# Integer Construction

• Aug 29th 2007, 07:37 PM
ThePerfectHacker
Integer Construction

Definition: A set $\displaystyle S$ is "finite" if and only if $\displaystyle \exists \ T\subset S$ and a bijective map $\displaystyle \phi: T\mapsto S$.

Warning: Everything from here might be considered Naive Set Theory. This is why I will use the term "proper class" instead. (And I will not be so careful, because I do not want to spend time being careful).

Defintion: Let $\displaystyle \Omega$ be the proper class of all finite sets. So that if $\displaystyle S_1,S_2 \in \Omega$ then $\displaystyle S_1\cap S_2 = \{ \}$.

Definition: Define an relation $\displaystyle R$ on $\displaystyle \Omega$ as $\displaystyle (X,Y)\in R$ for $\displaystyle X,Y\in \Omega$ iff there exists a bijection $\displaystyle \phi: X\mapsto Y$.

Theorem: The above relation is an equivalence relation.

Definition: Define $\displaystyle \Phi$ to be the proper class of $\displaystyle \Omega$ modulo $\displaystyle R$.

Definition: Define the cardinality of each element in the proper class of $\displaystyle \Phi$ to be an "natural".

Definition: For $\displaystyle n_1$ and $\displaystyle n_2$ naturals we define $\displaystyle n_1+n_2$ to be the cardinality of the union of the two equivalence classes.
(And this is well-defined).

Now with this with continue to define some properties ....
• Aug 29th 2007, 09:06 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker

Definition: A set $\displaystyle S$ is "finite" if and only if $\displaystyle \exists \ T\subset S$ and a bijective map $\displaystyle \phi: T\mapsto S$.

????

Isn't this the definition of a "non-finite" set?

RonL
• Aug 30th 2007, 06:06 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
????
Isn't this the definition of a "non-finite" set?

Yes, sorry. (When I was writing this I had a lot to drink). A infinite set is exactly as described above. So a finite set is a non-infinite set. The negation of that definition.