# Compact Sets of Metric Spaces Which Are Also Open

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

Edit: No discrete metric shenanigans either.

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## Answers and Replies

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."

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.

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

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.

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
Science Advisor
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
Science Advisor
Homework Helper
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.

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
Science Advisor
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.

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.