Infinitesimals are too small for countably infinite fair lotteries

Abstract

We show that infinitesimal probabilities are much too small for modeling the individual outcome of a countably infinite fair lottery.

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.$\square$