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Suppose $X$ is an $n$-dimensional for differentiable manifold for $n \geq 1$: in our definition this is a second countable Hausdorff space with a maximal differentiable atlas. If $p \in X$ is a point, we are asked to prove that $X \backslash \{p\}$ is not compact. I don't really see why. I have had a hint to look at a chart around $p$ which somehow can be related to $D^n$ minus a point, where $D^n$ is a compact $n$-dimensional ball. However, it is required to use the Hausdorff property for this, I cannot really see how to use it here though. Any hints?
Thanks in advance!

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  • $\begingroup$ What definition of compact are you using? There are several ways to show it. $\endgroup$ – Matt Samuel Nov 11 '14 at 19:04
  • $\begingroup$ "$X$ is compact if every open cover of $X$ has a finite subcover" $\endgroup$ – Steve Nov 12 '14 at 12:47
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    $\begingroup$ That's equivalent to "For any collection of closed sets, if every finite intersection of set in the collection is nonempty, then the intersection of all of them is nonempty." This is seen by taking complements. That's what is used in the answer below. $\endgroup$ – Matt Samuel Nov 12 '14 at 12:49
  • $\begingroup$ Thanks for your time, I just don't really see how this uses the Hausdorff property on $X$? $\endgroup$ – Steve Nov 12 '14 at 12:55
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    $\begingroup$ Hausdorff does tell you that the closed ball, being compact, is closed, and it remains relatively closed after removing a point. $\endgroup$ – Matt Samuel Nov 12 '14 at 13:02
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In a chart around $p$, consider the set of all closed balls centered at $p$ of radius $1/n$ for some integer $n$. Every finite intersection of sets in this collection is nonempty but the intersection of all of them is empty. Since the manifold is Hausdorff and closed balls in the chart are compact, they are closed in the manifold. Once a point is removed, the punctured balls are closed. Thus the intersection of all these sets being empty implies the punctured manifold is not compact.

As suggested by Najib, an equivalent formulation is to consider the collection of complements of the punctured balls in the manifold, which constitutes an open cover, and the open cover contains no finite subcover because all of the sets in a finite subcover would be contained in the largest one, which does not contain every point in the punctured manifold.

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  • $\begingroup$ You can also reformulate that to accommodate the definition of the OP dealing with finite subcovers, if you want (this is strictly equivalent): if $B_n = B(p,1/n) \setminus \{p\}$, then $\{X \setminus B_n\}_n$ is an open cover of $X$ that doesn't admit a finite subcover. $\endgroup$ – Najib Idrissi Nov 12 '14 at 13:58
  • $\begingroup$ @Najib Thanks. Edited. $\endgroup$ – Matt Samuel Nov 12 '14 at 14:03

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