# The integers $\mathbb{Z}$ are limit point compact with the symmetric topology.

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?

• Yes. This is 4-20 in "Introduction to Topological Manifolds" by John M. Lee. The question seems to imply that it is limit point compact, but from the above, I don't think that it is.
– user525033
Commented Nov 4, 2019 at 19:04
• The limit point is only required to have some other point from the set in any of its neighborhoods, not infinitely many. For example, $-1$ is a limit point of the your set $S$, actually it is a limit point of the set $\{-1\}$. Commented Nov 4, 2019 at 19:08
• @conditionalMethod That makes sense. I was only considering points inside the set.
– user525033
Commented Nov 4, 2019 at 19:10
• So, to prove that your space is limit point compact, let $A$ be an infinite set. It must contain elements other than $0$. Let $x\in A$ with $x\neq 0$. Then $-x$ is a limit point of $A$. Commented Nov 4, 2019 at 19:12
• To show that it isn't compact, consider the open cover $\{0\}\cup\bigcup_{x\in\mathbb{N}^*}\{-x, x\}$. Commented Nov 4, 2019 at 19:14

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.
The issue with the question is where you say "Then a limit point $$x \in S$$..." where I think you are implicitly assuming a limit point of a set $$S$$ must be a member of a said set. Limit points of sets need not be in the set, and in this example, because of the strangeness of this topology, the limit points of $$S = \mathbb{N}$$ are exactly the negative integers, and no element of $$\mathbb{N}$$ is actually a limit point of $$\mathbb{N}$$ under this topology. (every open set containing the number $$-n$$ must also contain $$n \neq -n \in \mathbb{N}$$ - a distinct point in $$S$$. Meanwhile every $$n \in \mathbb{N}$$ permits the open set $$\{n,-n\}$$ which doesn't contain any distinct points of $$\mathbb{N}$$).
While it is very strange for every limit point of $$S$$ to not be a member of $$S$$ for some sets in this topology, it is still true that every infinite $$S\subseteq \mathbb{Z}$$ has a limit point, so it is limit point compact.
Taking an easier example of the standard topology on $$\mathbb{R}^n$$, limit points of open sets $$U$$ can be in the boundary of the set, which are not included in the set itself. This example doesn't capture all of the weirdness of your example, because every member of a set in the standard topology is also a limit point of that set, but it is an obvious example of limit points of a set not contained in said set.