Re: Partitions of unity question

From: Ron Sperber (ronsperber_at_optonline.net)
Date: 02/01/05 

Date: Mon, 31 Jan 2005 23:36:21 -0500

Tony wrote:
> If M is a topological space with the property that for every open cover X of
> M, there exists a partition of unity subordinate to X, show that M is
> paracompact.
>
> I can't quite see how to do this. If {f_i} is a partition of unity for X =
> {X_i}, then certainly
>
> X = Union of {supp(f_i)} (over i), since if x is in M, then there exists a
> function f_k such that f_k(x) is not zero since by definition of parition of
> unity,
>
> sum f_i(x) = 1 for all x in M (where the sum is taken over all i).
>
> So if supp(f_i) was open for all i, then {supp(f_i)} would be a locally
> finite refinement of {X_i} since supp(f_i) is a subset of X_i and the supp's
> form a locally finite set.
>
> But they may not be open. So I'm not sure what to do. For a particular k,
> can I maybe write supp(f_k) as a union of a finite amount of open sets?
> Because it seems clear to me that a finite refinement of a locally finite
> set is locally finite. But it doesn't appear clear to me that any
> refinement of a locally finite set is locally finite.
>
> Any help is highly appreciated, Thank you,
>
> Tony
>
>
Assuming you mean supp(f_i)={x in M| f_i(x) not equal 0} then this is
open because it is the inverse image of an open set. That is
supp(f_i)=f_i^{-1}( (0,1]) and (0,1] is an open subset of [0,1].

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