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

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:

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

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.

• The set $M= \{ m \mid (m∈ \mathbb N) ∧ (m≠1) \}$ is the set $\{ 2,3,\ldots \}$. Thus, the set $E = \{ n-1 \mid n \in M \}$ is the set $\{ 1,2,\ldots \}$. Commented Dec 28, 2020 at 13:35
• @MauroALLEGRANZA, How does your comment relate to my question?
– RFZ
Commented Dec 28, 2020 at 13:42
• The comment is about "in my opinion I don't think that this formally correct proof". It is: the proof proves that $E = \mathbb N$. Thus, what does it mean "if $E= \mathbb N$ then it does not imply our statement" ? Our statement is $E = \mathbb N$. Commented Dec 28, 2020 at 13:45
• @MauroALLEGRANZA, By statement I meant $(n\in \mathbb{N}) \land (n\neq 1) \Rightarrow ((n-1)\in \mathbb{N}).$
– RFZ
Commented Dec 28, 2020 at 13:51
• If $n \in \mathbb N$ and $n \ne 1$ i.e. $n > 1$, then $n=2,3,4,\ldots$. Thus, $n-1=1,2,3,\ldots$. That's all Commented Dec 28, 2020 at 14:21

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

• Thanks a lot for your reply! I will take a look.
– RFZ
Commented Dec 30, 2020 at 8:07
• I read your reply very carefully and I would like to ask you some questions: 1) Why did you consider this specific map $\sigma$? 2) It is obvious that $\sigma$ is bijective. But have you ever used this bijectivity anywhere?
– RFZ
Commented Dec 30, 2020 at 22:32
• $\sigma$ is just the successor. It is obvious that it is bijective on the reals, but we need to show that it is a bijection $\mathbb{N} \to N_2$, in which case it follows that if $n \in N_2$ then $\sigma^{-1}(n) \in \mathbb{N}$ which is what was to be proved. Commented Dec 31, 2020 at 3:46
• Thanks a lot fòr your help! I really get your proof and it is much more clearer that Zorich's original proof! Many thanks!
– RFZ
Commented Jan 1, 2021 at 9:12