Question

Let $X$ be a set and let $f: X\longrightarrow X$ be a function on $X$. Introduce a topology on $X$ by the following basis of open sets: for any subset $S$ of $X$, let $B_S$ be the set of forward images of $S$ under $f$; i.e. $$B_S = \{f^n(s): s\in S, n\in \textbf{Z}^+\}.$$ My question is, is this topology well-known and well-understood? Is there a theory which relates properties of $f$ to the resulting topology?

Answer 1

This topology is rather combinatorial in nature. You certainly could call it well-understood, but I would not expect it to have a name in the context of topology. It is essentially a special case of a concept in combinatorics known as an order ideal, or if you like the set of all order ideals. The set of all order ideals of a partially ordered set is indeed a topology, but not one that looks very topological. On the one hand, it does not satisfy any of the separation axioms other than $T_0$. On the other hand, it satisfies an even stronger axiom than finite intersection: The intersection of any collection of open sets is open. The set of order ideals is better understood as a distributive lattice than as a topology, even though it is both.

We can clean things up a little as follows. First, if we switch to the image of $f$, we can let $B_S$ instead be the union of images of $S$ including $S$ itself. Second, in each periodic orbit of $f$, any open set that contains one point contains the other one, so we might as well collapse that orbit to a point. After that, we can say that $x \prec y$ when $x = f^n(y)$. This defines a partial ordering on $X$ with respect to which the topology is just the set of order ideals.