Sam, this thread has pulled me into places where I do not feel comfortable. Now that I am here, my obsessive compulsive nature keeps me here trying to make sense out of the various posts.

First, let me repeat what I said in a previous post: Mathematics relating to Cantor's work is just not an area in which I trust my knowledge. Having said that, I want to further state that I am very suspicious of your knowledge in this area.

I just do not trust either of us. That having been said, let me try to bring up some issues.

Yes, it seems that X-X = zero for all X, which suggests that infinity - infinity = zero. However, I am not sure that it is valid to just plug infinity into some equation without paying any attention to the many infinities available. It seems to me that infinity - infinity is likely to be as indeterminate as zero/zero and infinity/infinity.

For example, consider the infinity associated with all the real numbers. This is provably larger than the infinity associated with all the integers. If subtracting the latter from the former is a valid operation, the result is clearly not zero.

Now having warmed up with some non rigorous remarks, let me at least try to be rigorous. What about the following
1)N = Cardinal number of members in the set of all integers from 1 to N, inclusive. This is essentially a definition of Cardinal number.
2)Define E = Cardinal number of even integers in same set.
3)Define X = Cardinal number of odd integers in same set.
4)Now, for all N the following seem to be true. N = E + X, X = N - E, and E = N - X

When N is allowed to grow without bound, Cantor's Aleph0 (aleph Null) is the cardinal number associated with all three of the above sets, At least it can be proven that the same cardinal number is associated with the infinite version of all three of the above sets.

Hence Aleph0 = Aleph0 - Aleph0

I think I can develop a similar proof that Aleph0 - Aleph0 = zero.

As I said, perhaps (infinity - infinity) is indeterminate/ambiguous, just like 0/0, infinity/infinity.

I disagree with proof that "X - X = X" can only be valid if X is zero. I think that it is valid only if you assume that X is finite.

You posted the following.
one more rigorour was of showing that x - infinity does not work is by defining y = 1/x (which if x = infinity = 0

so 1/y - 1/y = 1/y
(1-1) / y = 1/y

multiply both sides by y (this only works if y is not infinity)

0 = 1 if y is a real number

and hence y cannot be a real number and so x must be 0.
First, you are merely stating "X - X = X" in an unnecessarily complicated fashion. The proof you are looking for should go as follows. Consider the equation "X - X = X"
1)Zero obviously satisfies the equation.
2)If X is any finite value other than zero, you can divide both sides by X.
3)This gives 0 = 1, a contradiction.
4)Hence, X cannot be any finite value other than zero.

Note that if X = infinity, X/X is ambiguous/indeterminate, so division by X is not allowed in this case. What you have proved is that zero is the only finite value which satisfies "X - X = X"
Further analysis is required for infinite values.

I am not certain of the above, only somewhat comfortable.

Next issue for discussion: I do not believe that 1^infinity = e, and I feel very sure about it. You posted the following.
Code:
lim(x->0)  exp(h) = lim(h->0) 1 + h
lim(x->0)    h    = lim(h->0) ln(1 + h)
                   1    = lim(h->0) ln(1 + h) * (1/h)

                  e   = lim(h->0)   (1 + h) ^ (1/h)
I agree with the final statement, but deny that it can be used to claim that 1^infinity = e.
First, the above is unnecessarily confusing. The exponential function (e^x) is defined as or proved to be the limit of [1 + x/n)^n] as n approaches infinity. This leads to e= limit[(1 + 1/n)^n]. Using h = 1/n and the rest of the above is merely confusing.

The general proof involves the expansion of (1 + 1/n)^n: The kth term is
Code:
C(n,k)*1^(n-k)*(1/n)^k
Where C(n,k) is the combination of n items taken k at a time. Note that each term can be shown to be a fraction with denominator containing n^k*k! (k factorial). The numerator contains a power of one times a polynomial of order k. Determining the limiting series involves the following logic

1)The power of one in each term equals one.
2)The first term of the polynomial is n^k. when divided by n^k, this becomes one.
3)In the limit, the n^k in the denominator causes all but the first term of the polynomial to vanish.
4)Hence, the general term is 1/k!, resulting in the following.
Code:
e = 1 + 1/1! +  1/2!  + 1/3! + 1/4! . . . . .
Note that assigning e= 1^n for n = infinity, would make e the first term of the above series. It would also make the second term e, and the third term would be e/2.