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?
Best Answer
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?