NakaCristo wrote:
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I have used the notation closer to wikipedia as possible under the restriction of the plaintext. Whenever I wrote <=_S, I meant the less or equal with subindex S used in wikipedia, and with <=_T the same but for the subindex T.
In LaTeX I could have written $\leq_S:=\leq$, $\leq_T:=\geq$, ending with $u\leq_S v \leftrightarrow f(u)\leq_T f(v)$.
Please, look at the first example of the wikipedia page, which I highlighted, in which two different orders are used, exactly like our case.
Let abs() be the absolute value function.
I did. And one of them was negation. The negative integers are order isomorphic to the natural numbers under the absolute value function since abs(+n) = n = abs(-n) and thus n < n+1 <-> abs(-n) < abs(-n-1). So as you can see there exists an order isomorphism between N and -N.
Discarding the Axiom of Foundation, I now define
epsilon as a quantity with the following properties:
1. For any nonzero real number r, abs(r) > epsilon.
2. SUM(n = 1, Omega) {epsilon} = 1. This sum is epsilon added to itself countably many times.
So now I define O to be the pair { (1,1/2,1/3,..............epsilon), <=} which IS well ordered because for any natural number n, epsilon < 1/n < n. And epsilon is the least element of O under the ordering "<=" .
Now if A and B are 2 transfinite ordinals, we define the product A*B to be the cartesian product (AxB) with the
lexicographic ordering defines as:
For any (a1,a2) in A and for any (b1,b2) in B,
(a1, b1) <= (a2,b2) iff b1<= b2 XOR b1=b2 AND a1<=a2
So now we define the product of Omega = { (1,2,3,.........), <=} and O = { (1,1/2,1/3,........), <=} as being the set of pairs:
{ (1/n, n),(n,1/n) <= } for every natural number n( where n+ = (n+1)
Now (1/n, n) <= (1/n+, n+) since n < n+ for all n. Now since for all n, both n and (1/n) are rationals, we can see that if phi is a function on each ordered pair such that phi: (n,1/n) -> N <- (1/n,n), then for all n phi(1/n,n)=phi(n,1/n)=1 because n/n=!
Now to show that n/n = 1. If you have a set of n elements, the number of subsets with 1 element is exactly n by the finite axiom of choice(which is provable by induction). Thus we have n/n = n* 1/n = 1/n * n = 1.
So if for all n, we map the pairs (1/n, n) and (n, 1/n) into countably many copies of the element (1), we have an order isomorphism. So using the finite ordinal product, when we map all elements of Omega * O and O * Omega into the natural numbers, all elements of these products map to (1). Thus, the ordinal product of Omega * O = O * Omega = 1
And just to clarify, we say that O = 1/Omega and so Omega/Omega = 1.
06 Sep 2012, 7:35 pm
