Question about defining a set by a recursive formula This is probably a trivial question, but I was a bit confused so I thought I would ask. Suppose for example take Fibonacci sequence defined by 
$F_0= 1$, $F_1 = 1$, and 
$$
F_k = F_{k-1} + F_{k-2} \ (k \geq 2).
$$
Does it then makes sense to define a set $S$ of these numbers?
Say $S = \{ F_k : k \geq 0 \}$. 
I am just confused (about probably nothing to be confused about...) because the set is defined recursively which goes on forever, but we can not go to infinity... Does this always make sense or do I need some sort of axiom for this to make sense?
Could someone possibly clarify this for me? Thank you!
 A: The essential fact needed here is that a recursive definition like that of $F_n$ can be replaced by an equivalent explicit definition.  Specifically, the Fibonacci sequence $W=\{\langle k,F_k\rangle:k\in\mathbb N\}$ can be defined as the set of all those ordered pairs $\langle k,x\rangle$ such that there exists a function $f$ with the following properties: (1) the domain of $f$ is $\{0,1,\dots,k\}$ (known in set theory as $k+1$), (2) $f(0)=1$, (3) if $k\geq1$ then $f(1)=1$, (4) for all $j$ such that $2\leq j\leq k$, $f(j)=f(j-1)+f(j-2)$, and (5) $f(k)=x$.  In view of this explicit definition, we can prove the existence of $W$ by applying the separation axiom (also called subset axiom, Aussonderung, and comprehension axiom) of ZF to the set $\mathbb N\times\mathbb N$.  And once we have $W$, we get the desired $S$ as the range of $W$, i.e., as the set of second components of the ordered pairs in $W$.
A: It does make sense.
Define $S_n = \{ F_0, F_1, F_2, \dots, F_n \}$. (I think you'll agree that this set exists for every $n$.)
Then let $$S = \bigcup_{n \in \mathbb{N}} S_n$$
Since each $S_n$ exists, and $\mathbb{N}$ certainly exists, the union does as well (we have an axiom, the Axiom of Unions, to tell us that).
