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.
Best Answer
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
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 $N=\mathbb N$.