I'm reading Armstrong's Basic Topology and have come across an example of limit points in the finite complement topology that I'm not sure I entirely understand.
This is Example 5 in Chaper 2.1, p. 29.
Take $X$ to be the set of all real numbers with the so called finite-complement topology. Here a set is open if its complement is finite or all of $X$. If we now take $A$ to be an infinite subset of $X$ (say the set of all integers), then every point of $X$ is a limit point of $A$. On the other hand a finite subset of $X$ has no limit points.
With the help of some results on Wiki Proof, here's how I follow this:
If $A$ is an infinite subset of $X$ and $O$ is any open set in the topology, then $A \cap O$ is infinite, and so for any neighborhood $N$ of $x \in X$, $A \cap N \not= \emptyset$.
If $B$ is a finite subset, then $B$ may be defined as $X \setminus U$ for some open set in the topology. It follows for any neighborhood of $M$ of $x \in X$, $B \cap M = \emptyset$ and so $B$ has no limit points in the topology.
Any help on pointing out where my understanding might be failing would be really helpful.