The
greater than theorem on some
sets and their complements:
Let
N be the set of
natural numbers. Let
A and
B be non-
empty proper subsets of
N such that
A is not a subset of
B nor is
B a subset of
A.
Let
f(x) be a
one-to-one function with
domain N and
range N where
f(x+1)>f(x) for all x. Let
R be the set of natural numbers in the range
[f(x)+1, f(x+1)]. Let
k(x)=f(x+1)-f(x), which is the total number of elements in the range
R. Let
a(x) be the number of elements in
A which are also in
R; similarly, let
b(x) be the number of elements in
B which are also in
R.
Let
Ac and
Bc be the complements of
A and
B in
N respectively, and define
ac(x) and
bc(x) similar to
a(x) and
b(x) for
Ac and
Bc respectively.
Then, a(x)>=b(x) for all x iff bc(x)>=ac(x) for all x.
Proof:
k(x) is the number of elements in
R (a finite number since it is a range with definite endpoints in
N).
0<=a(x)<=k(x) for all x since
a(x) is the number of elements in common between
A and
R.
0<=b(x)<=k(x) for all x since
b(x) is the number of elements in common between
B and
R.
By the definition of complement set, if an element t is in
A, then it is not in
Ac, so
ac(x)=k(x)-a(x). Similarly,
bc(x)=k(x)-b(x). Solving for
k(x),
k(x)=a(x)+ac(x)=b(x)+bc(x).
This means that
a(x)-b(x)=bc(x)-ac(x). Therefore,
a(x)>=b(x) for all x iff bc(x)>=ac(x) for all x.
This theorem is
obscure to say the least, but it may prove
useful very
soon.