Abstract
We show that infinitesimal probabilities are much too small for modeling the individual outcome of a countably infinite fair lottery.
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Notes
For any finite number \(x\) in an ordered field \(F\) extending the reals, there is a standard part \(\mathrm{St }x\), which is the unique real number such that \(x-\mathrm{St }x\) is infinitesimal (we can let \(\mathrm{St }x = \sup \{ y \in \mathbb R : y < x \}\)).
Let \(P^\gamma (A) = \mathrm{St }P(A) + \gamma (P(A) - \mathrm{St }P(A))\). Then it is easy to see that \(P^\gamma \) is a finitely additive probability measure if \(P\) is, and \(P^\gamma (\{ n \}) = \gamma P(\{n\}) = \gamma \delta \), since \(\mathrm{St }P(\{ n\}) = 0\) as \(P(\{n\})\) is infinitesimal.
For instance, we might let \(M\) be the unique hypernatural number such that \(M \le 1/\sqrt{\delta } < M+1\). Then \(\delta /(1/M) = M\delta \le \sqrt{\delta }\). But \(\sqrt{\delta }\) is infinitesimal if \(\delta \) is, so \(\delta \ll 1/M\).
The author is grateful to three anonymous readers for comments that have improved this paper.
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Appendix: The uncountability of \([M]\)
Appendix: The uncountability of \([M]\)
One of our arguments used the fact that \([M]\), i.e., the set of hypernaturals less than or equal to \(M\), is an uncountable set when \(M\) is infinite. This follows from standard facts about hyperreals [e.g., Lemma 11.2.2 of Corbae et al. (2009)], but it is worth giving a simple proof that works not just in the case of the hyperreals, but for any extension of the reals that has an analogue of the hyperintegers.
Let \(F\) be any totally ordered field containing the real numbers as an ordered subfield. Say that a subset \(Z\) of \(F\) is a set of \(F\)-integers provided that (a) \(0\in Z\) and (b) for every positive \(x\in F\), there is a unique \(n\in Z\) such that \(n \le x < n+1\). This unique \(n\) is called the integer part of \(x\) (see Mourges and Ressayre 1993), and we will write \(n=\lfloor {x}\rfloor _Z\). Write \([M]_Z = \{ n \in Z : 0 < n \le M \}\). Say that \(M\in F\) is infinite provided that \(x<|M|\) for all real \(x\) (thus, \(M\) is infinite if and only if \(1/M\) is infinitesimal).
Theorem
If \(M \in F\) is infinite and positive, then \([M]_Z\) has at least the cardinality of the continuum, and in particular is uncountable.
Proof
Write \((0,1)\) for the interval of all real numbers strictly between 0 and 1. Let \(A = \{ \lfloor {xM}\rfloor _Z : x \in (0,1) \}\). Observe that \((0,1)\) and \(A\) have the same cardinality. For let \(f(x) = \lfloor {xM}\rfloor _Z\). This is a function from \((0,1)\) onto all of \(A\). To get that \((0,1)\) and \(A\) have the same cardinality, we need only prove that \(f\) is one-to-one. To do this, suppose \(f(x)=f(y)\) for \(x\) and \(y\) in \((0,1)\). Then \(\lfloor {xM}\rfloor _Z=\lfloor {yM}\rfloor _Z\) and there is an \(n\in Z\) such that \(n \le xM < n+1\) and \(n \le yM < n+1\). It follows that \(|xM-yM|< 1\). Thus, \(|x-y|M <1\) as \(M\) is positive, and so \(|x-y| < 1/M\). But if \(M\) is infinite, then \(1/M\) is infinitesimal, and the non-negative real number \(|x-y|\) is less than that infinitesimal. But the only non-negative real number less than an infinitesimal is zero, so \(|x-y|=0\) and hence \(x=y\). Thus, \(f\) is one-to-one and onto, and so \(A\) has the same cardinality as \((0,1)\), i.e., the cardinality of the continuum.
But \(A\) is a subset of \([M]_Z\) as is easy to check [every member of \(A\) is a non-negative member of \(Z\) and less than \(M\); also, \(0\notin A\) since \(M\)’s being infinite guarantees that \(xM\) is infinite for all \(x\in (0,1)\), and hence that \(\lfloor {xM}\rfloor _Z\) is also infinite for all \(x\in (0,1)\)]. Hence, \([M]_Z\) has at least the cardinality of the continuum, and in particular is uncountable.\(\square \)
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Pruss, A.R. Infinitesimals are too small for countably infinite fair lotteries. Synthese 191, 1051–1057 (2014). https://doi.org/10.1007/s11229-013-0307-z
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DOI: https://doi.org/10.1007/s11229-013-0307-z