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I'm not a logician, so I apologize if what follows translates to nonsense. I would like to try to define a different theory of random choice. I hesitate to call it probability theory because I do not expect it to follow the usual rules of probability. I will however refer to it as a warped theory of probability.

For the sake of simplicity take $\mathbb{Z}$ or $[-\infty,\infty]$. It is an easy fact that one CANNOT define a discrete uniform distribution on either of these spaces in the usual sense. One of two things necessarly goes wrong: normalization or countable additivity. However, according to the axiom of choice, I can pick an element of $\mathbb{Z}$ or $[-\infty,\infty]$. Even though these sets are well ordered, I would still like to use the axiom of choice as explained below.

In fact, if I have infinitely many copies of $\mathbb{Z}$ or $[-\infty,\infty]$ then I can pick elements from each copy turning it into a product space. Since there is no real recipe for how the axiom of choice picks elements, I would like to think that if I want to define a "uniform" distribution on $\mathbb{Z}$ or $[-\infty,\infty]$, then I would invoke the axiom of choice to generate an element. Can any of this be formalized into a useful theory? In short, I would like to pick an element of each set with the underlying notion that I have no preference to the choice I make. I'm invoking the axiom of choice as a means to do so. That is, I am defining the notion of a uniform distribution through the process of saying "by the axiom of choice I can pick an element." Literally, I am thinking of the axiom of choice as a black box into which I feed an arbitrary collection of sets from which it spits out picked elements.

Of course this would not be in line with usual probability. In particular there would be huge issues with sets such as $[-1,1]$ and $[-\infty,\infty]$ where in the usual theory of probability I CAN define a uniform distribution on $[-1,1]$ but cannot on the latter, even though the two sets have a bijection. I am not sure how to reconcile this.

My experience with the axiom of choice has been mostly in proofs that say, involve some collection of equivalence classes so I can pick representatives from each one. The point is in this case I don't care which representatives I pick. So if I were to suddenly care, is there some warped notion of probability that one can invoke to make "inferences?"

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You do not need the axiom of choice to pick an element from a well-ordered set, just take the smallest element. In fact, this is how the proof that the axiom of choice is equivalent to the well ordering theorem goes. – Chris Janjigian Apr 29 '12 at 20:07
You don't even need the axiom of choice to pick finitely many elements from finitely many non-empty sets. Only when choosing from infinitely many elements we begin to thread on the grounds of the axiom of choice. – Asaf Karagila Apr 29 '12 at 20:09
The fact that you can define a uniform distribution on $[-1,1]$ means that you can on $[-\infty,\infty]$ via transfer of structure (use a bijection between the two to generate the probability on the latter). It will, however, be incompatible with what you would like to think is the real line (in terms that this is no longer guaranteed to play nice with translation or scaling). Regardless to that, if you simply taking copies of the same set you do not need the axiom of choice. Remember that you also fix a choice function, so once you fixed that you suddenly care what your elements are. – Asaf Karagila Apr 29 '12 at 20:36
You should read this paper and the Freiling paper mentioned in it. – Mr. F Apr 29 '12 at 20:59
Cox's Theorem prevents you from distributing "indifference" among infinitely many things. That's not a state of knowledge one can possibly have, and probability is in the mind – Mr. F Apr 29 '12 at 21:06
up vote 4 down vote accepted

It seems to me that the question is based on an unfortunate but quite common double-use of terminology. When people apply the axiom of choice, they often replace formally correct statements like "let $f$ be a choice function, $\dots$, apply $f$ to the set $\dots$" with more colloquial formulations involving words like "pick" or "choose arbitrarily". In probability theory, one also talks about "picking" or "choosing" elements of some space according to some probability distribution. The question asked here seems to be based on the idea that this coincidence of terminology might correspond to a coincidence or at least some close relationship in the underlying mathematics, so that something like the probabilistic "picking" can be obtained from the axiom of choice "picking". As far as I can see, the answer is no; these are two quite different uses of "choose" and "pick".

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You dig deep! :-) – Asaf Karagila Nov 10 '12 at 16:40
I still tend to see them as related, and I'll give you an example of why. Please let me know what is wrong in my construction.How I believe you can use AC to build a random number generator: Supposse we have $\omega$ well ordered copies of the set N. We fix a choice function that will pick one element of each set. Then put the chosen numbers in the same order than the copies of N, within a black box. Then by definition my "natural number generator" box will generate a sequence of natural numbers. If this were possible to imagine,do I have any expectations about the distribution probability? – Wolphram jonny Jul 14 '14 at 21:16
@julianfernandez: If I were fixing a choice function for $\omega$ copies of the set $N$, it would be the identically zero function. You want to fix a rather different choice function. How, exactly, do you want to do that? – Andreas Blass Jul 14 '14 at 21:31
Assume the choice function treats all the elements of N as unlabeled, or undiferentiated, so the function looks truly random. – Wolphram jonny Jul 14 '14 at 21:31
@julianfernandez Yes. What you're doing has little to do with AC. All AC tells you in this situation is that there exists a choice function. It doesn't guarantee that this function is uncorrelated with the hidden labels. A choice function is just a set of ordered pairs; it has nothing to do with what we know about the entities involved. – Andreas Blass Jul 14 '14 at 21:47

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