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I'm currently reading about Peano's axiom and the construction of the natural numbers. The fifth Axiom of Peano does not make sense to me:

  • If a subset $T$ contains $0$ as an element and for all $n \in T$, $s(n)$ is also in $T$, then $T$ is the set of Natural numbers $\mathbb{N}$.

I know that this axiom exists in order to exclude elements which could have the following property: $$s(a) = b$$ $$s(b) = a$$

These elements would not violate axioms $1$ to $4$ however they don't have the desired "natural number" properties hence axiom $5$ has been created to allegedly clean these elements up. I also know that axiom $5$ is supposed to work as follows:

  • If $0$ is in our set $T$ then $s(0)$ is also in $T$ but if $s(0)$ is in $T$ then $s(s(0))$ is also in $T$ and so on. Thus we declare this set to be the natural numbers.

Now to my actual question: suppose we throw in $a$ and $b$ and $0$ into $T$. Since we do not yet know what the natural numbers are, we can't exclude $a,b$ either (if my thinking is correct). Then for all $n$ in $T$ the succesor $s(n)$ is also in $T$. As previously stated $s(0), s(s(0)), ...$ are all in in $T$ (the actual natural numbers) but if such elements $a$ and $b$ are also in T then their successor $s(a)$ and $s(b)$ are also in $T$ which is a true statement since $s(a) = b$ and $s(b) =a$.

Axiom $5$ should exclude such elements but the way I am stating it, it seems like these elements do not violate axiom $5$.

Roger Crook
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    The fifth axiom, as usually stated, does not say that $T=\Bbb N$: it says that $T\supseteq\Bbb N$. The presence of $a$ and $b$ in $T$ is then not a problem. – Brian M. Scott Feb 13 '21 at 03:41
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    The line "Since we do not yet know what the natural numbers are" suggests to me that you're thinking of the PA axioms as building the natural numbers. Think of them instead as describing the natural numbers: $\mathbb{N}$ is "over there," and we're writing down some basic facts about it which will help us understand it better. Note, then, that the "subset" bit of the axiom rules out non-natural numbers from consideration. – Noah Schweber Feb 13 '21 at 03:46
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    What does "and so on" really mean? We cannot $define$ $\Bbb N$ as ${s^n(0): n\in \Bbb N}$ because this is circular: The term $\Bbb N$ in the expression ${s^n(0): n\in\Bbb N}$ cannot be used before it is defined. – DanielWainfleet Feb 13 '21 at 06:40
  • @NoahSchweber you write "Think of them instead as describing the natural numbers..." Is that why there is no axiom that says "the successor of a natural number n is n + 1" (as opposed to n + 0.5) ? I always felt that the Peano axioms do not explicitly state that the successor function is "+1", as if that is a given. – Charbel Bejjani Dec 24 '22 at 16:33
  • Can you cite the source of the first quote? Where did you get it from? I wonder if it should read “ If a subset T [of $\mathbb{N}$]contains 0 as an element and for all n∈T, s(n) is also in T, then T is the set of Natural numbers N.” – zeynel Dec 10 '23 at 11:01

2 Answers2

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Axiom 5, in conjunction with the other four axioms, tells you that:

  1. Every natural number other than $0$ is a successor; and
  2. You can “reach” every natural number by starting with $0$ and applying the successor function finitely many times.

To see that the five axioms prevent you from having “cycles” with the successor function, note first that we cannot have $s(a)=a$ for any $a$. To see this, consider $$T = \{n\in\mathbb{N}\mid n\neq S(n)\}.$$ By Peano’s third axiom, $0\in T$ (since $0$ is not a successor, so $0\neq S(0))$.

Assume $k\in T$. Then $k\neq S(k)$. By Peano’s fourth axioms, this implies $S(k)\neq S(S(k))$; hence $S(k)\in T$.

By Peano’s Fifth Axiom, $T=\mathbb{N}$. In particular, $S(a)\neq a$ for all $a\in\mathbb{N}$.

Similarly, if $S(a)=b$ and $S(b)=a$, then $S(S(a))=a$. Consider $$T=\{a\in\mathbb{N}\mid a\neq S(S(a))\}.$$ Again, $0\in T$, since $0$ is not a successor. Assume $k\in T$. Then $k\neq S(S(k))$, hence $S(k)\neq S(S(S(k))$, so $S(k)\in T$. Thus, $T=\mathbb{N}$, and you cannot have the two elements you propose.

You can similarly verify that there cannot exist elements $a_1,\ldots,a_m$ with $S(a_i) = a_{i+1}$ for $i=1,\ldots,m-1$, and $S(a_m)=a_1$.

