Intersection of a non-empty set of natural numbers (set-theoretic definition) is a natural number? This question is very similar to 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 a natural number?

 A: Let $x$ be $\bigcap E$.  All elements of $E$ are natural numbers and thus all elements of $E$ are sets of natural numbers.  So, $x$ is a set of natural numbers.
As $E$ is not empty, there's a natural number $n_0 \in E$.  Clearly, $x \subseteq n_0$.
If $y \in x$, then $y \in n$ for all $n \in E$.  As all $n \in E$ are transitive, we have $y \subseteq n$ for all $n \in E$, so $y \subseteq \bigcap E = x$, i.e. $x$ is transitive.
Now you have that $x$ is a transitive set of natural numbers which is a subset of some natural number $n_0$.  Does that suffice?
A: EDIT: As pointed out in Eliott's comment, the proof I have suggested is not correct:

I feel like this should suffice, but I still need to convince myself that "a transitive set of natural numbers which is a subset of some natural number is itself a natural number"

However, he also suggests how the proof can be corrected:

I ended up doing something like this: (1) prove by induction that if X is a non-empty subset of a natural number, then the intersection of X is a natural number; (2) find a set F such that the intersection of F is the same as the intersection of E, but F is a subset of a natural number.


We can prove by induction on $n$ that
$$a\in n \qquad \Rightarrow \qquad a\in\omega,$$
i.e., every element of a natural number is also a natural number.
We also know that for two natural number we have
$$m\in n \qquad \Leftrightarrow \qquad m\subsetneq n.$$

Now if you have $E\subseteq\omega$, $E\ne\emptyset$, and $a=\bigcap E$, then $a\subseteq n$ for each $n\in E$. 
If there is an $n\in E$ such that $n\ne a$, we get $a\in n$. So by the above claim, $a$ is a natural number.
The other possibility is that $E=\{a\}$. In this case we have $a\in E\subseteq\omega$ which, clearly, means that $a\in\omega$.
