Characterization of paracompactness

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In summary: Thanks for the clarification!In summary, the two conditions on a topological space X that are equivalent to say that X is paracompact are that every open cover U of X admits a refinement U' such that around every point x of X, there is an open nbhd A which intersects only finitely many elements of U', and that X is metacompact.
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quasar987
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A topological space X may be defined as paracompact by the condition that every open cover U of X admits a refinement U' such that every point of X intersects only a finite number of elements of U'.

A seemingly stronger condition on X would be that every open cover U of X admits a refinement U' such that around every point x of X, there is an open nbhd A which intersects only finitely many elements of U'.

I'm pretty sure that in fact these conditions are equivalent (at least for metric spaces) but I'm having trouble proving it.
 
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  • #2
I thought your second paragraph was the definition of paracompact. The first paragraph would be metacompact. Paracompactness implies metacompactness obviously. The converse is not true, and not obvious at all. Every metric space is paracompact, hence metacompact. Hopefully the additional term will give you something to search on. If I were Mary Ellen Rudin, I could prove these myself. But I'm not, so I'd look them up! :)
 
  • #3
Ok, thanks Billy bob!
 
  • #4
P.S.

I looked it up in Steen and Seebach. Here are two examples that are not excruciatingly difficult.

First is #54 Interlocking Interval Topology, where X = positive reals excluding positive integers. Base consists of the sets S_n, where S_n is the union of (0,1/n) and (n,n+1) where n=positive integer. The open cover {S_n} has no open refinement. Consider S_1 to see that X is not paracompact. On the other hand, metacompactness is not hard to show.

#54 is not Hausdorff. If you want Hausdorff, look at second example.

Second example is #64 Smirnov's Deleted Sequence Topology, which apparently the same as what Munkres (2nd edition) calls the K-topolgy. Let X=real line. Let K={1/n : n is a positive integer}. Let scriptB = usual open intervals, and let scriptB_K be scriptB union {sets of the form B - K where B is in scriptB}. Then scriptB_K is a basis for the K-topology. It looks like K-open sets are of the form U-L where U is usual open and L is a subset of K. Then the K-topology could be shown to be metacompact. It is not paracompact: consider for each positive integer n the set O_n = (reals-K) union {1/n}, and consider covering by {O_n}.
 
  • #5
I just take the definition of paracompact to mean there exists a partition of unity...
 
  • #7

Related to Characterization of paracompactness

1. What is paracompactness?

Paracompactness is a topological property of a space, which means that it is a characteristic of the way in which points and open sets are organized in that space. A paracompact space is one in which every open cover has a locally finite open refinement.

2. What is the significance of paracompactness in mathematics?

Paracompactness is an important concept in topology and analysis, as it ensures that certain operations and constructions can be carried out smoothly. For example, paracompact spaces have nice partition of unity properties, which are essential for defining integrals and solving differential equations.

3. How is paracompactness different from compactness?

While both paracompactness and compactness are topological properties, there are some key differences between the two. A space is compact if every open cover has a finite subcover, while a space is paracompact if every open cover has a locally finite open refinement. Essentially, compactness is a stronger condition than paracompactness.

4. Can you provide an example of a paracompact space?

Yes, the real line with the standard topology is a paracompact space. In fact, all metric spaces are paracompact. Other examples include any smooth manifold, any Hausdorff space, and any locally compact space.

5. How is paracompactness related to the Tychonoff theorem?

The Tychonoff theorem states that the product of any collection of compact spaces is compact. This theorem can be used to prove that all paracompact spaces are normal, which means that they satisfy a stronger separation axiom than Hausdorff spaces. Thus, paracompactness is closely related to the Tychonoff theorem and has important implications in topology and analysis.

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