2

In Zermelo–Fraenkel (ZF) set theory, the natural numbers are defined as $0 = \emptyset$ and $n + 1 = n \cup \{n\}$

$$\begin{alignat}{2} 0 & {} = \{\} && {} = \emptyset,\\ 1 & {} = \{0\} && {} = \{\emptyset\},\\ 2 & {} = \{0,1\} && {} = \{\emptyset,\{\emptyset\}\},\\ 3 & {} = \{0,1,2\} && {} = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\} \end{alignat}$$

My intuition would be, to define it as $0 = \emptyset$ and $n + 1 = \{n\}$

$$\begin{alignat}{0} 0 && = \emptyset,\\ 1 && = \{\emptyset\},\\ 2 && = \{\{\emptyset\}\},\\ 3 && = \{\{\{\emptyset\}\}\} \end{alignat}$$

Why is it defined that way? What makes it more straightforward than my definition?

Andrés E. Caicedo
  • 79,201
Iter Ator
  • 598

1 Answers1

2

Zermelo's original definition of the natural numbers was actually your "intuitive" definition – it's not a new idea. The standard definition (the first one) is due to von Neumann, and is used because it makes defining order on the natural numbers much easier using the subset relation.

For example, $3\le5$ because the von Neumann representation of $3$, $\{0,1,2\}$, is a subset of the representation of $5$, $\{0,1,2,3,4\}$. Doing the same with Zermelo ordinals, in which $3\equiv\{\{\{\{\}\}\}\}$ and $5\equiv\{\{\{\{\{\{\}\}\}\}\}\}$, is very finicky.

The von Neumann ordinals also have the nice property that each number is matched to a set containing that many elenents, so they serve as representatives for cardinalities.

Parcly Taxel
  • 103,344