Arturo Magidin
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  • Does there have to be a separate proof for each $m$ or is there a proof for all $m$. Probably need a proof by induction on $m$. – marty cohen Feb 13 '21 at 04:29
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    @martycohen: If you first prove that every number is either $0$ or a successor (which requires induction), then you can use the well-ordering principle (which is essentially induction): if there exist $m\in\mathbb{N}$ and $k\gt 0$ such that $S^k(m)=m$, let $m_0$ be the least such natural number, and let $k_0$ be the least positive integer such that $S^{k_0}(m_0)=m_0$. Then $m_0\neq 0$ because $0$ is not a successor. Then $m_0=S(r)$ for some $r$; therefore $r\neq S^{k_0}(r)$, and so $m=S(r)\neq S(S^{k_0}(r)) = S^{k_0}(S(r))) = S^{k_0}(m)$, which is impossible. Thus, no such $m$ and $k$ exist. – Arturo Magidin Feb 13 '21 at 04:47
  • @ArturoMagidin If we don't apply the Axiom of Induction (AI), then the set $\mathbb{N}$ satisfying the rest of the Peano axioms can contain a member $a$ such that $S(a) = a$. (AI) is supposed to exclude $a$ from $\mathbb{N}$. I don't think it does. Your proof, which I have found in Elliott's book, for the nonexistence of such $a$ in $P$ cleverly picks a set $T$ where there is no such member as $a$. Then it shows that, $T$ satisfies the conditions of (AI). Thus applying (AI), you have $T = \mathbb{N}$, proving $\mathbb{N}$ can't have such $a$.

    [Continued in the next comment.]

     – Mostafizur Rahman May 09 '23 at 16:47
  • But I could choose the set $T = {n \in \mathbb{N} : n \neq S(n)} \cup {a}$, which the OP tried to demonstrate. This set $T$ also satisfies the conditions of (AI). Thus applying (AI), we have $T = \mathbb{N}$, proving $\mathbb{N}$ doesn't contain such $a$. So, the Axiom of Induction doesn't exclude such "rogue" members from $\mathbb{N}$. – Mostafizur Rahman May 09 '23 at 16:47
  • @MostafizurRahman: You cannot apply AI to $T$ unless you first establish that $T\subset\mathbb{N}$. You have not, since that would require you to assume that $a\in\mathbb{N}$, which would make your argument circular: if you assume $\mathbb{N}$ contains "rogue" elements, then you can conclude that $\mathbb{N}$ has "rogue" elements. Big whoop. That does not establish the existence, or even the possible existence, of such elements; it just shows that if you assume your conclusion, you get your conclusion. My argument, on the other hand, has no such assumption. So, you are just wrong. – Arturo Magidin May 09 '23 at 16:51
  • @MostafizurRahman: Remember, the Axiom of Induction says: "If $T$ is a subset of $\mathbb{N}$ such that $0\in T$ and for all $n\in\mathbb{N}$, $n\in T$ implies $s(n)\in T$, then $T=\mathbb{N}$." You cannot invoke it for random sets, only for subsets of $\mathbb{N}$. – Arturo Magidin May 09 '23 at 16:54
  • @ArturoMagidin Aa.... that's some nasty trick! Thank you for clarifying it. I'm learning these axioms for the first time. This helps a lot. – Mostafizur Rahman May 09 '23 at 17:38
  • @MostafizurRahman It's not a "trick", it's what the Axiom of Induction says. – Arturo Magidin May 09 '23 at 17:46
  • @ArturoMagidin Yes. I understand. I was referring to the wit of the person who came up with the axiom. – Mostafizur Rahman May 09 '23 at 17:48
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(Posted after previous posting was accepted.)

I know that this axiom exists in order to exclude elements which could have the following property:

$~~~~s(a) = b$

$~~~~s(b) = a$

$\cdots$

Axiom $5$ should exclude such elements but the way I am stating it, it seems like these elements do not violate axiom $5$.

In this case, $a, b\neq 0$ and $\{a, b\}$ is an isolated, closed loop under the successor function $s$, i.e. neither of the elements of $\{a, b\}$ are accessible from any other natural numbers. This contradicts the fact that induction holds on $N$.

It can be shown that induction holds on any set $X$ with $x_0$ being the "first" element under the successor function $f$ if and only if there exists no such isolated (unconnected?) subsets $Y\subset X$ under $f$, with $x_0 \notin Y$. (See my formal proof using a form of natural deduction here, 228 lines) .

Dan Christensen
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