Editing Paracompact space (section)

=== Partitions of unity ===

The most important feature of paracompact [[Hausdorff space]]s is that they admit [[partition of unity|partitions of unity]] subordinate to any open cover. This means the following: if ''X'' is a paracompact Hausdorff space with a given open cover, then there exists a collection of [[continuous function (topology)|continuous]] functions on ''X'' with values in the [[unit interval]] [0, 1] such that:

* for every function ''f'':&nbsp;''X''&nbsp;→&nbsp;'''R''' from the collection, there is an open set ''U'' from the cover such that the [[support (mathematics)|support]] of ''f'' is contained in ''U'';
* for every point ''x'' in ''X'', there is a neighborhood ''V'' of ''x'' such that all but finitely many of the functions in the collection are identically 0 in ''V'' and the sum of the nonzero functions is identically 1 in ''V''.

In fact, a T<sub>1</sub> space is Hausdorff and paracompact if and only if it admits partitions of unity subordinate to any open cover (see [[#Proof that paracompact Hausdorff spaces admit partitions of unity|below]]). This property is sometimes used to define paracompact spaces (at least in the Hausdorff case).

Partitions of unity are useful because they often allow one to extend local constructions to the whole space. For instance, the integral of [[differential form]]s on paracompact [[manifold]]s is first defined locally (where the manifold looks like [[Euclidean space]] and the integral is well known), and this definition is then extended to the whole space via a partition of unity.

==== Proof that paracompact Hausdorff spaces admit partitions of unity ====

{{hidden
|(Click "show" at right to see the proof or "hide" to hide it.)

|A Hausdorff space <math>X\,</math> is paracompact if and only if it every open cover admits a subordinate partition of unity. The ''if'' direction is straightforward. Now for the ''only if'' direction, we do this in a few stages.

: '''Lemma 1:''' If <math>\mathcal{O}\,</math> is a locally finite open cover, then there exists open sets <math>W_{U}\,</math> for each <math>U\in\mathcal{O}\,</math>, such that each <math>\bar{W_{U}}\subseteq U\,</math> and <math>\{W_{U}:U\in\mathcal{O}\}\,</math> is a locally finite refinement.

: '''Lemma 2:''' If <math>\mathcal{O}\,</math> is a locally finite open cover, then there are continuous functions <math>f_{U}:X\to[0,1]\,</math> such that <math>\operatorname{supp}~f_{U}\subseteq U\,</math> and such that <math>f:=\sum_{U\in\mathcal{O}}f_{U}\,</math> is a continuous function which is always non-zero and finite.

: '''Theorem:''' In a paracompact Hausdorff space <math>X\,</math>, if <math>\mathcal{O}\,</math> is an open cover, then there exists a partition of unity subordinate to it.

: '''Proof (Lemma 1):''' 
:Let <math>\mathcal{V}\,</math> be the collection of open sets meeting only finitely many sets in <math>\mathcal{O}\,</math>, and whose closure is contained in a set in <math>\mathcal{O}</math>. One can check as an exercise that this provides an open refinement, since paracompact Hausdorff spaces are regular, and since <math>\mathcal{O}\,</math> is locally finite. Now replace <math>\mathcal{V}\,</math> by a locally finite open refinement. One can easily check that each set in this refinement has the same property as that which characterised the original cover.

