Intersection of a non-empty set of natural numbers (set-theoretic definition) gives an element of that set? Consider the following set-theoretic definition of natural numbers:


*

*$0$ is defined as $\emptyset$

*If $n$ is defined, then the successor of $n$ is defined as $n^+ = \{n\} \cup n$


Thus $1 = \{0\}$, $2 = \{0, 1\}$, and so on.
Let $\omega$ be the set of all natural numbers defined as above, and let $E$ be an arbitrary non-empty subset of $\omega$.

How can we show that that the intersection of $E$ is itself an element of $E$?

For context, I was working on the problem described in the link below, and got stuck after following the given hint:Prove: If $E$ is a nonempty subset of natural numbers , then there exists an element $k$ in $E$ such that $k\in$ m for any $m$ in $E$ and$m \ne k$
 A: There is a proof on proofwiki that the intersection of any set of ordinals is the smallest ordinal in that set.
It proceeds as follows:


*

*Let $m$ be $\bigcap E$. It is easy to show that every element of a natural number is a natural number, so by definition of intersection, the intersection of a set of natural numbers must be a natural number.

*By the definition of intersection, $m \subseteq a$ for every $a$ in $E$.

*Now to show that $m \in E$, we consider the successor of $m$, $m^+$.

*For every $a$ in $E$, either $m^+ \subseteq a$ or $a \in m^+$. If $m^+ \subseteq a$ for all $a \in E$, then $m^+$ is in the intersection of $E$, so $m^+ \subseteq m$, i.e. $m \in m$ which contradicts the axiom of foundation. Hence there is some $e \in E$ such that $e \in m^+$.

*Either $e \in m$ or $e = m$ by definition of $m^+$ and the fact that $e$ is an element of it.

*So $e \subseteq m$ as if $e \in m$ then $e \subset m$ (if this fact is not established for natural numbers, it's easy to do so by induction).

*$e \subseteq m$ and $m \subseteq e$ so $m = e$.

*$e \in E$ so $m \in E$ as required.
$\square$
