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.
Best Answer
For part 1) your explanation is perfect.
For part 2) you need to adjust the quantifiers:
If $B$ is a finite subset, then $B$ may be defined as $X \setminus U$ for some open set in the topology. It follows that for any $x$ in $X$ there is a neighbourhood of $x$, namely $M=U\cup \{ x \}$, such that $B\cap M=\{ x \}$ (in case $x\in B$), or $B\cap M=\emptyset$ (in case $x\not\in B$).
Remember that in the definition of limit point, for any open neighbourhood of $x$ you have to be able to find a point of $B$ other than $x$ itself.