- #1

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

Edit: No discrete metric shenanigans either.

Edit: No discrete metric shenanigans either.

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- Thread starter Poopsilon
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- #1

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Edit: No discrete metric shenanigans either.

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- #2

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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."

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- #4

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

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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.

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- #7

disregardthat

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- #8

HallsofIvy

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- #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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"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, 1/2

- #11

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"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, 1/2^{n}), 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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