Axiomatic construction of natural numbers and its properties from Zorich's book

Question

I am reading the book of V.A.Zorich Mathematical analysis and trying to follow the formal construction of natural numbers and its properties:

Definition 1: The set $X\subset \mathbb{R}$ is called inductive, if for any $x\in R$ the element $x+1 \in \mathbb{R}$.

Definition 2: The set of natural numbers is defined as the minimal inductive set containing $1$, i.e. $$\mathbb{N}:=\bigcap_{A\in \mathcal A}A,$$ where $\mathcal A$ is the family of all inductive sets containing $1$ and we see that $\mathcal A\neq \varnothing$ because $\mathbb{R}\in \mathcal A$.

Principle of Mathematical Induction: If $E\subset \mathbb{N}$ with $1\in E$ and $\forall x\in E (x+1\in E)$ then $E=\mathbb{N}$.

I was able to show that:

1. If $m,n \in \mathbb{N}$ then $m+n\in \mathbb{N}$ and $mn\in \mathbb{N}$.

But I have some issues to show the following:

2. $(n\in \mathbb{N}) \land (n\neq 1) \Rightarrow ((n-1)\in \mathbb{N}).$

Proof: Let $E:=\{n-1\in\mathbb{N}\mid (n\in \mathbb{N}) \land (n\neq 1)\}$ then he show that $E$ satisfies the principle of Induction and hence $E=\mathbb{N}$ and we are done.

But in my opinion I don't think that this formally correct proof because if $E=\mathbb{N}$ then it does not imply our statement.

Can anyone explain me please? Maybe I am misunderstanding smth.

Answer 1

I suspect there is a much cleaner way of showing this:

Zorich actually defines $ E=\{n-1\in\mathbb{R} | n\in \mathbb{N} \text{ and }n\neq 1 \} $. It is straightforward to show that $E$ is inductive and contains $1$. Then we have $\mathbb{N} \subset E$, of course. It remains to show that $E = \mathbb{N}$ (Zorich appears to have omitted this part).

First show that the $n \neq 1$ is equivalent to $ n \ge 2$:

Let $F=\{n \in\mathbb{N} | n = 1 \text{ or } n \ge 2 \}$. It is straightforward to show that $F$ is inductive, $1 \in F$ and since $F \subset \mathbb{N}$ we see that $F = \mathbb{N}$. Hence, if $n \in \mathbb{N}$ and $n \neq 1$ we have $n \ge 2$ and so $N_2 = \{ n | n\in \mathbb{N} \text{ and }n\neq 1 \} = \{ n | n\in \mathbb{N} \text{ and }n \ge 2 \}$. So, we have $E =\{n-1\in\mathbb{R} | n \in N_2 \} $.

Now show that $n \in N_2$ means that $n-1 \in \mathbb{N}$ (this is essentially 2.):

Define $\sigma:\mathbb{R} \to \mathbb{R}$ by $\sigma(n) = n+1$. From the properties of $+$ on the reals we know that $\sigma$ is a bijection on the reals. It is straightforward to show that $\{ n \in \mathbb{N} | \sigma(n) \in N_2 \}$ is inductive and contains $1$ hence it equals $\mathbb{N}$. In particular, $\sigma: \mathbb{N} \to N_2$ is well defined. To show that $\sigma$ is a bijection it is sufficient to show that $\sigma$ is surjective. Let $G = \{1\} \cup \{\sigma(n) | n \in \mathbb{N}\}$, it is straightforward to show that $1 \in G$ and that $G$ is inductive hence $G = \mathbb{N}$ and so $\{\sigma(n) | n \in \mathbb{N}\} = N_2$.

Finally, $E = \{ \sigma(n)-1 | n \in \mathbb{N} \} = \{ n+1-1 | n \in \mathbb{N} \} = \mathbb{N}$.