Construction of the addition function

Question

I am reading a book called Analysis I by Herbert Amann and Joachim Escher. I am currently stuck on page 33 where they construct the addition operator using functions. One property the addition operator should fulfil is the following $$n+ v(m) = v(n+m)$$

where $v$ is the succesor function. In order for the $+$ operator to obtain this property, the authors propose the following set:

$$ N:= \{n \in \mathbb{N}; \exists \ \varphi_n: \mathbb{N} \to \mathbb{N} \ \ \text{with } \ \varphi_n(0) = v(n) \ \ \text{and } \ \varphi_n(v(m)) = v(\varphi_n(m)) \ \forall \ m \in \mathbb{N}\} $$

I somewhat do see why they propose this set. It makes sense later on once you define the binary operator $+$. But before they ultimately define the addition operator they first prove that $N = \mathbb{N}$. As always, they prove it with Peano's induction Axiom. First they define $\varphi_0 := v$. It is clear to me that $0 \in N$ since $\varphi_0(0)= v(0)$ and $\varphi_0(v(m)) = v(\varphi_0(m))$. By Peano's axiom which states that there exists a successor function these two conditions are true statements hence $\varphi_0$ exists. From this point on I am not able to follow their thinking. The next step is to show that if $n \in \mathbb{N}$, then $v(n) \in N$. I don't really understand how they prove that if $n \in N $ then $v(n) \in N$.

Answer 1

They use induction to prove that $N=\mathbb N$.

By definition above, $N \subseteq \mathbb N$; thus, it is enough to prove that $N$ includes all naturals.

How to do this? By induction.

The base step: $0 \in N$, is Ok.

The next step is to show that if $n \in N$, then also $\nu(n) \in N$.

Thus, assume that $n \in N,$ i.e. there is a function $\varphi_n$ that satisfies the condition specified by the definition of $N$, and construct $\varphi_{\nu(n)}$.

They introduce an "auxiliary" function $\psi$ defined by:

$\psi(m)=\nu(\varphi_n(m))$, for every $m$.

And we have to check that is satisfies the defining condition.

Firstly, $\psi(0)=\nu(\varphi_n(0))$ and $\varphi_n$ is already defined in such a way that $\varphi_n(0)=\nu(n)$.

Thus, $\psi(0)=\nu[\nu(n)]$ and this "match" with our expectation that $\varphi_{\nu(n)}(0)=\nu[\nu(n)]$ (read it as: $\varphi_{n+1}(0)=\nu(n)=n+1$, because $0+(n+1)=n+1$).

Next, $\psi(\nu(m))=\nu (\varphi_n(\nu(m)))$.

Again, $\varphi_n$ is already defined; thus we know its value for every argument, $\nu(m)$ included.

But $\varphi_n$ satisfies the defining condition: $\varphi_n(\nu(m))=\nu(\varphi_n(m))$, and thus we can "commute" it to get: $\psi(\nu(m))=\nu (\nu(\varphi_n(m))$.

But $\psi(m)$ has been defined as: $\nu(\varphi_n(m))$, and thus we get:

$\psi(\nu(m))=\nu(\psi(m))$.

Thus, we have checked that the specified condition is fully satisfied, and we can use $\psi$ and $\varphi_{\nu(n)}$.

Conclusion: we have showed that $\varphi_0$ exists, and we have showed that, if $\varphi_n$ exists, also $\varphi_{\nu(n))$ exists.

Applying axiom (N1) we conclude that every $n \in \mathbb N$ is in $N$, i.e. that $N = \mathbb N$.

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The gist of the construction is to define "sum" through an infinite sequence of functions: $\{ \varphi_0, \varphi_1, \ldots, \varphi_n, \ldots \}$ where $\varphi_0$ is "add $0$" to every number $m$ [i.e. $\varphi_0(m)=m+0$] and $\varphi_{\nu(n)}$ is "add $\nu(n)$" to every number $m$ [i.e. $\varphi_{\nu(n)}(m)=m+\nu(n)=m+(n+1)$].

The result is that:

$\varphi_n(m)=m+n$, for every $n,m$.