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We have the following definitions of the rank of a set in the von Neumann hierarchy: $$\mathrm{rank}(x)=\sup\{(\mathrm{rank}\,y)^+:y\in x\}.$$ I now want to find the ranks of $\mathbb{Z},\mathbb{Q}$ and $\mathbb{R}$.

$\mathbb{Z}$ can be realised as a subset of $2\times \omega$, each element having finite rank (regardless of implementation details like pairing method). Hence I get that the rank of $\mathbb{Z}$ is $\omega$ as well.

Similarly, $\mathbb{Q}$ can be thought of as a subset of $\mathbb{Z}\times \mathbb{Z}$, and again, regardless of implementation details, each element in this set has finite rank, hence I get that the rank of $\mathbb{Q}$ is $\omega$.

We construct $\mathbb{R}$ as subsets of $\mathbb{Q}$ (e.g. the lower part of Dedekind cuts), the relevant subsets having rank $\omega+1$. Hence the rank of $\mathbb{R}$ as defined like this is $\omega+2$.

My questions are:

  1. Some other answers seemed to suggest that the ranks of $\mathbb{Z}$ and $\mathbb{Q}$ depend on implementation details - however, with the argument above, I find this hard to see. So are the ranks both $\omega$ in most cases, or have I missed something?
  2. Is it true that, defined like this, the rank of $\mathbb{R}$ is $\omega+2$?

Other relevant links:

The real numbers and the Von Neumann Universe

Rank of $\mathbb Z$ and $\mathbb Q$

mss
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  • Is $\mathbb{R}$ a well-founded set? As far as I remember ranks are defined for well-founded sets. Of course, they depend on implementation: cause you can (well-)order an infinite set in different ways. For example, $\mathbb{Z}$ as $\omega\times2$ - and the ranks differ. – Maxim Nikitin May 17 '20 at 18:57
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    @MaximNikitin Why do the ranks differ between the implementations $2\times \omega$ and $\omega\times 2$ of $\mathbb{Z}$? – mss May 17 '20 at 19:15
  • @NoahSchweber That makes sense! And I admit that my constructions are hardly the most practical ones - I was trying to keep things somewhat minimal. – mss May 17 '20 at 19:23

1 Answers1

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"Regardless of pairing method" What if we use a truly terrible pairing method, like $\langle a,b\rangle=\{\{\{a\},\{\omega_{17}\}\},\{\{a,b\},\{\omega_{17}\}\}\}$? Very few things are truly implementation-independent.

More seriously, at the second linked post a very different construction of $\mathbb{Z}$ (for example) is used: elements of $\mathbb{Z}$ are infinite sets of pairs of natural numbers, namely equivalence classes under the appropriate relation. Arguably (and I hold this opinion) this is actually more natural than your approach, since it captures the algebraic nature of the construction $\mathbb{N}\leadsto\mathbb{Z}$. And under this approach, each element of $\mathbb{Z}$ has rank $\omega$ and so $\mathbb{Z}$ itself has rank $\omega+1$.

And the rank issue only gets worse as we go further. The algebraic approach says that elements of $\mathbb{Q}$ should be sets of pairs of elements of $\mathbb{Z}$, so by this approach an element of $\mathbb{Q}$ is an infinite set of infinite sets of natural numbers - and so has rank $\omega+1$, leading to $\mathbb{Q}$ itself having rank $\omega+2$. If we construe a real as a set of Cauchy sequences of rationals this then kicks $\mathbb{R}$ all the way up to rank $\omega+4$ - its elements are infinite sets of infinite sequences of infinite sets of infinite sets of natural numbers.

Noah Schweber
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