What is the Bing Metrization Theorem?

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SUMMARY

The Bing metrization theorem states that a space X is metrizable if and only if it is regular and has a countably locally discrete basis. The proof of this theorem closely parallels that of the Nagata-Smirnov metrization theorem, utilizing similar lemmas regarding locally discrete collections. Key lemmas include properties of locally discrete collections and the refinement of open coverings in metrizable spaces. The discussion confirms that understanding the proof of the Bing metrization theorem requires familiarity with the Nagata-Smirnov theorem.

PREREQUISITES
  • Understanding of the Bing metrization theorem
  • Familiarity with the Nagata-Smirnov metrization theorem
  • Knowledge of regular and normal spaces in topology
  • Concept of locally discrete bases and Gδ sets
NEXT STEPS
  • Study the proof of the Nagata-Smirnov metrization theorem
  • Explore the concept of paracompactness in topology
  • Investigate properties of locally discrete collections in topological spaces
  • Learn about Gδ sets and their significance in regular spaces
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Mathematicians, particularly those specializing in topology, students studying metrization theorems, and researchers interested in the properties of regular and normal spaces.

radou
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Unless I'm mistaken about something, the Bing metrization theorem is more or less only a corrolary of the Nagata-Smirnov metrization theorem.

Theorem (Bing metrization theorem). A space X is metrizable iff it is regulat and has a countably locally discrete basis.

The proof is practically the same as the proof of the Nagata-Smirnov theorem, since analogous results of some lemmas regarding locally finite families hold for locally discrete families, too. I'll number these lemmas like in Munkres book, only with asterisk signs.

Lemma 39.1.* Let A be a locally discrete collection of subsets of X. Then:

(a) Any subcollection of A is locally discrete
(b) The collection of the closures of elements of A is locally discrete
(c) The closure of the union of elements of A equals the union of their closures

Lemma 39.2.* Let X be a metrizable space. If A is an open covering of X, then there exists an open covering E of X which refines A and is countably locally discrete.

In the proof of the analogue of this lemma for a countably locally finite refining, it is shown in a step that for any x in X, the 1/(6n)-neighborhood of x can intersect at most one element of En, which is exactly what we need.

Lemma 40.1.* Let X be a regular space with a basis B that is countably locally discrete. Then X is normal, and every closed set in X is a Gδ set in X.

Now, with these results laid out, it follows that the proof of the Bing metrization theorem is exactly the same as that one of the Nagata-Smirnov metrization theorem, as far as I've inspected. If I made any mistake, please correct me about this.
 
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You are correct. The proof of Bings metrization theorem is exactly the same as the proof of Nagata-Smirnov.

So I guess paracompactness is next? :smile:
 
micromass said:
You are correct. The proof of Bings metrization theorem is exactly the same as the proof of Nagata-Smirnov.

So I guess paracompactness is next? :smile:

OK, thanks a lot! Yes, paracompactness follows! Although I think I won't be entering 2011 with a knowledge about paracompactness.. :biggrin:
 

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