Trouble proving that $\bigcup \mathbb N = \mathbb N$

Question

Prove that $\bigcup \mathbb N = \mathbb N$

I have no idea how to even approach this question; I know that $\mathbb N$ is defined as the intersection of all inductive subset of any inductive set $y$, and intersection is related to union in the sense that $\bigcap x \subseteq \bigcup x$, where $x$ is any non-empty set - but that is all I can think of.

This question is an exercise of a chapter related to ZF 7-9 (ie. ax. of infinity, replacement and foundation), so I suspect I will have to utilise ax. of infinity at some point, but I just don't know where it fits in.

Ax. of infinity guarantees that $\mathbb N$ is a set; and ax. of union also guarantees that $\cup \mathbb N$ is a set - but I don't suppose that these are at all helpful.

Could anyone help or at least give me a sense of direction please?

Answer 1

It depends on how you define $\mathbb{N}$ to begin with. If you define it as the least inductive set containing $\emptyset$ as an element, then $\mathbb{N}$ is a (von Neumann) ordinal, hence it is $\in$-transitive: if $a\in b$ and $b\in \mathbb{N}$, then $a\in\mathbb{N}$. This proves $\bigcup\mathbb{N}\subseteq\mathbb{N}$.

Further, if $a\in\mathbb{N}$, then $a\in a+1$, so $\mathbb{N}\subseteq\bigcup\mathbb{N}$.

By the way, this is also true for every limit ordinal. If $\alpha=\beta+1$ is a successor ordinal, then $\bigcup\alpha=\beta$.

Answer 2

I think the issue will resolve itself if you go back to some definitions; taking from the wikipedia article on the axiom of infinity, you can have a very explicit set theoretic definition of the elements of $\mathbb{N}$ beyond the usual "intersection of all inductive subsets of any inductive set".

Once you have that, you can do a very straightforward proof that the union of all of the elements of $\mathbb{N}$ is itself $\mathbb{N}$. For instance, pick an element of $\mathbb{N}$, say $5$. Well, $5\in6\in\mathbb{N}$, so $5\in\cup\mathbb{N}$.