Is there an intuitive way to understand them? Let's use $ \mathbb N$ as an example. $2$ is defined as $1 \,{\cup}\,\{1\}$, but I fail to understand why is it so and how it connects with the intuitive meaning of $2$.

  • $\begingroup$ The "meaning" of $2$ is that of successor of $1$. If you understand why the successor of $1$ would be $1\cup \{1\}$,then the question is futile. Do you understand why the successor is defined as it is? $\endgroup$ – Git Gud Jan 5 '16 at 21:29
  • $\begingroup$ I don't understand the successor sets at all. $\endgroup$ – user285146 Jan 5 '16 at 21:32
  • $\begingroup$ I don't know how much flexibility you have in choosing a foundation for number theory, but there are other more intuitive approaches. If you are not stuck with ZFC, you could, for example, simply define $\mathbb{N}, S$ and $0$ such that (1) $0\in \mathbb{N}$ (2) $S: \mathbb{N}\to \mathbb{N}$ (3) $S$ is injective (4) $\forall x\in \mathbb{N}: S(x)\neq 0$ (5) $\forall P\subset \mathbb{N}:[0\in P \land \forall x\in P:[ S(x)\in P] \implies P= \mathbb{N}]$ I think you will find this approach or slight variations of it are the most widely used in analysis textbooks. $\endgroup$ – Dan Christensen Jan 6 '16 at 17:21

If you're trying to understand why the definition has to be that way, don't - there's no reason it has to be that way. But the definition you're talking about works out very nicely - it's one of the reasons Von Neumann got the big money:

Don't think of it in terms of $S(n)=n\cup\{n\}$, think of it as $n=\{0,1,\dots,n-1\}$.

We want to give definitions of $0,1,2\dots$ in terms of sets.

It seems reasonable that whatever set we choose to represent the natural number $n$, it should be a set with exactly $n$ elements.

So what should $0$ be? The only choice is $\emptyset$, since that's the only set with no elements.

Now what should $1$ be? It should be a set with one element. What should that element be? We could say $1=\{\{\{\emptyset\}\}\}$, since that set has exactly one element. But that seems needlessly complicated. Why not take that one element to be the simplest set possible? So we define $1=\{\emptyset\}$.

And now we notice that hmm, $1=\{0\}$ because of our previous definition.

Now $2$ should be a set with two elements. We already have the sets $0$ and $1$ lying around, and it might be cool or convenient to have $1$ be a subset of $2$, because after all $2$ is supposed to be "larger" than $1$. So why not say $2=\{0,1\}$?

And now we notice that $1=0\cup\{0\}$ and $2=1\cup\{1\}$...

It really does work out very nicely. Consider another possibility. We could have said $0=\emptyset$, $1=\{\emptyset\}$, $2=\{\{\emptyset\}\}$, etc. Let's say those are the ugly naturals. First, they're just not as pretty. For example, $0$ has no elements, while $1,2,\dots$ all have exactly one element.

More important, there's no ntaural way to extrapolate from the ugly naturals to infinite ordinals (saying $\omega=\{\{\dots\emptyset\dots\}\}$ or in other words $\omega=\{\omega\}$ messes up our set theory). But with the Von Neumann construction, as Akiva Weinberger pointed out, it's perfectly natural to say that the next thing after all the natural numbers is $\omega=\{0,1,2,\dots\}$. Looks just like a natural number except bigger. And then $\omega+1=\omega\cup\{\omega\}$ gives the next one.

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    $\begingroup$ It's also nice in that it leads nicely to the ordinals. $\endgroup$ – Akiva Weinberger Jan 5 '16 at 21:54
  • $\begingroup$ ... might be cool or convenient to have $1$ be a subset of $2 ...$ Maybe you meant "an element" rather than "a subset"? (If so, and you change it, I'll delete this comment so that no one will be the wiser!) $\endgroup$ – Dave L. Renfro Jan 5 '16 at 22:18
  • $\begingroup$ @DaveL.Renfro I actually meant subset. Not that it matters, makes just as much fuzzy sense with "element". $\endgroup$ – David C. Ullrich Jan 5 '16 at 22:23
  • $\begingroup$ Yeah, you're right, and I should (and probably used to) know better. For $n = \{0, 1, 2, \ldots, n-1\}$ and $n+1 = \{0,1,2,\ldots,n-1,\;n\} = \{0,1,2,\ldots,n-1,\; \{0,1,2,\ldots,n-1\}\},$ we have both $n \in n+1$ and $n \subseteq n+1.$ I'll leave my comments in case others might be helped. $\endgroup$ – Dave L. Renfro Jan 5 '16 at 22:47
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    $\begingroup$ @DaveL.Renfro Oh. Given that the topic was what was cool or convenient, I thought you were just making an aesthetic suggestion, didn't realize you were pointing out a (non-)error. Not that it matters, but I'll tell you why $n\subset n+1$ seems to me to be a cooler thing than $n\in n+1$: Because $n\subset n+1$ says that $n+1$ is larger than $n$, as it should be. $\endgroup$ – David C. Ullrich Jan 5 '16 at 22:55

"The intuitive meaning of $2$" in set theory is just a canonical way to define a set with exactly two elements (so we can use it as a "measuring rod": a set has 2 elements iff it is in bijection with the standard set that we designated as "2".

The Peano intuition is "just": the minimal set such that $0$ is in it, every $n$ in it has a unique successor denoted $n+1$, and the set is minimal with those properties.

Well, the successor set for $n$ is defined as $n \cup \{n\}$, which has one element more than $n$ (to be made more precise). We start with the empty set (that we denote as $0$, it's the unique set with no elements, so no contest about it being a standard measuring rod). Then $1 = \emptyset \cup \{\emptyset\} = \{\emptyset\}$, the set with one element (namely the empty set), which is in a way the simplest set we can make after the empty set (using only minimal axioms wrt construction of sets). Then $2$ is $1 \cup \{1\} = \{\emptyset\} \cup \{\{\emptyset\}\} = \{\emptyset, \{\emptyset\}\}$, which has 2 elements etc.

We only want two things of natural numbers: their cardinal structure: canonical ways to make $n$-element sets, and their order structure and the successor set construction has the nice property that $n \in m$ iff $n \subseteq m$ so we have a nice (well)-order structure on the "natural numbers" that we define in this way. And these do match our intuition w.r.t ordering and size.


As the $n$th successor of $0$: $1 = s(0), 2 = s^2(0), 3 = s^3(0)$, where $s^n(x) = \overbrace{s(\cdots s}^n(x))$.


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