First infinite von Neumann ordinal

Set

definiendum $ \omega_{\mathcal N}$
postulate $ \emptyset\in\omega_{\mathcal N}$
for all $ n\in\omega_{\mathcal N}$
postulate $ n = \emptyset\ \lor\ \exists (k\in\omega_{\mathcal N}).\ n = {\mathrm{succ}}\ k $
for all $ m\in n $
postulate $ m = \emptyset\ \lor\ \exists (k\in\omega_{\mathcal N}).\ m = {\mathrm{succ}}\ k $

As is common, I'll also use the symbol $\mathbb N$ to denote the set theoretic object $\omega_{\mathcal N}$.

infinite_jest.jpg

Idea

This is probably the most straightforward way to set up a countably infinite set.

Elaboration

The first requirement says that all elements of $\omega_{\mathcal N}$ are either $\emptyset$ or a successor of another set. The second guaranties that there are no superfluous sets in $\omega_{\mathcal N}$, apart from the ones which are e.g. required to making $\omega_{\mathcal N}$ an ordinal. Put together the axiom says that $\omega_{\mathcal N}$ contains $\emptyset$ and for each established element $m$, it also inductively contains all successors ${\mathrm{succ}}\ m\equiv m\cup\{m\}$.

Notation

To model the natural numbers, one can make the following identifications:

  • $0\equiv \emptyset$
  • $1\equiv {\mathrm{succ}}\ (0)=\emptyset\cup\{\emptyset\}=\{\}\cup\{0\}=\{0\}$
  • $2\equiv {\mathrm{succ}}\ (1)=1\cup\{1\}=\{0\}\cup\{1\}=\{0,1\}$
  • $3\equiv {\mathrm{succ}}\ (2)=2\cup\{2\}=\{0,1\}\cup\{2\}=\{0,1,2\}$
  • $4\equiv {\mathrm{succ}}\ (3)=\dots$

Using our common language conception of natural numbers we can say: Each number contains the numbers which are less than itself, i.e. $n$ is $\{0,1,2,3,4,\dots,n-1\}$. Being an ordinal, we can model the order relation of the natural numbers via set inclusion $k<n\equiv k\in n$. This also gives us the familiar ordinal order.

Reference

Requirements

Subset of

Element of

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