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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.

  • 1
    (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.2012-04-17

4 Answers 4

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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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    I don't see what you mean by $\mathbb{C} \smallsetminus \mathbb{N}^+$, The standard addition does not work, and even if you change to $\mathbb{C} \smallsetminus \mathbb{N}^-$ the standard multiplication does not work. [Edit: Oh I see you mean just the Peano axioms and not Peano Arithmetic.]2016-01-14
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As was noted in the question, the set $\mathbb{N}$ which we are trying to specify is infinite because we know $T \subseteq \mathbb{N}$ for the infinite set $T=\{ 0, S(0), S(S(0)), \ldots \}.$ I will rephrase your question as

How do we show that in fact $\mathbb{N}=\{ 0, S(0), S(S(0)), \ldots \}?$

Well we have to say that, it doesn't follow from the first four axioms because we could use $\mathbb{Q}$ in place of $\mathbb{N}$ and let $S(q)=q+1$ and everything so far works fine.

So we might say that " once you have $0,S(0),S(S(0)),S(S(S(0))),$ etc. that is everything!" That is pretty much what axiom 5 says.

There are some technical details but they are not directly relevant to why we want axiom 5. One detail is, how exactly do we specify what we mean by the etc.? The technical form of axiom 5 handles that, We say any set $A$ with $0 \in A$ and $S(n) \in A$ whenever $n \in A$ has all of $\mathbb{N} \subseteq A.$ This means $\mathbb{N} \subseteq T.$

Another detail is that we want to prove things about $\mathbb{N}$ and axiom 5 gives us proof by induction.

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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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    @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 $0$.2012-04-17
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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.