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

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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...
 
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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...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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