Under which formalization of the natural numbers does $\bigcup_{n \in \mathbb{N}} \{n\} = \mathbb{N}$?

Question

If we recall that the set-theoretic natural numbers (defined as transitive sets) have the property that $\{n\}\cup\{n+1\} = \{n+1\}$, then it seems to me clear that $\bigcup_{n \in \mathbb{N}} \{n\}$ must be a strict subset of $\mathbb{N}$. There is the alternative case in which $\{n\} \cup \{n+1\} = \{n,n+1\}$, and then it would make sense that $\bigcup_{n \in \mathbb{N}} \{n\} = \mathbb{N}$, although it seems to me that this is something like an axiom. Is there a formalization of the natural numbers that leads to $\bigcup_{n \in \mathbb{N}} \{n\} = \mathbb{N}$, or is this an axiom?

Answer 1

Uh first things first $\bigcup_{i\in\mathbb{N}} n$ as far as I know doesn it really mean much unless you meant $\bigcup_{i\in\mathbb{N}} i$(henceforth we will denote that set as $\mathbb{I}$) which I'm guessing you did. Besides that, the standard set theoretic construction of the naturals entails that $\{n\}\cup \{n+1\}=\{n,n+1\}$ and that $n\cup(n+1)=n+1$. Now for tackling $\mathbb{I}$.

Firstly, we will prove that if x is an element of $\mathbb{I}$ then x is a natural number(or equivalently that $\mathbb{I} \subseteq \mathbb{N}$). We can do this by simply using the properties of the union. If x is an element of $\mathbb{I}$ then there is a natural number n such that x is an element of n. I will assume we already know that every element of a natural number is natural number (this can be proved via induction). Therefore, x is a natural number or equivalently that $\mathbb{I} \subseteq \mathbb{N}$.

Secondly, we will prove that $\mathbb{I}$ contains every natural number (or that $\mathbb{N} \subseteq \mathbb{I}$ and by extension that $\mathbb{N}= \mathbb{I}$). My first instinct would be to use induction. We know that $0\in 1$ and thus $0\in \mathbb{I}$. Next we assume $n\in \mathbb{I}$, again we can use the fact that $S(n)\in S(S(n))$ and thus $S(n)\in \mathbb{I}$. Therefore every element of $\mathbb{N}$ is an element of $\mathbb{I}$ and vice versa. Therefore, $\mathbb{N}= \mathbb{I}$ or $\mathbb{N}=\bigcup_{i\in\mathbb{N}} i$