Numbers as sets

Question

As you can see in the questions in my profile I am struggling with mathematical notation. Now my objective is to define numbers and I find out that can be defined as sets:

$$0=\varnothing,1=\{\varnothing\},2=\{\varnothing,\{\varnothing\}\}\cdots $$ $$0=\{\},1=\{0\},2=\{0,1\}\cdots $$

This definition was gave by von Neumann to define ordinal numbers. My question is, can these numbers be used as normal? For example: $$\mathbb{N}=\{\varnothing,\{\varnothing\},\cdots\}$$ Is this possible? Because we would be saying that $\mathbb{N}$ is another ordinal, not a set of numbers. $$1+2=\{\varnothing\}+\{\varnothing,\{\varnothing\}\}=\{\{\varnothing,\{\varnothing\}\},\{\varnothing,\{\varnothing\}\}\}=3$$

I don't know if this is equalities make sense (probably not) but I want to know if there's a way to formalize numbers to give them a formal definition

NEW QUESTION

If we take this definition of 1 does this work? $$A=\{x,y,z\}$$ $$A\cup\{0\}=A\cup1=\{x,y,z,0\}=\{x,y,z,\varnothing\}$$ It looks weird but should be right. Am I wrong?

Answer 1

We can build mathematics based on these von Neumann ordinals, but we have to do some ground work first, like Peano did to axiomatize arithmetic:

• For a finite ordinals $a$ and $b$, we define $a+0\stackrel{def}{=}a$, and $a+(b\cup\{b\})\stackrel{def}{=}(a+b)\cup\{a+b\}$. This definition of addition is complete by induction on the right-hand argument. For example, $3+1=(3+0)\cup\{3+0\}=3\cup\{3\}$, which we call $4$.
• Similarly, we define $a\times 0\stackrel{def}{=}0$ and $a\times(b\cup\{b\})\stackrel{def}{=}a+(a\times b)$. This time, the induction is also done on the right-hand argument, but the inductive step assumes $+$ has already been defined for all $b$, or at least all $b\leq a\times b$.

These definitions are clunkier than the intuitive operations on sets like taking unions (as far as I know this clunkiness of von Neumann ordinals is a popular argument for using simpler type-theory-based definitions as a foundation for arithmetic.) However as pointed out in the main comments, they're necessary, as taking unions fails to perform addition. (In fact, $a\cup b$ for two ordinal numbers is only $\textrm{max}(a,b)$, since each ordinal number is the set of ordinals less than itself.)