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

**A**^{c} and

**B**^{c} be the complements of

**A** and

**B** in

**N** respectively, and define

`a`_{c}(x) and

`b`_{c}(x) similar to

`a(x)` and

`b(x)` for

**A**^{c} and

**B**^{c} respectively.

`Then, `**a(x)>=b(x) for all x iff b**_{c}(x)>=a_{c}(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

**A**^{c}, so

`a`_{c}(x)=k(x)-a(x). Similarly,

`b`_{c}(x)=k(x)-b(x). Solving for

`k(x)`,

`k(x)=a(x)+a`_{c}(x)=b(x)+b_{c}(x).
This means that

`a(x)-b(x)=b`_{c}(x)-a_{c}(x). Therefore,

`a(x)>=b(x) for all x iff b`_{c}(x)>=a_{c}(x) for all x.

This theorem is

obscure to say the least, but it may prove

useful very

soon.