: Now we define <math>W_{U}=\bigcup\{A\in\mathcal{V}:\bar{A}\subseteq U\}\,</math>. The property of <math>\mathcal{V}\,</math> guarantees that every <math>A\in\mathcal{V}</math> is contained in some <math>W_U</math>. Therefore <math>\{W_{U}:U\in\mathcal{O}\}\,</math> is an open refinement of <math>\mathcal{O}\,</math>. Since we have <math> W_{U} \subseteq U </math>, this cover is immediately locally finite.
:
: Now we want to show that each <math>\bar{W_{U}}\subseteq U\,</math>. For every <math>x \notin U</math>, we will prove that <math> x \notin \bar{W_U} </math>. Since we chose <math> \mathcal{V} </math> to be locally finite, there is a neighbourhood <math> V[x] </math> of <math> x </math> such that only finitely many sets in <math> \mathcal{V} </math> have non-empty intersection with <math> V[x] </math>, and we note <math> A_{1},...,A_{n},... \in \mathcal{V} </math> those in the definition of <math> W_U </math>.  Therefore we can decompose <math> W_U </math> in two parts: <math> A_{1},...,A_{n} \in \mathcal{V} </math> who intersect  <math> V[x] </math>, and the rest <math> A \in \mathcal{V} </math> who don't, which means that they are contained in the closed set <math> C:= X\setminus V[x] </math>. We now have <math> \bar{W_U} \subseteq \bar{A_1} \cup...\cup \bar{A_n}\cup C </math>. Since <math> \bar{A_i} \subseteq U </math> and <math> x \notin U </math>, we have <math> x \notin \bar{A_i} </math> for every <math> i </math>. And since <math> C </math> is the complement of a neighbourhood of <math> x </math>, <math> x </math> is also not in <math> C </math>. Therefore we have <math> x \notin \bar{W_U} </math>.
{{NumBlk|1=|2=|3=<math>\blacksquare\,</math> (Lem 1)|RawN=.}}

: '''Proof (Lemma 2):''' 
:Applying Lemma 1, let <math>f_{U}:X\to[0,1]\,</math> be continuous maps with <math>f_{U}\upharpoonright\bar{W}_{U}=1\,</math> and <math>\operatorname{supp}~f_{U}\subseteq U\,</math> (by Urysohn's lemma for disjoint closed sets in normal spaces, which a paracompact Hausdorff space is). Note by the support of a function, we here mean the points not mapping to zero (and not the closure of this set). To show that <math>f=\sum_{U\in\mathcal{O}}f_{U}\,</math> is always finite and non-zero, take <math>x\in X\,</math>, and let <math>N\,</math> a neighbourhood of <math>x\,</math> meeting only finitely many sets in <math>\mathcal{O}\,</math>; thus <math>x\,</math> belongs to only finitely many sets in <math>\mathcal{O}\,</math>; thus <math>f_{U}(x)=0\,</math> for all but finitely many <math>U\,</math>; moreover <math>x\in W_{U}\,</math> for some <math>U\,</math>, thus <math>f_{U}(x)=1\,</math>; so <math>f(x)\,</math> is finite and <math>\geq 1\,</math>. To establish continuity, take <math>x,N\,</math> as before, and let <math>S=\{U\in\mathcal{O}:N\text{ meets }U\}\,</math>, which is finite; then <math>f\upharpoonright N=\sum_{U\in S}f_{U}\upharpoonright N\,</math>, which is a continuous function; hence the preimage under <math>f\,</math> of a neighbourhood of <math>f(x)\,</math> will be a neighbourhood of <math>x\,</math>.
{{NumBlk|1=|2=|3=<math>\blacksquare\,</math> (Lem 2)|RawN=.}}

: '''Proof (Theorem):''' 
:Take <math>\mathcal{O}^*\,</math> a locally finite subcover of the refinement cover: <math>\{V\text{ open }:(\exists{U\in\mathcal{O}})\bar{V}\subseteq U\}\,</math>. Applying Lemma 2, we obtain continuous functions <math>f_{W}:X\to[0,1]\,</math> with <math>\operatorname{supp}~f_{W}\subseteq W\,</math> (thus the usual closed version of the support is contained in some <math>U\in\mathcal{O}\,</math>, for each <math>W\in\mathcal{O}^*\,</math>; for which their sum constitutes a ''continuous'' function which is always finite non-zero (hence <math>1/f\,</math> is continuous positive, finite-valued). So replacing each <math>f_{W}\,</math> by <math>f_{W}/f\,</math>, we have now — all things remaining the same — that their sum is everywhere <math>1\,</math>. Finally for <math>x\in X\,</math>, letting <math>N\,</math> be a neighbourhood of <math>x\,</math> meeting only finitely many sets in <math>\mathcal{O}^*\,</math>, we have <math>f_{W}\upharpoonright N=0\,</math> for all but finitely many <math>W\in\mathcal{O}^*\,</math> since each <math>\operatorname{supp}~f_{W}\subseteq W\,</math>. Thus we have a partition of unity subordinate to the original open cover.
{{NumBlk|1=|2=|3=<math>\blacksquare\,</math> (Thm)|RawN=.}}

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