I am typing up a proof that establishes an isomorphism between two countably infinite graphs. One establishes such an isomorphism inductively. So if we have $f_{n}$, whose domain is the first $n$ vertices of the first graph, we can extend it to a map $f_{n+1}$ to include one more vertex. In the source I am working with it says taking $f$ to be the union of all these partial maps is the required isomorphism, however taking a union over functions seems foreign to me. Does anyone have any suggestions for stating this differently? I considered $f = \lim_{n \rightarrow \infty} f_{n}$ but am open to others.
I know all of the details are omitted, but didn't see them to be necessary for this question. If people disagree though, I will edit accordingly.
If $v_n$ is the $n$-th vertex of the first graph, you could define $f$ by the rule $$f(v_n) := f_n(v_n).$$ (Of course, $f_n(v_n)$ is also equal to $f_m(v_n)$ for any $m > n$, but we've picked one index that's good enough to work.)