0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $\{1,2,3\}$ 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. $\endgroup$ – Jeremy Upsal Nov 12 '13 at 23:10
5
$\begingroup$

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 $n$.

$\endgroup$
  • $\begingroup$ 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 $\{n\}$. ) $\endgroup$ – Cheerful Parsnip Nov 13 '13 at 0:07
  • $\begingroup$ @GrumpyParsnip: I know. However, they’re all equivalent for Hausdorff spaces, hence for discrete spaces, and I deliberately chose a statement that easily implies all of the usual competing definitions. $\endgroup$ – Brian M. Scott Nov 13 '13 at 0:10
  • $\begingroup$ Yeah, I figured you knew, but I wanted to have it recorded for anyone who might be confused. $\endgroup$ – Cheerful Parsnip Nov 13 '13 at 2:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.