set of natural numbers Prove each set of natural numbers is the set of all smaller natural numbers.
IE $ n =$ { $m \in N | m < n$ } 
Hint: use induction to prove that all elements of a natural number are natural numbers.
The natural numbers are defined as : 0, 1, 2, ...etc.
The textbook I am using (Hrbacek and Jech) further defines:
$ 1 = \{ 0 \} = \{ \emptyset \}$
I am not sure if I am going about this correctly.
Step ONE: show it is true for the second smallest natural number 1
it is a given that $ 1 \in $ {$0$} 
Step TWO: assume it is true for the natural number $k$
  IE: $k =$ { $m \in N | m< k$ } 
Step THREE: prove it is true for natural number $ k + 1 $
since $ k = \{ m \in N | m < k \}$
and since $ k < k + 1$
then $m < k + 1 $
therefore $ k + 1 = \{ m \in N | m < k + 1 \}$
am I correct?  
 A: I suspect that you’ve actually defined $N$ to be the smallest set such that $0\in N$ and for all $n\in N$, $S(n)\in N$ or something very similar, and that you’ve defined $n+1$ to be $S(n)=n\cup\{n\}$ for $n\in N$. To show by induction that $n=\{k\in N:k<n\}$ for each $n\in N$, you must show two things:


*

*$0=\{k\in N:k<0\}$, and  

*if $n\in N$ and $n=\{k\in N:k<n\}$, then $n+1=\{k\in N:k<n+1\}$.


Note that for (1) you want to start at $n=0$, not at $n=1$. (And what you have for $n=1$ isn’t correct: $1\notin 0$. In fact $1=0\cup\{0\}=\{0\}$, so $0\in 1$.) In order to prove (1) you must show that there is no $k\in N$ such that $k<0$. How you do this will depend on you you’ve defined the relation $<$; I don’t know your definition, so I can’t help here.
For (2), assume that $n=\{k\in N:k<n\}$. Then 
$$n+1=S(n)=n\cup\{n\}=\{k\in N:k<n\}\cup\{n\}\;,$$
so to finish the induction step you must show that
$$\{k\in N:k<n\}\cup\{n\}=\{k\in N:k<n+1\}\;.$$
It’s very easy to see that
$$\{k\in N:k<n\}\cup\{n\}\subseteq\{k\in N:k<n+1\}\;,$$
so all that remains is to show that
$$\{k\in N:k<n+1\}\subseteq\{k\in N:k<n\}\cup\{n\}\;.$$
For this you’ll want the result from this question.
