Axioms for The Integers

Question

In his 1961 article Successor Axioms for The Integers (The American Mathematical Monthly, Vol. 68, No. 5 (May, 1961), pp. 441-444) Angelo Margaris proposed the following axioms to define the integers:

2. The axioms. The axioms for the set $Z$ of integers are

1. $Z$ is not empty.
2. To each integer $x$ there is associated a unique integer ${x}'$ (called the successor of $x$).
3. For all integers $x$ and $y$, if ${x}'={y}'$, then $x=y$.
4. For each integer $y$ there is an integer $x$ such that ${x}'=y$. (This $x$, which is unique by Axiom 3, is called the predecessor of $y$, and is denoted by ${}^\backprime{y}$.)
5. If $M$ is a set of integers such that (i) $M$ is not empty ; (ii) for every integer $x$, $x$ is in $M$ if and only if ${x}'$ is in $M$; then $M=Z$.
6. There is a subset $Q$ of $Z$ such that (i) $Q$ is not empty; (ii) for every integer $x$, if $x$ is in $Q$, then ${x}'$ is in $Q$; $Q\ne Z$.

Sounds quite reasonable to me. What I don't understand is this peculiar way to define the set of positive integers $P$.

4. Order. In this section we show that $Z$ is an ordered ring, and its set of positive elements satisfies the Peano axioms.

For the set $Q$ of Axiom 6, there is an $x$ such that ${x}'$ is in $Q$ and $x$ is not in $Q$. Otherwise, $Q=Z$ by Axiom 5, contradicting Axiom 6. Let $0$ be such an $x$ and let $1={0}'$. The $0$ and $1$ of Section 3 are now fixed to be this $0$ and $1$.

I completely agree with that. If the statement $\exists x\in Z\left( {x}'\in Q\wedge x\notin Q \right)$ was false then $\forall x\in Z\left( {x}'\notin Q\vee x\in Q \right)$ would be true. The statement $\forall x\in Z\left( x\in Q \right)$ implies that $Z\subseteq Q$ - an obvious contradiction. The same can be said about $\forall x\in Z\left( {x}'\notin Q \right)$. Since each integer $x$ has its predecessor $y$, no integer can be contained in the set $Q$, therefore, $Q=\varnothing $ - a contradiction. So far, we've established that there is at least one $x\in Z$ such that ${x}'\in Q\wedge x\notin Q$ . Shouldn't we at first prove that there is precisely one such $x\in Z$ and only then assign the symbol $0$ to it?

Let $P$ be the intersection of all sets $M$ such that

I. $1$ is in $M$;

II. For every $x$, if $x$ is in M, then ${x}'$ is in $M$;

III. $0$ is not in $M$.

$P$ exists because the set $Q$ of Axiom 6 is such an $M$. Then $P$ has properties I-III, and is a subset of every set satisfying I-III.

Why do we need this intersection in the first place? Isn't just $P=Q$?

Answer 1

1. No, you don't need to prove it is unique: Let define $B(x)={x}'\in Q\wedge x\notin Q$, we know that $\{x∈Z\mid B(x)\}≠\emptyset$, so let $y$ be a arbitrary element of $\{x∈Z\mid B(x)\}$, and define $0:=y$.
2. Yes, $P=Q$, but to prove this, you will have to prove that $y$ is indeed unique, i.e. $\{y\}=\{x∈Z\mid B(x)\}$. Why? If you have $0_0$ and $0_1$, then we have $\{0_0',0_0'',0_0''',\ldots\}≠\{0_0',0_0'',0_0''',\ldots\}∪\{0_1',0_1'',0_1''',\ldots\}$, but both sets satisfy $1,2$ and $3$.

So you either prove it is unique, or use the intersection. To prove it is unique we need to prove that if $a≠b\in Z$, then either $b\in\{a',a'',\ldots\}$ or $a\in\{b',b'',\ldots\}$(from now one, I denote the set $\{x',x'',\ldots\}$ as $N_x$, and $\{\ldots,''x,'x,x\}$ as $N_x^*$)

Assume $a∉N_b$ and $b∉N_a$, then $a∉N_b^*$ and $b∉N_a^*$, but notice that both $N_b∪N_b^*$ and $N_a∪N_a^*$ satisfy the integer's axioms, so both must be equal to $Z$, but clearly $N_a∪N_a^*≠N_b∪N_b^*$