Say that a subset $U \subseteq \mathbb{Z}$ is symmetric if it satisfies the following condition: for each $n\in\mathbb{Z}, n\in U$ if and only if $-n \in U.$ Define a topology on $\mathbb{Z}$ by declaring a subset to be open if and only if it is symmetric.
So something like $U = \{-1, 0, 1\}$ would be open, while $V = \{-1, 0, 1, 2\}$ would not be open.
My goal is to show that $\mathbb{Z}$ with this topology is limit point compact but not compact. I've tried showing that it is limit point compact, but I think I am missunderstanding some of the definitions. I know that a set is limit point compact if every infinite subset $S \subseteq \mathbb{Z}$ has a limit point. Then a limit point $x \in S$ is a point where every open set $U$ containing $x$ must contain another distinct point of $S.$ As an example, I'm thinking about $$S = \{1, 2, 3, \dots\}.$$ Then any open set containing say $1$ would also have to contain $-1,$ but this doesn't intersect the set $S$ other than at $1.$ Wouldn't $S$ not have any limit points then? What part of this am I misunderstanding?
Best Answer
If $A$ is infinite and $x \in A$ with $x \neq 0$ (which must surely exist), then every open neighbourhood $O$ of $-x$ also contains $x$ by symmetry and $x \in O \cap A\setminus \{-x\}$, so $-x$ is a limit point of $A$.
And the open cover $\{0\}, \{n,-n\}, n \in \Bbb N$ is a disjoint infinite cover of $\Bbb Z$ from which we cannot spare a single element so this cover has no finite subcover, showing that $\Bbb Z$ is not compact.