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As some of you may have noticed, I've been coming at Peano from a few different angles. This time I'm stuck on what the Schaum's Outline Abstract Algebra version might mean. Here's Schaum's Peano:

  1. $1 \in \mathbb{N}$.
  2. For each $n \in \mathbb{N},\:$ $\exists!\; n^* \in \mathbb{N}$ called the successor of $n$.
  3. For each $n \in \mathbb{N}$ we have $n^* \neq 1$.
  4. If $m, n \in \mathbb{N}$ and $m^* = n^* \implies m = n.$
  5. Any subset $K$ of $\mathbb{N}$ having the properties (i) $1 \in K$, (ii) $k^* \in K$ whenever $k \in K$ is equal to $\mathbb{N}.$

It then says

  1. states that the first natural number is $1$

which I can only assume is because 2. says there is a successor to a member of $\mathbb{N}$ called $n^*$ and nothing in $\mathbb{N}$ may have $1$ as a successor. That implies that $1$ cannot have as a successor $1$. To me this only establishes that $1^*$ of $1$ is something else in $\mathbb{N}$, but yes, there is an order to $\mathbb{N}$ where $1$ must be first. Then,

  1. states that distinct natural numbers $m$ and $n$ have distinct successors $m + 1$ and $n + 1$.

Not sure where it is getting $m + 1$ and $n + 1$. This seems like an unwarranted supposition based on a semantic understanding of the word successor, as yet not stated or proven.

Then it goes on with a proposition aimed at establishing induction.

$P(m): m^* \neq m, \quad \forall\, m \in \mathbb{N}$

Then with the five postulates above, it wants to establish the $P$ proposition above. So, define

$K = [k : k \in \mathbb{N}, P(k) \;is\: true]$

Then $1 \in \mathbb{N}$ by 1., and $1^* \neq 1$ by 3. Thus, $P(1)$ is true and $1 \in K$. Next let $k$ be any element of $K$, then (a) $P(k): k^* \in k$ is true.

I guess this comes from simply repeating the proposition using items from $K$?

Now if $(k^*)^* = k^*$ it follows from 4. that $k^* = k$, a contradiction of (a). Hence $P(k^*): (k^*)^* \neq k^*$ is true and so $k^* \in K$. Now $K$ has the two properties stated in postulate 5, thus, $K = \mathbb{N}$ and the proposition is true for all $m \in \mathbb{N}$.

What's eating me is whether the $k^* \neq k$ then $(k^*)^* \neq k^*$ really means "next", i.e., is this enough to really establish a successor as "the next number?" I don't see the original postulates really establishing this -- or are they?

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    $\begingroup$ The $m+1$ and $n+1$ is premature, it is fine once we define addition. The states" statements are informal, connecting the formal stuff with intuitively known properties of the natural numbers. $\endgroup$ – André Nicolas Apr 8 '16 at 22:53
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    $\begingroup$ You are not reading the best exposition of Peano axioms $\endgroup$ – G Cab Apr 8 '16 at 23:04
  • $\begingroup$ @G Cab Yes, possibly not. But my problem is how the successor comes about. From the basic Peano Axioms I'm not seeing how the successor is truly a "next one" function. The proposition P does seem to build up a successor in that it seems to allow for, e.g., $(((...(k^*)^*)^*...)$, after it establishes that $1^* \neq 1$. This would seem to "shift" everything "up", but how can we say it shifts them in an orderly (i.e., increments of $+1$) fashion? After $1^* \neq 1$, the successor "function" might be mapping successors all over the place. $\endgroup$ – 147pm Apr 9 '16 at 0:10
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    $\begingroup$ Any exposition of Peano axioms that starts by including $1$ instead of $0$ should be immediately returned to the library. $\endgroup$ – neth Apr 9 '16 at 1:27
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  1. $1 \in \mathbb{N}$.
  2. For each $n \in \mathbb{N},\:$ $\exists!\; n^* \in \mathbb{N}$ called the successor of $n$.
  3. For each $n \in \mathbb{N}$ we have $n^* \neq 1$.
  4. If $m, n \in \mathbb{N}$ and $m^* = n^* \implies m = n.$
  5. Any subset $K$ of $\mathbb{N}$ having the properties (i) $1 \in K$, (ii) $k^* \in K$ whenever $k \in K$ is equal to $\mathbb{N}.$

Maybe it would help to visualize the set of natural numbers with arrows indicating succession.

$1\to 2 \to 3 \to 4 \to \cdots$

In this chain of numbers:

  • There is exactly one outgoing arrow from each number, i.e. the successor relation is a function (axiom 2).

  • $1$ has no incoming arrow, i.e. no number has a successor of $1$ (axiom 3).

  • Every number has at most one incoming arrow, i.e. the successor function is one-to-one or injective (axiom 4).

  • There are no natural numbers that will not be included in the chain of numbers pictured above. In other words, you would be able to reach any number if you start at $1$ and repeatedly went from one number to the next (axiom 5).

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  • $\begingroup$ Sorry for the late response. As the axioms are stated above, no numerically ascending order can be inferred. However, the Schaum's does seem to acknowledge this by later stating that for addition, the successor function is $n+1$. But again, we cannot assume the successor function is $n+1$ from the basic axioms. Even from the $P(m):m∗ \neq m$, we cannot really assume your arrow analogy of well-ordered ascending numbers except in the most abstract, general way. The problem, IMHO is books tend to hand-wave Peano, not really sorting this out properly. $\endgroup$ – 147pm Apr 17 '16 at 18:56
  • $\begingroup$ Using the axioms of set theory, you can construct the addition functions starting with the 5 Peano Axioms, i.e. you can prove the existence of an add function $+$ such that $x+0=x$ and $x+s(y)=s(x+y)$. Shaum's no doubt skips over this point, but there is nothing "hand wavy" about Peano's axioms. $\endgroup$ – Dan Christensen Apr 18 '16 at 2:04

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