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I am unable to understand the motivation behind the set theoretic definition of a natural number. The definition given in the book by Goldrei is as follows:

First he defines an inductive set:

A set $y$ is inductive if $\emptyset \in y$ and $x^{+} \in y$ whenever $x \in y$, where $x^{+} = x \cup \{x\}$.

So far fine.

But then the set of natural number is defined as follows:

The set of natural numbers $\mathbb{N}$ is the intersection of all inductive subsets of any inductive set $y$, i.e.

$\mathbb{N} = \cap \{z:z \text{ is an inductive subset of }y\} = \{x: x \in z, \forall \text{ inductive }z \subseteq y\}$.

I am actually lost with this definition. I feel any inductive set is isomorphic to $\mathbb{N}$. What is the reason/motivation for defining $\mathbb{N}$ as the intersection of all the inductive subsets of an inductive set $y$? What are we missing if we were to define $\mathbb{N}$ as just an inductive set?

If my statements/questions don't make sense, it is because I am confused. I would appreciate if someone could throw light on this.

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I'm not an expert on set theory, but surely there are inductive sets that are larger than $\mathbb{N}$. Think of infinite ordinal numbers. – Hans Lundmark Dec 20 '10 at 12:26
@Hans: An inductive set is one which contains the empty set and contains its $x^+$ whenever it contains $x$? The set of natural numbers is also defined in the same right? I am not completely familiar with infinite ordinal numbers. Ordinal numbers are to be discussed later in the book I am reading. – user17762 Dec 20 '10 at 12:41
@Hans: Oh Ok. I understand now. I did some browsing and read about it. We want to define $\mathbb{N}$ so that it is the smallest such inductive set. Other inductive sets could contain more than one element which do not have a successor which in turn would give rise other elements. Is my naive understanding right? – user17762 Dec 20 '10 at 12:46
Every element in an inductive set has a successor, and $\emptyset$ never has a predecessor. What you lose in larger inductive sets is that every element can be obtained by taking a finite number of successors of $\emptyset$. – Carl Mummert Dec 20 '10 at 12:56
@Carl: Ok. Makes sense. Thanks. – user17762 Dec 20 '10 at 12:58
up vote 11 down vote accepted

Usually in set theory we write $\omega$ instead of $\mathbb{N}$. A concrete example of an inductive set other than $\omega$ is the ordinal $\omega + \omega$, which is the closure of $\omega \cup \{\omega\}$ under ${}^+$. This set has the structure of two copies of $\omega$, one after the other.

Given any set $x$, if you take the closure of $x \cup \{\emptyset\}$ under ${}^+$ then you get an inductive set containing every member of $x$. So it is not true that any inductive set is even isomorphic to $\omega$. For example there are uncountable inductive sets.

What you really want is a $\subseteq$-*minimal* inductive set. There is only one of these, which is $\omega$ (which, as before, you are identifying with $\mathbb{N}$).

The most likely reason that Goldrei says "the intersection of all inductive subsets of a fixed inductive set" rather than "the intersection of all inductive sets" is that the latter of those is an intersection of a proper class worth of sets, which he may not have defined formally. But the result is the same: if you take the intersection of all inductive sets, or the intersection of all inductive subsets of an inductive set, either way you get the unique $\subseteq$-minimal inductive set.

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Thanks. It makes sense. I was missing the fact that we want a minimal inductive set. Though now thinking about it, it is something which I should not have missed. Thanks again. – user17762 Dec 20 '10 at 12:51

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