I have spent some time looking for a rigorous, set-theoretic definition of the complex numbers. I have read the book Elements of Set Theory by Herbert Enderton (1977) which does an excellent job of constructing numbers from sets including the natural numbers, integers, and rational numbers, but stops at the real numbers.

So far, I have only found two comparable constructions of complex numbers

  • The set of all $2 \times 2$ matrices taking real-valued components
  • The set of all ordered pairs taking real-valued components

I favor the second construction better, because I feel it has a stronger geometric interpretation because of its similarities to Euclidean vector spaces. That is, define \begin{equation*} \mathbb{C}=\{(x,y):x,y \in \mathbb{R}\}, \end{equation*} which also is exactly how the Euclidean plane, $\mathbb{R}^2$, is defined.

This leads me to my question. With $\mathbb{C}$ defined exactly the same as how one defines $\mathbb{R}^2$, how does one distinguish the elements of these two sets? For example, how does one distinguish the ordinary vector $(0,1) \in \mathbb{R}^2$ from what we define to be $i$, namely the number $i=(0,1) \in \mathbb{C}$, when they are set-theoretically identical? In set theory, these two very different "numbers" -- the vector $(0,1)$ and the number $i$ -- are exactly the same set!

Thanks for your thoughts!


Consider these two ordered sets:

  1. $(\{0,1\},<)$ where $<$ is the usual order, $0<1$.
  2. $(\{0,1\},\prec)$ where $\prec$ is the discrete order, $1\nprec 0$ and $0\nprec1$.

How do you distinguish between $0$ in the first and in the second? It's the same set, $\{0,1\}$! And indeed you cannot distinguish between them. If it's the same set, then it's the same set. Period.

But $\Bbb C$ and $\Bbb R^2$ have additional structure, they are not just sets. They have addition, multiplication, and so on defined on them. How do you distinguish between $0$ in the first order and in the second, you don't.? How distinguish between the two ordered sets? They are different ordered sets, one is linear and the other is not.

If you define $\Bbb C$ as a field whose underlying set is $\Bbb R^2$, and you can do that, then you do not distinguish $(0,1)$ from $i$. They are defined to be the same object. But you distinguish $\Bbb C$ and $\Bbb R^2$ by the fact they have different multiplication defined on them, one is a field and the other is a ring with zero divisors.

Of course, you can define $\Bbb C$ as a quotient of $\Bbb R[x]$ instead, which gives a completely different underlying set.

  • $\begingroup$ I am not used to seeing Euclidean 2-space as a ring. Is that a standard definition? $\endgroup$ – MSIS Jul 24 at 23:17

If you define the set $\mathbb{C}$ as $\mathbb{R}^2$, then you obviously cannot distinguish these sets. This follows from the reflexivity of the = symbol in set theory: if $a=a$ then $a$ "is" $a$. The difference is that you have a defined multiplication in $\mathbb{C}$. In other words, the difference is hidden in the structure with which you endow the set, not in the underlying sets.

I appologize if this looks like a stupid or trivial answer and if I'm missing something deep.


There is no difference in the usual definitions of $\mathbb C$ and $\mathbb R^2$. There is little reason for any difference. The important differences are the operations and relations that we consider to be significant, not the sets themselves.

In fact, what exactly do you mean by the vector $(0,1)$ in $\mathbb R^2$? Kuratowski defined it as the set $\{\{0\}, \{0,1\}\}$. These items are equal, but we do not care since there is little reason to care. We could take it further and substitute the definitions of $0$ and $1$, and some books define $\{0\}$ as $\{0,0\}$, but you get the idea.


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