# Finding limit points of subsets of the cofinite topology on $\mathbb{Z}$

Is my reasoning correct?

Problem:

Let $(Z,\tau)$ be the cofinite topology on $Z$. Find the limit points of the sets:

1. $A = \{1,2,\dots,10\}$
2. $E$, the even integers

My solution:

1. $A$ is closed (finite), so it contains its limit points. Let $x$ in $A$, then $\{x\} \cup (Z-A)$ is a neighborhood of $x$ containing no points of $A$ other than $x$, so $A$ has no limit points.

2. Every open neighborhood of a point $x$ in $Z$ must contain infinitely many points of $E$ for the complement to be finite. So every point in $Z$ is a limit point of $E$ (i.e., $E$ is dense in $Z$).

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Yep. ${}{}{}{}$ – anon Jul 24 '13 at 1:55
Thanks${}{}{}{}$ – saadtaame Jul 24 '13 at 1:59
Do you see how you can generalize both cases? The generalization allows you to determine the limit points of any set $S\subseteq \bf Z$. For that matter you can put the cofinite topology on any countably infinite set $X$ and get the same results. – Karl Kronenfeld Jul 24 '13 at 2:02
Hmm, nice observation! I think it has to do with finiteness. Right? – saadtaame Jul 24 '13 at 2:06
Right, you can make that an answer if you want--it's perfectly fine to answer your own questions. – Karl Kronenfeld Jul 24 '13 at 2:07

The proof actually applies to arbitrary sets. If $A$ is finite, the open neighborhood $\{x\}\cup(Z-A)$ contains no points of $A$ other than $x$ ($\forall x\in A$) thus $A$ has no limit points. If $A$ is infinite, then every open neighborhood of $A$ must contain infinitely many points of $A$ for the complement to be finite, so every point of $Z$ is a limit point of $A$. A subset of $Z$ is either dense or has no limit points.