Homework Question: Understanding a Lemma on Open Coverings in Regular Spaces

  • Thread starter Thread starter radou
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around a lemma in topology concerning open coverings in regular spaces, specifically focusing on the relationship between locally finite and countably locally finite coverings. The original poster expresses confusion regarding a part of the proof that claims the implication is trivial.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the nature of the proof and its implications, with some suggesting that the proof is straightforward while others express confusion about the reasoning behind it. There is a discussion about the definitions of locally finite and countably locally finite coverings.

Discussion Status

The conversation includes attempts to clarify the proof's simplicity, with some participants finding the proof trivial while others are still grappling with understanding it. There is an acknowledgment of differing perspectives on the complexity of the lemma.

Contextual Notes

Participants mention the lemma's implications and related concepts, such as paracompactness and connections to measure theory, indicating a broader context of study within topology.

radou
Homework Helper
Messages
3,149
Reaction score
8

Homework Statement




OK, just a short question about a lemma I'm going through in Munkres and a part of its proof.

I won't quote the whole lemma (it's a few statements which are equivalent), but only the part I don't get:

Let X be regular. If every open covering of X has a refinement that is an open covering of X and locally finite, then every open covering of X has a refinement that is an open covering of X and countably locally finite.

In the proof it states that this is trivial, so I'm missing something obvious apparently. But I just can't see what.

Thanks in advance...
 
Physics news on Phys.org
I think you're thinking to far here. The proof is really simple: take an covering \mathcal{B} which is open and locally finite. Then the covering is also countable locally finite. Thus we have found an open cover which is countable locally finite...
 
micromass said:
I think you're thinking to far here. The proof is really simple: take an covering \mathcal{B} which is open and locally finite. Then the covering is also countable locally finite. Thus we have found an open cover which is countable locally finite...

Ahhhh this is utterly trivial ! Since this cover (call it A) can be written as a union A = U A (this doesn't make sense, but I think it's clear what I'm trying to say) consisting of a single element, that cover itself, which is locally finite! :rolleyes: :smile:

Sorry, sometimes I get confused on a really stupid base.
 
Haha, looks like somebody partied a little too hard last night :biggrin:
 
micromass said:
Haha, looks like somebody partied a little too hard last night :biggrin:

Yes, one could say... :cool:

Btw, the proof of the implication (3) ==> (4) in the same lemma (i.e. if every open covering of X has a refinement that is a closed locally finite covering of X, then every open covering of X has a refinement that is an open locally finite covering of X) is beautiful, at least to me. At a first glance, it seems a bit complicated, but when you actually think about what's going on, it's great!
 
Yeah, it is quite beautiful. It reminds me a bit of measure theory: you do some complicated things for complicated reasons, and it all ends really nice :smile:

When I first learned about it, I found paracompactness quite complicated. But it has it's beauty...
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
2K
Replies
7
Views
3K
  • · Replies 0 ·
Replies
0
Views
3K