The fifth of Peano's axioms states the following: If $S\subset \mathbb{N}$ such that $1 \in S$ and $n \in S \Rightarrow \sigma(n) \in S$, then $S = \mathbb{N}$ (where $\sigma(n)$ is the successor function).

My question is, simply, is it of any significance if we alter this to: $S = \mathbb{N}$ if and only if $S\subset \mathbb{N}$ such that $1 \in S$ and $n \in S \Rightarrow \sigma(n)$.

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Sure, this works. In fact, one might note that, at least as the axioms are listed here, your statement is equivalent to the conjunction of the 1st, 2nd and 5th axioms which read:

  1. $1$ is a number.
  2. If $n$ is a number, then $\sigma(n)$ is a number.


  1. If $S\subset \mathbb N$ is a set, $1\in S$, and $n\in S \Rightarrow \sigma(n)\in S$ then $S=\mathbb N$.

One might notice that the first two axioms are the converse of the fifth - that is, the converse of the fifth is that

If $S=\mathbb N$, then $1\in S$ and $n\in S\Rightarrow \sigma(n)\in S$ and $S\subset\mathbb N$.

Given that we have both the statement and its converse, we can naturally write:

$S=\mathbb N$ if and only if $S\subset\mathbb N$, $1\in S$, and $n\in S\Rightarrow\sigma(n)\in S$.

which is precisely the same as the three axioms listed.

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Well, sure. After all, starting a sentence with "iff" makes no sense.

Added: You are quite correct that the axiom of induction (in that form) is equivalent to $$\bigl(S\subseteq\Bbb N\wedge 1\in S\wedge(n\in S\implies\sigma(n)\bigr)\iff S=\Bbb N.$$

The form given is the less trivial direction of this biconditional.

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  • $\begingroup$ That's not a very helpful response. You can just rearrange the sentence, can't you? $S = \mathbb{N}$ if and only if $S\subset \mathbb{N}$ such that $1 \in S$, and $n \in S \Rightarrow \sigma(n) \in S$ $\endgroup$ – YouKnowNothing Dec 6 '15 at 1:34
  • $\begingroup$ Ah! I see what you're getting at, now. $\endgroup$ – Cameron Buie Dec 6 '15 at 1:44
  • $\begingroup$ How would you go about proving that they are equivalent? Because in the given form it is not a biconditional. $\endgroup$ – YouKnowNothing Dec 6 '15 at 1:57
  • $\begingroup$ Since "[statement] and [true statement]" is equivalent to "[statement]," then you need only prove the converse true. $\endgroup$ – Cameron Buie Dec 6 '15 at 2:07

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