Tag Archives: von Neumann


This is about the ordinal numbers, which (except for the finite ones) are less well known than the real numbers, although theoretically simpler.

The numbers of either kind compose a linear order: they can be arranged in a line, from less to greater. The orders have similarities and differences:

  • Of real numbers,

    • there is no greatest,

    • there is no least,

    • there is a countable dense set (namely the rational numbers),

    • every nonempty set with an upper bound has a least upper bound.

  • Of ordinal numbers,

    • there is no greatest,

    • every nonempty set has a least element,

    • those less than a given one compose a set,

    • every set has a least upper bound.

One can conclude in particular that the ordinals as a whole do not compose a set; they are a proper class. This is the Burali-Forti Paradox.

Diagram of reals as a solid line without endpoints; the ordinals as a sequence of dots, periodically coming to a limit Continue reading


Animation: circles within circles

From the poster depicting a few von Neumann natural numbers, I created this animation. The moving image no longer depicts natural numbers in the sense of the poster, since there is no infinite descending chain of natural numbers. There is an infinite ascending chain of them; but the poster does not actually depict such a chain as nested circles. So running the animation in reverse would not give a correct suggestion of the original poster, even if it were of infinite size. Continue reading

The von Neumann natural numbers: a fractal-like image

See the next article, “Self-similarity,” for an animation of the image here.

I have long been fascinated by von Neumann’s definition of the natural numbers (and more generally the ordinals). In developing axioms for set theory, Zermelo used the sets 0, \{0\}, \{\{0\}\}, \{\{\{0\}\}\}, \{\{\{\{0\}\}\}\}, and so on as the natural numbers. Here 0 is the empty set. Zermelo’s method works, but is not so elegant as von Neumann’s later proposal to consider each natural number as the set of all natural numbers that are less than it is, so that (again) 0 is the empty set, but also n+1=\{0,1,\dots,n\}. Continue reading