Suppose I have a pair of functions $f$ and $g$, and some number $x_0$. How in general do I define the vertices of the directed graph created by $f$ and $g$ as successors to $x_0$?
The rules I have in mind are that from every point $x$, both $f(x)$ and $g(x)$ define a successor of $x$.
For the sake of argument let $f(x)=x+1$, let $g(x)=\frac{x}{2}$ and let $x_0=1$.
We can see from $f^{n\to\infty}(x_0)$ that every positive integer not $1$ is a successor of $1$ - these could be plotted directly above $1$.
We can see from $g^{m\to\infty}(\mathbb{N})$ that every $$\frac{n+1}{2^m}: n,m\in\mathbb{N_{\geq0}}$$ can be connected and plotted to the right of the points in the first vertical column.
But there are also infinitely many successors to every one of those points. Correct me if I'm wrong but I imagine in this example every non-negative real number is a limit point of this graph. And I would guess the non-negative $2-$adic integers are its points within finite distance.
Is there a well-studied maths of this, and is there a general method for defining which points are connected to such graphs?
These kinds of problems are usually just algebra. You might be able to put graph theory into it, but it's mostly algebra.
The general method for finding all the reachable points is to guess the rule. You might be able to find some simpler methods for specific instances of the problem, but when you just have a general function, that could be anything, you can't really say much.
For example in the mentioned example, one could find all the connected points using the following method:
For positive integers $n$ and non-negative integers $m$, any number of the form $$\frac{n}{2^m}$$ can be reached, and any number that can be reached is of that form.
To prove that we can reach any number of this form, notice that the value of $$g^m(f^{n-1}(1)) = g^m(n) = \frac{n}{2^m}$$
Now, suppose you have any value of the form $n/2^m$, then adding one results in the value $(n+2^m)/2^m$, which is still of this form, and dividing by two gives $n/2^{m+1}$ which is also of this form. Since $1 = 1/2^0$ is of this form, we can only reach numbers of this form.
Any base-2 number with finitely many digits is representable in this form, simply remove the decimal separator from the binary representation and let that be $n$, then divide with $2^m$ where $m$ is the location of the decimal separator.
This means we can approximate any real number arbitrarily well using these numbers, since if we want to be within $2^{-p}$ of the actual value of the real number, simply write out the first $p+1$ binary digits of the real number and you have a number in your graph.
As far as I can see this doesn't have much to do with 2-adic integers.