# Limit Points and the Finite Complement Topology on $\mathbb{R}$

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.

• Your explanation seems fine to me – Ishan Levy Feb 5 '18 at 1:10
• @IshanLevy no, the second part needs to be adjusted a bit, see below. – Arnaud Mortier Feb 5 '18 at 11:47

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.