Another paracompactness problem

  • Thread starter radou
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In summary, if X is regular and a countable union of compact subspaces, then X is paracompact. The proof for this involves showing that any open cover for X can be refined to a countable subcollection that covers X. This can be done by taking finite subcovers for each compact subspace in X and then combining them into a countable subcollection.
  • #1
radou
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Homework Statement



Let X be regular. If X is a countable union of compact subspaces of X, then X is paracompact.

The Attempt at a Solution



Let X = U Ci. Let an open cover for Ci be given. Denote its finite subcover with {Ci1', ... , Cin'}. Clearly, X can now be expressed as the countable union of the elements Cij'. Denote this collection with C. Now, let U be an open cover for X. Then the collection {A[tex]\cap[/tex]Cij' : A is in U, Cij' is in C} covers X and is an open refinement of U. But nothing more.

The idea is to

a) either conclude that X is Lindeloff, since paracompactness would then follow
b) find a refinement of U which is an open cover for X and locally finite, paracompactness would be satisfied by definition
c) either find a refinement of U which is open, covers X and countably locally fininte, or simply covers X and is locally finite, or is closed, covers X and is locally finite, since then paracompactness would follow again from a lemma

Am I on the right track here? Any discrete hints are welcome...
 
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  • #2
Let's not make it harder then it is: go for (a) :smile:
 
  • #3
micromass said:
Let's not make it harder then it is: go for (a) :smile:

Ha, OK, I'll do that. It turns out to he an elegant set-up. OK, I'll post a bit later...
 
  • #4
Ahh, this turns out to be very easy, actually.

Let A be an open cover for X. The A is an open cover for every compact subspace of X, Ci. For every Ci, let {Ai} be the finite subcover of A covering Ci. Obviously the collection of all the fininte subsovers Ai is countable and a subcollection of A covering X.
 
  • #5
That is a perfect solution!
 
  • #6
micromass said:
That is a perfect solution!

Thanks, btw I don't see any other one. :smile:
 

1. What is "Another paracompactness problem"?

"Another paracompactness problem" is a mathematical concept that deals with the properties of topological spaces. It is a generalization of the paracompactness property, which states that every open cover of a space has a locally finite refinement. In "Another paracompactness problem", the refinement is required to be a closed cover, meaning that the sets in the refinement must be closed in the original space.

2. Why is "Another paracompactness problem" important in mathematics?

"Another paracompactness problem" is important in mathematics because it helps to understand the structure of topological spaces and their properties. It has applications in various areas of mathematics, such as differential geometry, functional analysis, and algebraic topology. Additionally, it has connections to other important concepts in mathematics, such as compactness and metrizability.

3. What are some examples of spaces that satisfy "Another paracompactness problem"?

Examples of spaces that satisfy "Another paracompactness problem" include metric spaces, locally compact spaces, and paracompact spaces. These spaces have the property that every open cover has a locally finite closed refinement, making them good candidates for studying "Another paracompactness problem".

4. What is the relationship between "Another paracompactness problem" and the Urysohn lemma?

The Urysohn lemma is a result in topology that states that every normal space can be embedded in a paracompact space. This result is closely related to "Another paracompactness problem" because it can be used to show that every normal space satisfies "Another paracompactness problem". In fact, "Another paracompactness problem" can be seen as a generalization of the Urysohn lemma.

5. Are there any known counterexamples to "Another paracompactness problem"?

Yes, there are known counterexamples to "Another paracompactness problem". One example is the Michael line, which is a topological space that is not paracompact but satisfies "Another paracompactness problem". Another example is the Sorgenfrey line, which is a space that is not paracompact and does not satisfy "Another paracompactness problem". These counterexamples show that "Another paracompactness problem" is a more strict property than paracompactness.

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