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More precisely, when we define the set of natural numbers $\mathbb{N}$ using the Peano axioms, we assume the following:

  1. $0\in\mathbb{N}$
  2. $\forall n\in\mathbb{N} (S(n)\in\mathbb{N})$
  3. $\forall n\in\mathbb{N}(0\neq S(n))$
  4. $\forall m,n (m\neq n\to S(m)\neq S(n))$
  5. If $P(n)$ denotes the fact that $n$ has property $P$, then $\Big(P(0)\wedge \forall n\in\mathbb{N}\big(P(n)\to P(S(n))\big)\Big)\implies \forall n\in \mathbb{N} (P(n))$

I understand that using these axioms we can derive everything about the natural numbers, but I also think it's helpful to know why the axioms were chosen the way they are. So my question is why we choose to accept the axiom of induction ((5.) above), which in a way makes this more of a metamathematical question.

For example in Tao's Analysis I, it says that the axiom of induction keeps unwanted elements (such as half-integers) from entering the set.

Wikipedia says, "Axioms [1], [2], [3] and [4] imply that the set of natural numbers is infinite, because it contains at least the infinite subset $\{ 0, S(0), S(S(0)), \ldots \}$, each element of which differs from the rest. To show that every natural number is included in this set requires an additional axiom, which is sometimes called the axiom of induction. This axiom provides a method for reasoning about the set of all natural numbers."---But I find this tautological: $\mathbb{N}$ is defined as the set of natural numbers so "$n$ is a natural number" means "$n\in\mathbb{N}$", right? So isn't every natural number included in $\mathbb{N}$ by definition?

Suppose we want to show $\mathbb{N}=\{0,1,2,3,\ldots\}$ using all five of the Peano axioms.

If we let $P(n)$ denote $n\in\{0,1,2,3,\ldots\}$, then $P(0)$ is true. Suppose $n$ is in $\{0,1,2,3,\ldots\}$. Then (informally) the dots indicate that $S(n)$ is in $\{0,1,2,3,\ldots\}$. So $\mathbb{N}\subseteq\{0,1,2,3,\ldots\}$, i.e., our defined set contains no "extra" elements (as in Tao's Analysis I).

Yet I still do not see how to show $\{0,1,2,3,\ldots\}\subseteq\mathbb{N}$ (in order to complete the "proof" that $\mathbb{N}=\{0,1,2,3,\ldots\}$) without just assuming it. (I think this is what the Wikipedia article was doing(?))

Thanks in advance for any help and I apologize if this kind of question is unsuitable for this site.

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(a) You can't deduce "everything" about the natural numbers using the axioms you stated. (b) Without the axiom of induction, you cannot preclude the existence of "numbers" such as $\omega$, where $\omega \ne 0$, $\omega \ne 1$, \omega \ne 2$, etc. –  Zhen Lin Apr 17 '12 at 11:30

3 Answers 3

You are trying to deduce something about the importance of 5th axiom only from the axioms itself, which is impossible. The axioms by itself do not carry any significance. What is important is that the set defined by the axioms is (a) unique and (b) isomorphic to some real-world object (real-world natural numbers, as in "two apples" and "three oranges").

Giving away the fifth axiom means that:

a) Peano axioms no longer define a set of natural numbers, as there could be two non-isomorphic (and even not of the same cardinality) sets, both compatible with Peano axioms; rather they define a class of sets, each containing a subset, isomorphic to $\{0, 1, 2, 3, \ldots\}$. For example, let us consider the set $\mathbb N$ in its usual sense, and the set $\mathbb C \setminus {\mathbb N}^+$. Both sets fulfill the Peano axioms (except for the fifth one), but they are quite different in itself.

b) Of course, the second set from the previous paragraph has nothing in common with a real=world object called "the natural numbers".

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Peano arithmetic (including the induction axiom) certainly has non-isomorphic models. See e.g. Godel's incompleteness theorem. Any (1st order) theory admitting an infinite model has models of arbitrarily large cardinalities (Lowenheim-Skolem). (The axioms stated in the question are a bit vague; one should be at least carefull about what "property" means. The usual formal axioms of Peano arithmetic look somewhat different.) –  user8268 Apr 17 '12 at 9:07
    
@user8268 I can construct such an isomorphism easily: given some model M with a starting element Z and successor function S, let $f(Z) = 0$ and $f(S(n)) = f(n)+1$. This, by the axiom of induction, gives us a function from M to $\mathbb N$ (defined for Z => defined everywhere on M). From $f(X) = 0$ it follows that there is no $Y$ such that $S(Y)=X$, which means $X = Z$. From $f(X) = n$ it follows $f(S^{-n}(X)) = 0$ and $X = S^n(Z)$. So f is an injection. By the axiom of induction, it is a bijection as well. Obviously such a bijection is a isomorphism as well. Where am I wrong? –  penartur Apr 17 '12 at 9:36
    
@penartur: There is no axiom stating that every nonzero element is of the form $S(n)$ for some $n$, hence, your definition of $f$ need not have domain all of $M$, but only a portion of $M$. And indeed, every nonstandard model of Peano arithmetic has a standard model inside of it, as your proof shows. –  Jason DeVito Apr 17 '12 at 12:19
    
@penartur: The Peano axioms as described in the post are imprecise. The problem is that "property" is not defined. If "property" is defined narrowly as expressible by a formula of the (first-order) language that has the usual symbols, we have the first-order version of the Peano axioms, many models, of many cardinalities. Furthermore, the theory is incomplete. If we formalize "property" using (first-order) set theory, then most of the same problems, but any model of set theory has a unique up to isomorphism set of naturals, a proof like yours works. –  André Nicolas Apr 17 '12 at 14:27
    
In my school they defined axiom of induction using the set notation: given the model M (with the zero Z and successor function S), if X is a set such that $Z \in X$ and $x \in X \Leftrightarrow S(x) \in X$, then X = M. –  penartur Apr 18 '12 at 5:19

Informally, $\{0,1,2,3,\ldots\}\subseteq\mathbb{N}$ comes from the first and second axioms.

Of course you would need to define what $\{0,1,2,3,\ldots\}$ is. Perhaps writing it $\{0,S(0),S(S(0)),S(S(S(0))),\ldots\}$ makes it clearer. Then you can take a particular member of this set and use the first and second axioms to show it is in $\mathbb{N}$.

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If $\{0,1,2,3,\ldots\}\subseteq\mathbb{N}$ follows from the first and second axioms (informally) then why does the Wikipedia article say that induction is needed to show that every natural number is in $\mathbb{N}$? –  russell11 Apr 17 '12 at 8:32
    
@russell11: because "$\ldots$" does not mean anything formal. I meant that you can use the first and second axiom to show for example $42 \in \mathbb{N}$ and similarly with any other particular identified member of the set of successors of $0$. –  Henry Apr 17 '12 at 16:45

Way of proving some theories by indcution is very importand technique without it mathematics would not be the same so it was added to formalize our intuition. From what I know suprise was that this axiom was not consequence of rest of the axioms but is needed to be stated explicite.

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