How to show inductive principle doesn't work for this system?

Question

I have to prove why each Peano axiom that doesn't hold for a set $\mathbb{N}$ with base case $1$, and $S(1) = 2, S(2) = 3, S(3) = 1$, and $S(n) = n + 1$ for all $n \geq 4$.

Obviously $S(3) = 1$ breaks one of the axiom's rules because the base case is not allowed to be a successor. I'm not sure if this requires a rigorous proof.

The induction principle does not hold (intuitively) because the number $4$ is not a successor of anything, even though it has to be a successor of something in order for the induction principle to work. However, I don't know how to rigorously show that the induction principle in general does not work for this entire system, and not just for the special case of the number $4$.

Reading the title, I thought you were taking up the challenge in this answer (are you by any means called Nelson?). Fortunately this impression appears to be wrong. — Marc van Leeuwen, Sep 23 '14 at 06:16
See post and Qiaochu's answer: "One point of view is that 'the natural numbers satisfy induction' is part of what we mean when we're talking about the natural numbers; that is, part of the definition of the natural numbers should be 'those things that satisfy induction.' This is just a slightly more sophisticated version of "the natural numbers are what you get when you start with 1, then add 1 to it, then add 1 to it, then..." In your model you can never reach 4 iterating the basic operation "+1". — Mauro ALLEGRANZA, Sep 23 '14 at 06:33

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

I have to prove why each Peano axiom that doesn't hold for a set $\mathbb{N}$ with base case $1$, and $S(1) = 2, S(2) = 3, S(3) = 1$, and $S(n) = n + 1$ for all $n \geq 4$.

Assuming you have $S(4)=5, S(5)=6, S(6)=7, ...$

Some of the Peano Axioms do hold:

$1\in \mathbb{N}$ (edit)

$S: \mathbb{N} \to \mathbb{N}$

$S$ is injective

The two remaining axioms do not hold. (edit)

As you point out, $\forall x\in \mathbb{N}:S(x)\ne 1$ is falsified since you have $S(3)=1$. (edit)

The induction principle does not hold (intuitively) because the number 4 is not a successor of anything, even though it has to be a successor of something in order for the induction principle to work.

Peano's axioms do not require that every number but $1$ has a predecessor, although induction can be used to prove this.

Hint as to the actual reason why this function $S$ does not satisfy the induction axiom:

Let $P=\{1, 2, 3\}$.

$1\in P$.

$k\in P \implies S(k)\in P$

ADDED:

That was not very clear, I'm afraid. Induction will not hold on $\mathbb{N}$ with your successor function $S$ because we have a subset $X=\{4, 5, 6, ...\}$ the elements of which are not connected in any way to the subset $P=\{1, 2, 3\}.$

We have:

$1\notin X$

$\forall x\in X:\neg \exists y\in \mathbb{N}:[y\notin X\land S(y)=x]$

If we have $P=\{1,2,3\}$, then it can be shown that:

$$1\in P \land \forall x\in P: S(x)\in P\land \exists x\in N: x\notin P]$$

Generalizing...

$$\exists P\subset N : [1\in P \land \forall x\in P: S(x)\in P\land \exists x\in N: x\notin P]]$$

Or equivalently...

$$\neg\forall P\subset N:[1\in P \land \forall a\in P: S(a)\in P \implies P=N]$$

This the negation of the induction axiom.

Also see my answer at Are there natural numbers that are not the descendant of 0?

score 0 · Answer 2 · answered Sep 23 '14 at 06:27

It seems you have already established that one of the axioms is not made true by the model, so you don't anything more to see that you don't have a valid model of Peano's axioms.

But you can easily show that that the induction principple fails as well, for instance for the predicate $P(n):n\neq 4$, or if you want something more interesting looking $P(n):n\in\{1,2,3\}\lor n>99$.

How to show inductive principle doesn't work for this system?

2 Answers2