definition of $\mathbb{N}:=\bigcap Ind$

Question

--- let $A$ a set, $A^+=A \cup \{A\}$

--- let $B$ a set, B is inductive if $\emptyset \in B \wedge \forall A \in B(A^+ \in B )$

--- let $Ind:=\{C|C \text{ is inductive }\}$

is correct this definition:

$\mathbb{N}:=\bigcap Ind $, with $\bigcap Ind=\{x|\forall C \in Ind (x \in C) \}$

????

Thanks in advance!!

Answer 1

I don't think your definition is correct. I don't know how to give a meaning to the formula $\bigcap \text{Ind}$ if $\text{Ind}$ is not a set, which it isn't.

A solution is to take inductive set, let's call it $I$, (it exists) and to consider the set $\Bbb N:=\{x\in I\colon \forall y(y \text{ is inductive}\longrightarrow x\in y)\}$, (it is indeed a set due to the Axiom schema of specification). This set is what one would end up with if one could consider an entity such as $\bigcap \text{Ind}$.

Answer 2

Well, a definition isn't really correct or incorrect. We could define $\mathbb N$ to be whatever we want. If you're asking if the above definition matches with what we expect or want the natural numbers to be, then yes, it does. In your definition, we are essentially defining $\mathbb N$ as the smallest inductive set. To see this, note that if we identify $\emptyset$ with $0$, $\{ \emptyset\}$ with $1$, and so on (I assume you've seen this previously) then clearly we see that $\mathbb N$ is inductive. As for why it's the smallest, note that any inductive set contains $\emptyset, \{\emptyset\}$ and so on, and so any inductive set contains $\mathbb N$. Thus the intersection of all inductive sets is exactly $\mathbb N$.

Answer 3

There are two issues. First of all, a priori, there might not be any inductive sets; Ind might be empty! In ZFC set theory, the axiom of infinity states that there exists an inductive set. When you define $\mathbb N$, you should explicitly call out that you're using this axiom.

Second, in ZFC, Ind is not a set. It's too large; it contains every limit ordinal. Since you're only taking the intersection, though, this is a minor issue. To work around it, take the inductive set that you got from the axiom of infinity, and then define $\mathbb N$ as a subset using the subset axiom scheme.