$X\subseteq Z^+$ defined recursively as:
$1)$ $3\in X$; and
$2)$ If $a,b\in X$, then $a+b\in X$.
Prove that $X=\{3k|k\in Z^+\},$ the set of all positive integers divisible by $3$.
Induction on the size of set $X$. If $X$ has one element it must be $3$ by $i)$. This forms the base case. Assume true for $|X_n|=n$. $X_n=\{3,3i_1,3i_2,...,3i_{n-1}\}$. By $ii)$ $X_{n+1}=\{3,3i_1,3i_2,...,3i_{n-1},3i_j+3i_k=3(i_j+i_k)\}$. By Principle of Mathematical Induction $X=\{3k|k\in Z^+\},$ the set of all positive integers divisible by $3$.
$\{3k\mid 0< k \in \mathbb{N}\} \subseteq X$ is clear, further $\{3k\mid 0< k \in \mathbb{N}\} $ satisfies the induction assumptions so equality holds. When sets are defined recursively in this way it always means the smallest such set.