Prove that if a set A of natural numbers contains $n\_0$ and that whenever A contains k it also contains k+1. 

**Prove that A contains all natural numbers $ \geq n\_0 $**

This is rather similar to a question already on this site but its not quite the same and the only answer to that question is incomplete. 

**So if someone could tell me if this proof is OK i would appreciate it.**

I could not for the life of me figure out how to turn this into an induction proof i think it may be possible to use strong induction on how i set this up to prove it or prove it by induction in a different way but i couldn't figure it out for the life of me even with the hint from the similar problem.

Case 1) In Case 1 set A=D but in D we shall define $n\_0 =1 $ 

Now since $ 1 \in D $ and k+1 is $\in D$ we know that $D= \mathbb{N}$ 

technically this is the only part of the proof that relies on induction and i couldn't even figure out how to prove that since $ 1 \in D$ that $ D= \mathbb{N}$

Case2) now we know that $n\_0 > 1$ so i will define a set $A\_1$ such that $ a\in A\_1 \forall a \in A$ such that $a \geq n\_0$ and a set B where $ 1\in B$ and $(k+1) \in B$ whenever $(k+1) < n\_0$ 

Since $ A\_1 \cup B = D $ Where D is the set from Case 1 and that this is true $\forall n\_0 \in \mathbb{N} $ thus we know that $A\_1 = \mathbb{N} - B$
Which by definition defines $A\_1$ as the set of natural numbers $ \geq n\_0$ lastly since $ A\_1 \subset A$ ( as A could contain numbers smaller than $n\_0$ ) We know that A contains all natural numbers $ \geq n\_0 $