Confused about the definition of $\mathbb{N}$

Question

To summarize my confusion briefly, what about inductive subsets that have negative integers? Here's the definition I'm going off of:

The set of natural numbers is defined to be the intersection of all inductive subsets of $\mathbb{R}$.

The definition of an inductive set that I have is:

A subset S of the $\mathbb{R}$ is inductive if 1 $\in$ S and $\forall x \in S, x \in S \Rightarrow x + 1 \in S.$

Like, what if $S \subseteq \mathbb{R}$ is inductive and $S = \{-10, -9, ... , 1, ... \}$? Surely then, $1 \in S$ and $\forall x \in S, x \in S \Rightarrow x + 1 \in S$, but $\{-10, -9 \} \subseteq S$ and $-10, -9 \not \in \mathbb{N}$. It seems that those two numbers would be included in the intersection.

Answer 1

It is useful to remark that the definition in the OP is equivalent to the fact that $\mathbb N$ is the smallest inductive subset of $\mathbb R$ (see below for a proof). This means that

1. $\mathbb N$ is inductive
2. if $S$ is an inductive subset of $\mathbb R$, then $\mathbb N \subseteq S$.

In particular, if $S\neq \mathbb N$ is an inductive subset of $\mathbb R$, then $S$ is bound to have elements that are not in $\mathbb N$.

---

Let's prove the two facts above. $1.$ is clear since an arbitrary intersection of inductive subsets of $\mathbb R$ is also inductive. As for $2.$, let $S$ be an inductive subset of $\mathbb R$. Then, $\mathbb N=\cap_{T\text{ inductive}} T$ is a subset of $S$ since an intersection is always a subset of every set of the intersection.

Conversely, if a subset $N$ of $\mathbb R$ satisfies the two properties above, then $N=\mathbb N$. Indeed, $\mathbb N \subseteq N$ since $\mathbb N$ is the intersection of all inductive subsets of $\mathbb R$ and $N$ is inductive. But $\mathbb N$ itself is inductive so $N \subseteq \mathbb N$ by $2.$. By double inclusion, we have $N=\mathbb N$.

Answer 2

They would not be included in the intersection, as the intersection is the set of elements which are included in all inductive subsets of $\mathbb{R}$, and not the set of elements which are included in some inductive subsets of $\mathbb{R}$. The definition you have given means that if

$$\mathcal{I}=\{A\subseteq\mathbb{R}: A\text{ is inductive}\},$$

then we define

$$\mathbb{N}=\bigcap_{A\in\mathcal{I}}A=\{x\in\mathbb{R}:x\in A \text{ for all } A\in\mathcal{I}\}.$$

In particular, if $S$ is as in your example, then $S\in\mathcal{I}$, but we also have that $S\setminus\{-10,-9\}\in\mathcal{I}$, and so even though $-10,-9\in S$, we still have that $-10,-9\notin\mathbb{N}$, as we can find an inductive set where they are not included.