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I know that it is possible to represent each natural number using the concept of successor. But how do you assign meanings into the representation? I mean, when presented as a set, it's just a set unless we agree on some definition that some form of sets refer to natural numbers. Or is it something impossible at ZFC?


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ZFC usually includes an axiom of Infinity, which basically says that there exists are inductive set, aka $\mathbb{N}$. Or maybe I'm not really understanding the question. –  you Mar 24 '12 at 4:28
@you I know that inductive sets exist. The question is how do one assign some sets as natural numbers. –  user27515 Mar 24 '12 at 4:31
@user27515 : " Assign some sets as natural numbers "? Do you mean something like $0 \overset{def}= \varnothing$, $1 \overset{def}= \{\varnothing\}$, $2 \overset{def}= \{1\}$, \dots , $\mathrm{successor}(n) \overset{def}= \{n\}$? –  Patrick Da Silva Mar 24 '12 at 4:33
You mean like $0:=\emptyset,1:=\{\emptyset\}, 2:= \{\emptyset,\{\emptyset\}\},\dots, n+1:= n\cup\{n\},\dots$? –  you Mar 24 '12 at 4:34
"0", "1", "2", etc. are just convenient labels for $\varnothing$, $S(\varnothing)$, $S(S(\varnothing))$, etc. The "names" occur in the metalanguage (what we are using to talk about the sets in question), rather than "in" set theory. As for singling out the natural numbers, you can see one construction here. –  Arturo Magidin Mar 24 '12 at 5:28
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up vote 5 down vote accepted

I am turning my comment into an answer. The best way I know to output the set $\{0,1,2,\dots \}$ from the set-theoretical construction goes as follows : $$ 0 = \varnothing, 1 = \{\varnothing, 0\}, 2 = \{\varnothing, 0, 1\}, 3 = \{\varnothing, 0, 1,2\}, \dots, \mathrm{successor}(n) = n \cup \{n\}. $$ In other words, to create the positive integer $n$, you consider the "set that contains the set that contains the set that $\dots$ that contains $\varnothing$ + the set that contains the set that $\dots$",and so on. Again in other words, the integer $n$ is the union of the sets that contains $\varnothing$ at $i$ levels of deepness, $i$ ranging from $0$ to $n-1$ (All this is only in familiar terms).

This construction can also be used to inductively define addition : $$ n+0 = n, \quad n + 1 \overset{def}= \mathrm{successor}(n),\quad n+2 \overset{def}= \mathrm{successor}(\mathrm{successor}(n)), \quad \dots \quad n+(m+1) \overset{def}= (n+m)+1. $$ Try to understand what I exactly said in the last definition.

You can also define multiplication using this definition : $$ n \cdot 0 \overset{def}= 0, \quad n \cdot (m+1) \overset{def}= (n \cdot m) + n $$ It is an exercise to show that those definitions have all the properties we know over positive integers : associativity, commutativity, etc.

Hope that helps,

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I made an almost identical post, but was 33 seconds too late :) –  you Mar 24 '12 at 4:46
Heh ; your comment was also one minute too late... it happens, don't worry. You'll have your chance on another one! –  Patrick Da Silva Mar 24 '12 at 4:48
Another definition lets 0 be the empty set and n +1 be the union of n and {n}. Yet another (I think older) option is to define each n as the equivalence class of sets with cardinality n. The idea underlying these definitions is that a natural number is simply a member of the domain of a model of a certain theory. That is, it doesn't matter or make sense to ask what the objects themselves are; only how they relate to each other (i.e., the relations defined on them) is relevant. –  Rachel Mar 24 '12 at 5:03
For set theorists, it is more common to define $\text{successor}(A) = A\cup\{A\}$ for any set $A$; a set $B$ is inductive if $\varnothing\in B$ and $x\in B\Rightarrow \text{successor}(x)\in B$. The Axiom of Infinity guarantees the existence of at least one inductive set. From there one constructs $\mathbb{N}$. The natural numbers are then $0=\varnothing$, $1=\{0\}$, $2=1\cup\{1\} = \{0,1\}$, and more generally, $n+1=\{0,1,\ldots,n\}$. See, e.g. this answer. –  Arturo Magidin Mar 24 '12 at 5:27
@Arturo : Yes, now I remember it and it makes more sense. It also helps when building an order over $\mathbb N$, because you can say $n \le m$ if $n \subseteq m$ (because integers are now sets). I'll edit my post. –  Patrick Da Silva Mar 24 '12 at 20:13
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In the ZFC set theory everything is a set. There are no non-set elements. Indeed there can be several interpretations of the natural numbers in a given model of ZFC.

Furthermore, if you constructed one model of the natural numbers and used it to generate the complex numbers you can identify the natural numbers as the canonically embedding of the natural numbers there, and of course these would be different sets.

It is common to take the finite von Neumann ordinals as the natural numbers. However even if we take different sets, since the addition and multiplication is not a part of the language of set theory we define them within the model. We can also define those differently (again, depending on the interpretation of the numbers).

The important thing is that the chosen representation will satisfy certain axioms, and have binary operations of addition and multiplication.

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How can everything be a set? Haven't you heard of $\{ u \, | \, u \notin u \}$? –  Patrick Da Silva Mar 24 '12 at 20:22
@Patrick: I didn't know that proper classes were elements of the universe of ZFC. –  Asaf Karagila Mar 24 '12 at 20:23
Well, when you say "everything"... =P –  Patrick Da Silva Mar 24 '12 at 20:25
@Patrick: Yes. Everything. Every element of the universe is a set. Things which are not in the universe cannot be quantified with "every ...". I don't see the problem with my statement. –  Asaf Karagila Mar 24 '12 at 20:28
"Everything in a universe" and "Everything" just felt different to me. I thought you didn't make the distinction. If you do then I'm okay with that. =P –  Patrick Da Silva Mar 24 '12 at 21:14
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