# Intersection of subset of a natural number a member of the subset?

I am working with the set theoretic natural numbers and trying to do an exercise from Halmos's Naive Set Theory. The exercise is to show that if $E$ is a nonempty subset of a natural number, then there is $k\in E$ such that for every $m\neq k$ in $E$, we have $k\in m$.

I am trying to show that $\bigcap E$ is the $k$ we seek. I have already shown that $\bigcap E$ has the "minimality" property, and am now just trying to prove that $\bigcap E \in E$, but am stuck. How should I go about this?

• @Mauro: This was the essence of my answer, but the OP remarked that Halmos doesn't define the order at that point, so this is not the way to go (and therefore I have deleted my answer). Jul 14, 2014 at 21:45

Use induction to prove that $$\forall n\in\omega\left[E\subseteq n\wedge E\neq\emptyset\Rightarrow\bigcap E\in E\right]$$
If $n=0$ then it is vacuously true.
If $E\subseteq n+1\wedge E\neq\emptyset$ then $E\subseteq n\wedge E\neq\emptyset$ or $n\in E\subseteq n+1$.
In the first case $\bigcap E\in E$ by induction.
In the second case $E=E'\cup\left\{ n\right\}$ with $E'\subset n$. If $E'\neq\emptyset$ then $\bigcap E=\bigcap E'\in E'\subset E$ again by induction and if $E'=\emptyset$ then $\bigcap E=\bigcap\left\{ n\right\} =n\in E$.
• Why is $\bigcap E = n$ in the second case? Jul 15, 2014 at 0:17