whether the thought had been previously adumbrated, perhaps confusedly, i know not, but the name of Bertrand Russell has become associated with the assertion that:

the number $2$ is the set of all sets which contain exactly two elements.

i think there is even today some adherence to this rather neat definition.

but my elementary knowledge of these matters suffices to give rise to two questions:

a) should not Russell have said: the proper class of sets having exactly two elements?

b) does this imply that to understand the meaning of the symbol $2$ it is necessary to have a model of the entire class of transfinite cardinals, and to accept something like the Axiom of Choice? and is the theory of these infinities in fact research into the Kolmogorov Complexity of the concept of a finite cardinal?

  • $\begingroup$ Russell wasn't working within $\mathsf{ZF}$, but rather within a version of the theory of types. In this system, the collection defining $2$ may as well be called a set. And no, of course the answer to part b) is no. $\endgroup$ – Andrés E. Caicedo Dec 17 '13 at 19:04
  • $\begingroup$ What axiom(s) would you use to prove the existence of 2? Or is the existence of 2 an axiom? $\endgroup$ – Dan Christensen Dec 17 '13 at 19:04
  • $\begingroup$ In ZF, instead of defining the cardinality of $X$ as the class of all equipotent sets, we usually define it as the set of all equipotent sets of minimal rank (among all equipotent sets). This is called Scott's trick. (This is a side remark; of course Russel did nothing of the sort.) $\endgroup$ – tomasz Dec 17 '13 at 19:24
  • $\begingroup$ Or you can use NF and have the Frege-Russel sets again (and a slew of other problems). As for (b) above, I can't think of why the question would even come up... $\endgroup$ – Malice Vidrine Dec 17 '13 at 19:35
  • $\begingroup$ @tomasz that is interesting as i thought a first line of defence, as it were, would be to find some satisfactory way of "shrinking the domain". but even a rank beginner like myself can see that any limit ordinal and its successor give you a set with two elements, and even without the obvious transpositions there are going to be "quite a lot" of those. or should one not just say that "2 is the cardinality of any set in bijection with $\{1,2\}$" and abandon Russell's obscurely-motivated program? $\endgroup$ – David Holden Dec 17 '13 at 22:13

Let me answer your first question, since I'm not quite sure I understand the second one at its current form.

Russell's definition was given much before von Neumann introduced the term "proper class".

In addition to that, von Neumann suggested we choose a representative from each equivalence class, and using the axiom of choice and the definition of ordinals he suggested, we have the modern definition of cardinals instead.

One of the first serious mathematical definitions of cardinal was the one devised by Gottlob Frege and Bertrand Russell, who defined a cardinal number |A| as the set of all sets equipollent to A. (Moore 1982, p. 153; Suppes 1972, p. 109). Unfortunately, the objects produced by this definition are not sets in the sense of Zermelo-Fraenkel set theory, but rather "proper classes" in the terminology of von Neumann.

(Weisstein, Eric W. "Cardinal Number." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CardinalNumber.html)

  • $\begingroup$ thx, Asaf. yes, as i recall Russell evolved a "theory of types", though posterity has not been kind to it, so probably few now remember what he was attempting (I certainly don't). but it seems to an outsider that the notion of a "proper class" is a conceptual idempotent, a place set conveniently beyond reason, where no questions are permitted. i know everyone needs a lumber room, but it seems somehow disrespectful to house the trans-trans-finite in such an undignified location. $\endgroup$ – David Holden Dec 17 '13 at 22:23

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