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I find it hard to quickly determine this. For instance,

$D^n/\{0\}=\{x \in \mathbb{R}^n| 0 < ||x|| \leq 1\}$ is non-compact.

$S^{n-1}=\{x \in \mathbb{R}^n | ||x||=1\}$ is compact.

So, okay, the first set is a punctured disc, it seems. So a disc, like a CD, with a hole at point $0$. The disc includes the boundaries, yes? but, why is it non-compact?

I'm not looking for a rigorous proof(at least for now). To simplify, let me set $n=2$. So I have a disc in a plane. With a hole of course.

The radius of the disc is $1$ so, well, how about I cover it with an ... open disc with radius $2$? That should suffice to cover(i.e. contain the punctured disc) and well it's just $1$ subset of $\mathbb{R}^n$ so it's also finite i.e. I have a finite subcover.

I can tell the second set is compact, since I can define a bunch of circles, i dunno, radius $1$ at points $1,-1$ on each $x,y$ axis. That would cover the entire circumference $S^{n-1}$ where $n=2$.

So, why is my cover of the punctured disc invalid?

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John Trail
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    You can use Heine-Borel theorem. – Henricus V. Apr 05 '16 at 22:53
  • I think that'll lead to a concrete proof, which I'm not looking for just now(i'm not all too familiar with the theorem actually). Can you argue back that my allegation is invalid above? That I have a finite subcover of the punctured disc in 2D? – John Trail Apr 05 '16 at 22:54
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    "The disc includes the boundaries, yes?" - but in the punctured disc, the puncture is an additional part of the boundary. – user2357112 Apr 06 '16 at 04:11

3 Answers3

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The flaw with your proof is that you showed that one particular open cover has a finite subcover. To show compactness, you have to show that every open cover has a finite subcover. Suppose $A_n=\{x\in\mathbb R^n:1/n<\Vert x\Vert<2\}$. This is an open cover with no finite subcover, so $D^n\setminus\{0\}$ is not compact. Alternatively, you can see that $0$ is a boundary point, so $D^n\setminus\{0\}$ does not contain all of its boundary points, so it is not closed. By Heine-Borel, it is not compact.

The proof that $S^{n-1}$ is compact is flawed for the same reason. You showed one cover has a finite subcover, but all open covers have to have a finite subcover. I would suggest using the Heine Borel Theorem for this one too. The Heine Borel Theorem is very powerful, and works in $\mathbb R^n$ for all $n$. It proves immediately that $D^n\setminus\{0\}$ is not compact, while $S^{n-1}$ is.

Plutoro
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  • Hi, thanks, so I was looking at the Heine Borel theorem. But is the "closed" supposed to mean in terms of metrics? So are we talking about open balls that defines open sets and so that the entire set minus open sets are closed etc? Or is it in terms of topological spaces? And I'm not exactly sure how to show it's bounded in some general $n$ :/ – John Trail Apr 06 '16 at 11:20
  • We can talk about closed in the sense of the metric or the topology, it is the same. Indeed, $\mathbb R^n\setminus A$ is closed if and only if $A$ is open. A set $A$ is bounded if there exists a number $M$ large enough so that the set ${(x\in\mathbb R^n:\Vert x\Vert<M}$ contains $A$. – Plutoro Apr 06 '16 at 14:34
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I am assuming that you are interested in compacts of $\Bbb R^n$. The Heine-Borel theorem tells us that the compact sets in $\Bbb R^n$ are precisely the closed and bounded ones, so this is an easy test.

As to why your proof that $D^2 \setminus \{0\}$ is compact is invalid, this is because you are confusing the definition of compactness. If $X$ is a topological space and $A \subset X$, $A$ is said to be compact if for any open cover of $A$, there exists a finite subcover of $A$. What you showed is that one particular open cover of $A$ has a finite subcover.

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The other answers do a good job in explaining the flaw in your argument.

The quick way to determine if a subset in $R^n$ is compact is to use the Heine-Borel theorem, which says that a subset of $R^n$ is compact if and only if it is closed and bounded. This is usually much easier than using the definition of compactness.

So in your first case you clearly see that the punctured disk $D^n\setminus \{0\}$ is bounded by a disk with radius $2$. Also, $D^n\setminus \{0\}$ is not closed, so by the Heine-Borel it is not compact.

Similarly we easily see that since $S^{n-1}$ is closed and bounded, it is compact.

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