Suppose F is an ordered field extending the reals. Then a subset Z of F closed under addition, subtraction and multiplication can be called a set of hyperintegers provided that for any x in F there is a unique n in Z such that n≤x<n+1. Every real closed field has a set of hyperintegers. Given a set of hyperintegers, we can call the non-negative ones "hypernaturals". Introduce the notation [x] for the set of all the hypernaturals smaller than x, for x in F. If x is itself a hypernatural, then [x] = {0,...,x−1}.
Here's an amusing and perhaps surprising little fact:
- If F has hyperintegers Z, and M is an infinite element in F, then [M] has at least the cardinality of the continuum, and in particular is uncountable.
If F is a field of hyperreals, this follows from the fact that [M] is an internal set and not finite.
The proof is very simple. For x in F, let floor(x) be the unique n in Z such that n≤x<n+1. Let A be the set of numbers of the form floor(xM) for x a real number in [0,1). Observe that A is a subset of [M]. Moreover, A has the same cardinality as the set of real numbers in [0,1), since the function f(x)=floor(xM) from the real numbers in [0,1) onto A is one-to-one. To see that it's one-to-one, observe that if f(x)=f(y) for real x and y, then |xM−yM|<1. Unless x=y, then M<1/|x−y|, and M is finite. So x=y. So A has the cardinality of the continuum as the set of real numbers in [0,1) does. Thus, [M] has at least the cardinality of the continuum, since A is a subset of [M].
This helps improve on and slightly generalize the argument here that infinitesimals are too small to model outcomes of infinite fair countable lotteries. Suppose we say that the probability of getting an outcome n in a lottery with possible outcomes 0,1,2,... is an infinitesimal u in an ordered field F extending the reals that has hyperintegers. But u is of the right size (if it's the reciprocal of a hyperinteger) or just slightly too small (otherwise) for being the probability of getting outcome n in a fair lottery on the set [1/u]. But the set [1/u] is much much bigger than the set {0,1,2,...}, since [1/u] is uncountable while {0,1,2,...} is countable.
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ReplyDeleteSlight generalization, assuming Choice. Suppose Z is any subset of F such that every element of F is within a real distance of Z, with the distance perhaps dependent on the element (i.e., for all x in F, there is a z in Z and a real r such that |x-z|<r). Then for any positive infinite M, { z in M : z < M } has at least continuum cardinality.
ReplyDeleteWe can call such a Z an R-core of F.
At first I assumed what you were saying was that hyper-integers are not numbers at all, but that if they are countable then they cannot reach an infinite set. Then you seemed to be saying that hyper-integers are something like irrational numbers or perhaps infinitely repeating numbers, so I got confused (I'm not much of a mathematician). At least, it seems that some of what you are saying relates to ratios more than to the numbers themselves. Why not say that 'for some ordered unusual set' if the set is exhaustible it does not consist of a numbered series at all? Yet in typology, just such a type of set can exist, simply stated to be an exclusive set, in which the series of numbers is defined as exhausted, and sufficient. It may be helpful to note that some 'qualia' properties remove the necessity of irrational extension of a number sequence, if it is merely stated that the sequence is 'specific' in the same way that whole numbers are specific. In this sense the only difference is technicalism. (This reminds me of the debate on M-Phi about nominalism and Quine's argument as a refutation of science, in which it might be stated that mathematics must be a kind of nominalism, or the thesis that the world consists of mathematics must be accepted or rejected, often leading to partial, incomplete answers much in the manner of your writing above).
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