Compact Sets of Metric Spaces Which Are Also Open

[Poopsilon]

Main Question or Discussion Point

Are there any down to earth examples besides the empty set?

Edit: No discrete metric shenanigans either.

 

[spamiam]

I think $\mathbb{Z}_p$, the set of p-adic integers, qualifies. They are the closed unit disk in the p-adic numbers, hence closed. Every closed disk with nonzero radius in the p-adics is clopen, and $\mathbb{Z}_p$ is sequentially compact.

Actually, I think any disk in $\mathbb{Q}_p$ would work.

I'm not sure if you'd count that as "down-to-earth."

 

[deluks917]

A compact subset of a metric space is closed. A closed open subset of a connected metric space is the whole space or the empty set. You need to look at metric spaces that are not connected.

 

[Poopsilon]

I'm confused, so there are such open compact sets, but only in non-connected metric spaces?

 

[spamiam]

I'm confused, so there are such open compact sets, but only in non-connected metric spaces?

I think that's correct. Suppose we have a nonempty proper subset S of a metric space X with S open and compact. As deluks said, then S is closed, hence clopen. Then S and $X \setminus S$ are both open, disjoint, and nonempty and $X = S \cup (X \setminus S)$. Therefore X is disconnected.

Incidentally, the p-adic numbers are totally disconnected.

 

[deluks917]

What is the simplest disconnected metric space you can think of? I'd be willing to bet it works. There are very simple examples.

 

[disregardthat]

The union of two disjoint compact subspaces of a metric space will work. Metric spaces are normal, so this union will be disconnected. The simplest would be a two-point metric space if you don't accept the empty or singleton metric space, but any metric space will do. Just consider sufficiently small closed balls around two different points.

 

[HallsofIvy]

For example, if X is $[0, 1]\cup[2, 3]$, with the topology inherited from R, then both [0,1] and [2, 3] are compact and open in X.

 

[micromass]

A compact subset of a metric space is closed.

Only in Hausdorff spaces. There are non-Hausdorff spaces with compact sets which are not closed.

For example, take X an infinite set and equip it with the finite complement topology:

$$\mathcal{T}=\{A\subseteq X~\vert~X\setminus A~\text{is finite}\}$$

then all subsets of X are compact, and so are the open subsets. But of course, not all subsets are closed.

This is an example of a noetherian space (= a space in which all open subsets are compact). They arise naturally in algebraic geometry.

EDIT: I'm sorry, I saw now that they were asking for a metric space. But I'll leave my answer here because it might be useful to some.

 

[Bacle2]

Deluks917 wrote, in part:

"A compact subset of a metric space is closed"

Just a quick comment, to refresh my metric topology: an argument for why compact subsets X of metric spaces M are closed: as metric subspaces, compact metric spaces are complete. Now, use closed set version that a subset is closed iff (def.) it contains all its limit points. Now, assume there is a limit point m of X that is not in X. Then you can construct a Cauchy sequence in X that would converge to m, e.g., by taking balls B(m, 1/2ⁿ), but then X cannot be complete, since m is not in X.

 

[micromass]

Deluks917 wrote, in part:

"A compact subset of a metric space is closed"

Just a quick comment, to refresh my metric topology: an argument for why compact subsets X of metric spaces M are closed: as metric subspaces, compact metric spaces are complete. Now, use closed set version that a subset is closed iff (def.) it contains all its limit points. Now, assume there is a limit point m of X that is not in X. Then you can construct a Cauchy sequence in X that would converge to m, e.g., by taking balls B(m, 1/2ⁿ), but then X cannot be complete, since m is not in X.

Indeed. Do notice that you have used here that limits of sequences are unique!! The argument does not hold for pseudo-metric spaces.