Shriking a point-finite open cover

[radou]

Homework Statement

Lemma. Let X be a normal space, let {U1, U2, ... } be a point-finite indexed open covering for X. Then there exists an indexed open covering for X {V1, V2, ...} such that Cl(Vi)\subseteqUi, for every i.

The Attempt at a Solution

Now, something I want to clear up.

Obviously, the "point-fininte" condition on the cover {Ui} is necessary, because we couldn't prove it witout this? If we look at Munkres' Theorem 36.1., Chapter 36, which proves the existence of a partition of unity, and first proves the same lemma but for a finite number of sets, couldn't we prove this the same way, but inductively for the infinite case? I guess we could if our sets would be indexed by a countable set? But since this set is not specified, it can be an arbitrary set, so we can't proceed that way, am I correct?

Edit, the title should be "Shrinking"

 

[ystael]

If an indexed family is presented to you in the format \{U_1, U_2, \dots\}, the index set is presumed to be the positive integers and thus the indexed family is presumed to be countable. To convey that the index set is unspecified one typically writes something like \{U_\lambda\}_{\lambda\in\Lambda}.

 

[radou]

If an indexed family is presented to you in the format \{U_1, U_2, \dots\}, the index set is presumed to be the positive integers and thus the indexed family is presumed to be countable. To convey that the index set is unspecified one typically writes something like \{U_\lambda\}_{\lambda\in\Lambda}.

Hm, then it could be proved inductively? But then I don't even see where the point-finite requirement comes in...

To illustrate the "inductive" proof I'm referring to for a finite family {U1, ..., Un}, I'll display the first step:

Let A = X\(U2 U ... U Un). A is closed in X. Because {U1, ..., Un} covers X, A is contained in U1. Now use normality to find a neighborhood V1 of A such that Cl(V1)\subseteqU1. Then {V1, U2, ..., Un} covers X. Etc.

 

[ystael]

Point-finiteness is required to ensure that the final collection \{V_1, V_2, \dots\} covers X. If some point x lies in U_n for arbitrarily large n, then it never gets tossed over into the closed set around which you draw a V_n.

 

[micromass]

Hmm, I've thought about this for quite some time. It is really a very good question.

I'm certain that you can't do this for arbitrary collections (even countable collections). You need point-finiteness. But a counterexample is very hard to give (I didn't find one yet).

Anyway, what you do in theorem 36.1, you essentially do the following:

You take \{U_1,U_2,...\} (I take a countable collection instead of a finite one), and you transform it in \{V_1,U_2,...\}. This is still a covering
In general, you take \{V_1,...,V_n,U_{n+1},...\} and you change the U_{n+1} into V_{n+1} and this remains a covering.

But the problem is the following: you have showed that the following collections are coverings: \{V_1,...V_n,U_{n+1},...\}. Thus a collection with FINITELY many V_n are coverings.
But there is no way you can show that \{V_1,...,V_n,V_{n+1},...\} is a covering! Since this are infinitely many V_n, this does NOT follow from our induction! (indeed: the induction only proves it for FINITELY many V_n.

To show that \{V_1,...,V_n,...\} is a covering, you'll need point-finiteness!

At least, that's what I think the problem is. But I'm still searching for a counterexample, which is not easy to find since you need to find normal spaces which are not paracompact, and these are hard to find...

 

[radou]

Point-finiteness is required to ensure that the final collection \{V_1, V_2, \dots\} covers X. If some point x lies in U_n for arbitrarily large n, then it never gets tossed over into the closed set around which you draw a V_n.

Oh, I see! Excellent, thanks a lot!

 

[radou]

micromass, I just saw your reply. The problem with the infinite case basically is what ystael pointed out, right? Since if an x would be in Un for every n, nothing would guarantee that it is in the closed set we construct!

 

[micromass]

Yes, yes. ystael is correct! He was just a little faster then me

 

[radou]

OK, so basically the proof of this lemma is the same as that of Theorem 36.1. Boy, this exercise section should be called "the corollary section".