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Assuming we have already defined the natural numbers $\mathbb{N}$ and function iteration $f^n$ for a natural number $n$, a set $A$ and a function $f:A\rightarrow A$, do the following axioms (based off the Peano axioms) define the set of integers?

  1. There exist an element $0$ in $\mathbb{Z}$
  2. There exists a function $S:\mathbb{Z}\rightarrow\mathbb{Z}$.
  3. There exists a function $P:\mathbb{Z}\rightarrow\mathbb{Z}$ such that for every element $x$ in $\mathbb{Z}$, $P(S(x)) = S(P(x)) = x$.
  4. For all $n$ in $\mathbb{N}$ where $n\neq 0$ and for every element $x$ in $\mathbb{Z}$, $S^n(x) \neq x$ and $P^n(x) \neq x$.
  5. If $A$ is a set such that $0$ is in $A$ and, for all $n$ in $\mathbb{N}$, $S^n(0)$ being in $A$ implies that $S(S^n(0))$ is in $A$ and $P^n(0)$ being in $A$ implies that $P(P^n(0))$ is in $A$, then $A$ contains every integer.
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  • $\begingroup$ I think you want $\quad ... S^{n+1}(0)$ is in $A ... P^{n+1}(0)$ is in $A ... $ $\endgroup$ – CopyPasteIt Dec 23 '19 at 1:24
  • $\begingroup$ Also, you will have to define addition and multiplication for this to work. $\endgroup$ – CopyPasteIt Dec 23 '19 at 1:26
  • $\begingroup$ Addition hasn't been defined yet, though. The $S$ in the exponent refers to the successor function in the natural numbers. Perhaps I could just use $S(S^n(x)$ and $P(P^n(x))$ instead. $\endgroup$ – 1103_base_6 Dec 23 '19 at 1:26
  • $\begingroup$ Why do you need to define addition and multiplication for this to work? Especially multiplication - the original definition of the integers by the Grothendieck construction does not need multiplication to be defined. $\endgroup$ – 1103_base_6 Dec 23 '19 at 1:28
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    $\begingroup$ @CopyPasteIt defining the negation unary function $-:\mathbb{Z}\rightarrow\mathbb{Z}$ as $-0 = 0$, $S(-x) = -P(x)$, and $P(-x) = -S(x)$ should work as well. $\endgroup$ – Madeleine Birchfield Dec 23 '19 at 1:50
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With these questions you want to make sure that you are constructing a unique object identical to the integers and not a class of objects.

The parallels to the Peano axioms are clear:

  • Axiom 1 is the same as the equivalent for the natural numbers, and rules out the empty set as a trivial object to the set of axioms.
  • Axiom 2 is the same as the equivalent for the natural numbers, and asserts that the function $S$ is closed over the set.
  • Axiom 3 corresponds to its equivalent for the natural numbers, and asserts that the function $S$ is bijective and has a unique inverse function $P$, (while the original for the natural numbers only asserts that the function $S$ is injective). It follows that the function $P$ is also closed over the set.
  • Axiom 4 asserts that while $S$ is a permutation, the cardinality of the set is not finite. This disallows finite cycles from being defined.
  • Axiom 5 is mathematical induction over the natural numbers on $S$ and $P$ and corresponds to its equivalent over $S$ for the natural numbers, and rules out any other elements other than those elements accessible from $0$ through function composition of $S$ and $P$, just as the original did for the natural numbers. Or equivalently, there does not exist a partition of the set into multiple mutually disjoint sets. This axiom also establishes that the set is countably infinite.
  • For $n = 2$, axiom 4 asserts that neither $S$ nor $P$ are involutions, and with axiom 3, it follows that for every element $x$ in the set, $S(x) \neq P(x)$, and the two functions $S$ and $P$ are not the same function. Therefore, the set is not identical to the natural numbers.

These axioms together construct a structure with an initial element $0$ and two countably infinite arms branching off from $0$ in distinct directions, defined by function iteration of $S$ and $P$; this uniquely forms the integers.

One can check indeed that this is isomorphic to the usual construction of the integers by equivalence classes of natural numbers (which does not need addition either if function iteration is defined):

We define the equivalence class $[a, b] \equiv [c, d]$ if $S^a(d) = S^c(b)$. We define $0 = [k, k]$ for all natural numbers $k$, which satisfies axiom 1. For the integer $x = [a, b]$ for natural numbers $a$ and $b$, we define $S(x) = [S(a), b]$ and $P(x) = [a, S(b)]$. Axiom two is satisfied because the successor function $S$ is closed over the natural numbers. Axiom three is satisfied because $P(S(x)) = S(P(x)) = [S(a), S(b)] \equiv x$. Axiom four is satisfied because the natural numbers are not a cycle. And axiom five is satisfied because the construction does not construct any elements inaccessible from $0$.

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  • $\begingroup$ One can define similar axioms by using negation in place of the predecessor, where negation is defined as $-:\mathbb{Z}\rightarrow\mathbb{Z}$, $-0 = 0$ and $S(-S(x)) = -x$ for all $x\in\mathbb{Z}$. $\endgroup$ – Madeleine Birchfield Dec 23 '19 at 4:04

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