1
$\begingroup$

With the usual discrete topology, $\mathbb{Z}$ is non-compact.

I wonder if it is possible to make $\mathbb{Z}$ compact; I think there are many different ways to do so.

I've heard about the following idea:

Consider the stereographic projection of the closed lower semi-circle as illustrated in the Picture below, where $\mathbb{Z}$ is embedded in the $x$-axis and where the Points $(\pm 1,1)$ represent $-\infty$ and $+\infty$, respectively. I guess

1) Is this a well known idea?

2) Why is this now compact?

3) As a metric one can use the arclength metric?

4) I guess this has nothing to do with one-point compactification?

5) I do not really see how the Points $(x,1)$ on the closed lower semi-circle which are different from $(\pm 1,1)$ are projected.

enter image description here

$\endgroup$
1
$\begingroup$

Observe that this is similar to embedding $\mathbb R$ in the extended reals $\overline{\mathbb R}$. The extended reals are just $\mathbb R$ with $-\infty$ and $\infty$ joined to each end. The extended reals are of course compact (since the space turns out to be homeomorphic to the interval $[0,1]$). Basically, what your construction does is considers $\mathbb Z$ first as a subspace of $\overline{\mathbb R}$ and then takes the closure - which includes $-\infty$ and $\infty$. Then, a closed subspace of a compact space is compact.

Note that, you can show your construction gives a compact space rather directly, since it identifies $\mathbb Z$ with a closed and bounded subset of the plane, which must be compact. You can also directly show that, if we let $\overline{\mathbb Z}$ be the new topology, then a subset $U$ is open if and only if it satisfies two conditions:

  • If $\infty\in U$, then there is some integer $N$ such that for all $n\geq N$ we have $n\in U$.

  • If $-\infty\in U$, then there is some integer $N$ such that for all $n\leq N$ we have $n\in U$.

Then, note that, if you have an open cover of $\overline{\mathbb Z}$, then all the large enough elements are covered by any open set containing $\infty$ and all the small enough elements are covered by any open set containing $-\infty$, so only finitely many elements remain to be covered, implying compactness directly.

I don't know if I'd call this "well-known", but it is implicit in a lot of places - for instance, if you take a limit $$\lim_{n\rightarrow\infty}(1+1/n)^n$$ then, you are implicitly using this topological space to define $\infty$.

The arc length metric would work fine.

This is somewhat related to the one-point compactification. Actually, if you use a stereographic projection to wrap the integers around a circle and then add the top point, you get the one point compactification. You seem to have conflated this projection with the gnomonic projection in your question.

| cite | improve this answer | |
$\endgroup$

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.