In my general topology textbook there is the following exercise:
Let $(\mathbb Z , \tau)$ be the set of integers with the finite-closed topology. List the set of limit points of the following sets:
1 – $A=\{1,2,3,…,10\}$
2 – The set, $E$, consisting of all even integers
My approach
1.
The set $A$ is closed hence it must contain already all of it's limit points, so let $p \in A$ be a limit point of $A$. The set $A \setminus \{p\}$ is closed because it is also finite, so we have that $B := \mathbb Z \setminus(A \setminus \{p\}) \in \tau$. We have that $p \in B \in \tau$, but there isn't a different member of $A$ in the set $B$ thus $p$ is not a limit point. So we conclude that the set $A$ does not have limit points in this topological space.
2.
Let $E'$ be the set of all limit points of the set $E$. Then we have that $\bar E = E \cup E'$. The closure of a set is closed, so we have that $\bar E$ is a finite set, because we are working with the finite-closed topology. But because $\bar E$ is the closure of $E$: $E \subseteq \bar E$ this is a contradiction because $E$ is infinite but $\bar E$ is finite. Thus the set $E'$ does not exists so the set $E$ does not have any limit points as well.
Are these two arguments right? Because I find it a little odd that the exercise asked me to find all of the limit points and in both cases it turned out that are none. Am I making something wrong?
Edit:
$\mathbb Z$ is closed in $(\mathbb Z, \tau)$ and because it is the only infinite set that is closed we have that $\bar E = \mathbb{Z}$, this leaves us with $\mathbb Z \setminus E \subseteq E'$. Now I just need to find out if any member of $E$ is in $E'$ before I can affirm that $E' = \mathbb Z \setminus E$.
Best Answer
With your edit you’re almost done. The fact is that if $S$ is any infinite subset of $\Bbb Z$, then every $n\in\Bbb Z$ is a limit point of $S$, whether it’s in $S$ or not.
Let $U$ be any open nbhd of $n$. Clearly $U\ne\varnothing$, so by definition $\Bbb Z\setminus U$ is finite; for convenience let $F=\Bbb Z\setminus U$. $S$ is infinite, so clearly $S\nsubseteq F$. In fact it’s clear that $S\setminus F$ must be infinite, so it must contain some point $s$ different from $n$. Then $s\in S\setminus\{n\}$ and $s\notin F$, so $s\in\Bbb Z\setminus F=U$, and $n$ is therefore an accumulation point of $S$.