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 $m \ne k$ 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 $m \ne k$
This is an exercise of Paul Halmos' text, Naive set theory, page 49.
I proved that when $0 \in E$ as $k=0$ , but  when  $0 \notin E$, I faced difficulties.
As I understand, the exercise asks me to prove that for any nonempty of subset of natural number , there is a number which is less than any other number (minimal number) which is the principle of well-ordering. So I have to use induction.
I look for a hint ?
I think, I should find a method which enable me to get a minimal element of the subset $E$ from arbitrary element $n$ in $E$
 A: The exercise in Halmos' book is actually a little different:

Prove that if $E$ is a non-empty subset of some natural number, then
there exists an element $k$ in $E$ such that $k \in m$ whenever $m$ is an
element of $E$ distinct from $k$.

I could not prove it at first either. That's why I found the question here.
But it is easy to prove by induction. Let $S$ be the set of all natural numbers $n$ such that if $E$ is a non-empty subset of $n$, then there exists an element $k$ in $E$ such that $k \in m$ whenever $m$ is an element of $E$ distinct from $k$. Since $0=\emptyset$ does not contain any non-empty subset, $0 \in S$. Assume that $n \in S$ and that $E$ is a non-empty subset of $n^+$. ($n^+=n \cup \{ n \} $.) If $E-\{n\}\ne \emptyset $, it contains a $k$ such that $k \in m$ whenever $m$ is an element of $E-\{n\}$ distinct from $k$. If $n \in E$, then, since $k \in n$, $k \in m$ for all $m \ne k$ in $E$. If $E-\{n\}=\emptyset$, then $E=\{n\}$, so there is no other element not equal to $n$ in $E$. Thus $n^+ \in S$. The principle of mathematical induction gives $S= \omega$.

The change of words in the question here from "subset of some natural number" to "subset of natural numbers" gives a statement that is harder to prove. It is equivalent to an exercise on page 52 in the book:

Prove that if $E$ is a non-empty set of natural numbers, then there
exists an element $k$ in $E$ such that $k \leq m$ for all $m$ in $E$.

Halmos defines $m<n$ as $m \in n$.
To prove it we need

For all natural numbers $m$ and $n$, $m \in n$ or $m=n$ or $n \in m$.

which is proved on page 51 in Halmos' book, that is, after the exercise on page 49.
The proof:
Let $E$ be a non-empty set of natural numbers. Let $n$ be some number in $E$. If $n$ is the least element in $E$ we are done. Otherwise $E$ can be divided into two parts, numbers less than $n$, and numbers greater than or equal to $n$. The theorem proved above gives a $k$ in $E$ less than all elements of $E$ less than $n$, which is also less than $n$ and all elements in $E$ greater than $n$.
