Okay, I think I finally understand what you were doing. You're not computing Aleph-0 at all, you're taking it as a basic entity, isomorphic to another dimension. Looking back at your previous post "I take it you have no objections to using ordered pairs of natural numbers? Then you should have no objection to using the ordinal omega^2, which is exactly the same thing!" No serious objection, though it seems (to me) sloppy to conflate trees and numbers like that. Though seeing as you're drawing on existing work, I can hardly blame you.Deedlit wrote:It is "infinite" because (1,0) > (0,n) for any natural number n

What I mean is that, for any natural number n, we can call the function H_n or I_n and it will be defined. Are you saying that if one is concerned with computation, it is not normal to consider the set of functions H_n defined? Sure there is a cost of O(f(n)) to compute the function - that just confirms that the function is in fact defined for all n.

You have the first pass definitions of all of H_n and I_n (by virtue of all H_n sharing the same first pass definition), but not the second pass definitions. Or to rephrase, you effectively have two functions metaH(n) and metaL(n) that return H_n and L_n. You have the full definitions of metaH and metaL, but you don't have the second pass of H_i until you execute metaH on i, and so on with metaL and l_n. So in addition computing the result of H_n(X), one must compute the function H_n itself, so in the strictest sense you only "have" four functions so far (two meta functions and two base case functions). I'd say for pure math it's perfectly fine to call H_n and I_n "defined" (first past definitions and computable, if not computed, second pass definitions).

That's fine then.C(x,y,z) is a finite set whenever y and z are finite, which they always are when I call m(x,y,z). So we simply take the largest element of a finite set.

By begging the question, I meant that you were defining transfinites as ordered tuples. When you originally said "...of the form omega*a + b, which we can think of as simply an ordered pair (a,b)" you seemed to be substituting the referent for the symbol; though now it seems (in context) omega represents "(1,0)", not the other way around.No, it's not begging the question at all. I don't mean sticking in aleph0 into your function; I mean that your function is an ordered function on triplets of natural numbers, which is the same thing as omega^3.

Yes, but the use of omega can be removed entirely. Which goes back to my original question "...that number is computable? You reference many uncomputable "numbers" and sets." It now seems that your number can be defined without dragging in the concept of infinity, although not preferable to one familiar with the same work as you.Is the latter H function computable or not?

Thank you for your explanations Deedlit.