set diversities is when two Dedekindian
infinite sets of equal Cardinality are disjoint! Here it is difficult to coin
a general criterion for comparing their diversities. Here in this account those
would be considered incomparable.
Here is a try to capture all of this formally.
Define: F is an unordered injection from A to B <->
For all x in F Exist a in A,b in B (x={a,b})
For all x in A Exist! y in B ({x,y} in F)
For all a,b in A for all x in B ({a,x} in F ^ {b,x} in F -> a=b)
Define: F(x)=y <-> exists p (p={x,y} ^ p in F)
Definitions:
1) A is subdiverse to B <->
exists F (F is unordered injection from A to B) ^
not exists G (G is unordered injection from B to A)
2) A is superdiverse to B <-> B subdiverse to A.
3) A is equidiverse to B <->
exists F (F is unordered injection from A to B ^ F is unordered injection from B to A) ^
not exists G (G is unordered injection from A to B ^ not G is unordered injection from B to A) ^
not exists H (H is unordered injection from B to A ^ not H is unordered injection from A to B)
/
Theorem 1. No unordered injection exists between a set and a proper subset of it.
Proof: Let B be a proper subset of A, let x be in A and not in B, and suppose
that an unordered
injection F from A to B exists, then F(x)=y in B and so in A,
and so F(y) would be in B and in A, we already have x=/=y, now if F(y)=y, then
F(y)=F(x), then F is not an unordered injection, so we must have x=/=y=/=F(y),
but we have F(F(y))=y! thus F(F(y))=F(x) and so F is not an unordered injection.
A contradiction. QED
Theorem 2: any set is equidiverse to itself.
Proof: Let X be any set, then the set of all singletons of elements of X would
be an unordered injection from X to X in both directions. Now the next two
conditions of equidiversity are fulfilled since otherwise theorem 1 would be
violated.
Theorem 3. If two sets overlap then to have the same diversity, the disjoint
portions of them must be of equal Cardinality.
Proof: Let's take any sets A,b,c that are pairwise disjoint, now let D = A U b
and E = A U c, so D and E overlap at A.
let f be an unordered injection from D
to E. Let b1 be the set of all elements of b that are sent by f to elements of c,
let c1 be the set of all elements of c that f sends elements of b to them, now
let b2 be the complementary set of b1 with respect to b, and likewise for c2, now
let x in b2 then f(x) in A, let f(x)=k, now by the argument above f(k) cannot be
an element of A! so we must have f(k) in
c, this mean that for each element x of
b2 there must exist an element y of c2 such that y=f(f(x)),
but this would mean
that b2 is injective to c2, and since to have the same diversity between D and E
there must be such an injection f in both directions, so c2 must be also
injective to b2 and so
c and b would be bijective. QED
As a result of theorem 3 suppose that b and c were Dedekind finite (whether
Tarski finite or not), then
this would obviously lead to situations where D and E
can be of equal Cardinalities and yet having different diversities like for
example N U {-1,-2} and N U {-3}.
Zuhair Al-Johar
May 14 2017