# ordered topology on integers and local compactness

Hi I am trying to make sure my logic is sound,

Let's suppose that we declared the discrete topology on $\mathbb{Z}$. Let us consider the set $\{1,2,3\}$. This set is open. However, this set is also closed because it has no limit points (the open set $\{3\}$ contains 3 but does not intersect the set $\{1,2,3\}$, the case is similar for $\{1\}$ and $\{2\}$) , also, under the discrete topology the $\mathbb{Z}$ is hausdorff so any finite point set has to be closed.

Is what I said correct?

Also, consider $\mathbb{Z}$ in the order topology. I would guess that order topology on $\mathbb{Z}$ would equal the discrete topology on $\mathbb{Z}$. However, I considered the order topology for the sake that I was reading the Munkres Topology book (p.183 Example 3) where it stated that "Every simply ordered set $X$ having the least upper bound property is locally compact. Given a basis element for $X$ it is contained in a closed interval in $X$, which is compact.

I guess what I am asking, if we define the order topology on $\mathbb{Z}$ (because it is simply ordered) , it surely is locally compact because any closed subset of it, is also open, correct?

thank you.

• $\{1,2,3\}$ is also closed in $\mathbb{Z}$ because it is the complement of the set $U=\mathbb{Z}\smallsetminus \{1,2,3\}$ which is open as all subsets are open. Yes, $\mathbb{Z}$ is Hausdorff under the discrete topology. Every set is Hausdorff under the discrete topology. – Jeremy Upsal Nov 12 '13 at 23:10

The order topology on $\Bbb Z$ is indeed the discrete topology, and it is locally compact. It’s locally compact because for each $n\in\Bbb Z$, $\{n\}$ is a compact open nbhd of $n$ that is contained in every open nbhd of $n$.
• There are different definitions of local compactness. Munkres states that $X$ is locally compact if every point has a compact set which contains a neighborhood of that point. (Which is of course true of $\{n\}$. ) – Cheerful Parsnip Nov 13 '13 at 0:07