Peano axioms-Mathematical Induction

Question

This is from the book we're using in my Analysis class:

The Peano Axioms of the set $\Bbb N$ are:

$1.$ Every natural number has a successor, i.e. $\forall n\in\Bbb N, \exists!s(n)\in\Bbb N$ and if $s(n)=s(m)$ then $n=m$

$2.$ There exists the first element of the set $\Bbb N$ denoted as $1$, i.e. $1\in\Bbb N$. That is the only element that is not the successor of a natural number.

$3.$ The axiom of Mathematical Induction is valid:

Let $S\subseteq\Bbb N$ such that

$1)$ $1\in S$

$2)$ $\forall n\in\Bbb N,n\in S\Rightarrow(s(n)\in S)$.

Then $S=\Bbb N$

I don't understand the third axiom. Doesn't it just repeat the first two axioms but for set $S$? What happens if we leave out the third axiom? Would that mean that there exists a set $\neq\Bbb N$ for which the first two axioms are true?

Answer 1

Consider the set $\mathcal{N}$ of complex numbers with the following properties:

• Both the real part and the imaginary part are integer.

• The imaginary part is non-negative.

• If the imaginary part is $0$, the real part is positive. Otherwise, it's an arbitrary integer.

Further, be $s(n) = n+1$.

Claim: $\mathcal N$ fulfils the first two axioms,

Proof:

Axiom 1:

Assume $n\in\mathcal N$.

• $s(n)$ doesn't change the imaginary part. And $s(n)$ increases the real part by $1$, giving an integer again. So obviously $s(n)$ has real and imaginary part integer.

• Since the imaginary part is unchanged, it stays non-negative.

• If the real part of $n$ is positive, so is the real part of $n+1$. This especially holds for zero imaginary part.

Therefore $s(n)\in\mathcal N$.

Axiom 2:

• Clearly $1\in\mathcal N$. Furthermore, $0\notin\mathcal N$, therefore there is no element $k\in \mathcal N$ such that $s(k)=1$

• If $n\in\mathcal N$ and $n\ne 1$, then either its imaginary part is zero and the real part is larger than $1$, thus $n-1\in\mathcal N$. Or the imaginary part is positice, then again $n-1\in\mathcal N$ as $n-1$ of course again has an integer real part. But $s(n-1)=n$, therefore we have a $k\in\mathcal N$ such that $s(k)=n$, namely $k=n-1$.

So we see, $\mathcal N$ fulfils both axioms.

However, $\mathcal N$ does not fulfil the induction axiom. In particular, the subset $S=\{n\in\mathcal N|\Im n=0\}$ (which is easily seen to be the usual $\mathbb N$) fulfils the conditions of the axiom, as it contains the number $1$ and for each $n\in S$, also $s(n)\in S$, but clearly $S\ne\mathcal N$, as e.g. the imaginary unit $i\in\mathcal N$ but $i\notin S$.

Answer 2

With only your first two axioms (usually stated as four axioms), you may have one or more "junk terms" thrown in along with the natural numbers. Without your third axiom, you would not, for example, be able to rule out the possibility that, for some junk term $x\in N$, you might have $s(x)=x$. Your third axiom has the effect of filtering out these terms. Remaining would be only $1$ or those terms that can be reached by starting at $1$ and repeatedly going from one number to its immediate successor. (See "What is a number again?" at my math blog)