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In Munkres topology, he defines paracompactness as a generalization of compactness, as follows:

$X$ is paracompact if every open cover of $X$ has locally finite open refinement that covers $X$

But why it should be "refinement" of open cover? In compact, we need "subcollection". Surely subcollection is more strong assumption, but then , is there any paracompact space that has refinement but not subcollection?

If then, what important difference between refinement and subcollection does make difference of definition in paracompactness?

Kim Seong Hyeon
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2 Answers2

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Take the open interval $(0,1)$ and consider its open sets $$U_n=(1/n,1-1/n)\qquad n\geq2.$$ Then $\mathcal{U}=\{U_n\}_{n\geq2}$ is a countable open cover of $(0,1)$ which has no locally-finite subcover. To see this observe the stronger fact that any locally-finite subcover of $\mathcal{U}$ is finite, and $\mathcal{U}$ has no finite subcover.

Of course $(0,1)$ a separable, locally compact metric space homeomorphic to $\mathbb{R}$. Evidently it is paracompact and $\mathcal{U}$ has a locally-finite refinement.

Note that in general a space $X$ is compact if and only if every open cover of it has a finite subcover if and only if every open cover of it has a finite refinement. If you consider this second condition to be the definition of compactness, then paracompactness is its logical extension.

Tyrone
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The property "every open covering has a locally-finite subcovering" is equivalent to compactness. So any paracompact space that is not compact fails to have this property.

In fact, even the property of having point-finite subcoverings is equivalent to compactness. (Pick $U_0 \in \mathcal{U}$ and take a point-finite subcovering of $\{U \cup U_0 | U \in \mathcal{U} \}$.)

David Hartley
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  • It's not true that 'locally-finite subcovers exist' is equivalent to compactness. A disjoint sum of infinitely many compact spaces has this property. – Tyrone Jan 08 '23 at 21:04
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    No it doesn't. Consider the disjoint sum of ${X_n|\ n \in \mathbb{N} }$. Let $U_n = X_1 \cup X_n$. ${U_n |\ n \in \mathbb{N}, n \neq 1 }$ is an open covering with no proper subcovering. It is not locally-finite, every point of $X_1$ is in every member. – David Hartley Jan 08 '23 at 22:35
  • Right! That was even the idea I used above, so you'd think I would understand it ;). Given the op is only meeting paracompactness for the first time, they'd probably find it useful if you wrote some more details into your argument. – Tyrone Jan 08 '23 at 22:49