Compact Sets of Metric Spaces Which Are Also Open

  • Thread starter Poopsilon
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  • #1
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Main Question or Discussion Point

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

Edit: No discrete metric shenanigans either.
 
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Answers and Replies

  • #2
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I think [itex] \mathbb{Z}_p [/itex], 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 [itex] \mathbb{Z}_p [/itex] is sequentially compact.

Actually, I think any disk in [itex] \mathbb{Q}_p [/itex] would work.

I'm not sure if you'd count that as "down-to-earth."
 
  • #3
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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.
 
  • #4
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I'm confused, so there are such open compact sets, but only in non-connected metric spaces?
 
  • #5
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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 [itex] X \setminus S [/itex] are both open, disjoint, and nonempty and [itex] X = S \cup (X \setminus S)[/itex]. Therefore X is disconnected.

Incidentally, the p-adic numbers are totally disconnected.
 
  • #6
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What is the simplest disconnected metric space you can think of? I'd be willing to bet it works. There are very simple examples.
 
  • #7
disregardthat
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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.
 
  • #8
HallsofIvy
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For example, if X is [itex][0, 1]\cup[2, 3][/itex], with the topology inherited from R, then both [0,1] and [2, 3] are compact and open in X.
 
  • #9
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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:

[tex]\mathcal{T}=\{A\subseteq X~\vert~X\setminus A~\text{is finite}\}[/tex]

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.
 
  • #10
Bacle2
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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.
 
  • #11
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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.
 

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