A set $S$ is defined recursively by:
Basis step: $0\in S$
Recursive step: if $a\in S$ then $a+3\in S$ and $a+5\in S$
Determine the set $S\cap\{a\in Z\;|\;0<a<12\}$
My thinking is since $a$ is $0$ member of $a$. Since $a>0$ my first term is assuming $a=1$ then $a + 3 = 4$, $a+5 =6$ so the next term is $a+7 =8$ then $a+9 =10$