Espacio paracompacto

A Hausdorff space ${displaystyle X,}$ 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 ${displaystyle {mathcal {O}},}$ is a locally finite open cover, then there exists open sets ${displaystyle W_{U},}$ for each ${displaystyle Uin {mathcal {O}},}$, such that each ${displaystyle {bar {W_{U}}}subseteq U,}$ and ${displaystyle {W_{U}:Uin {mathcal {O}}},}$ is a locally finite refinement.

Lemma 2: If ${displaystyle {mathcal {O}},}$ is a locally finite open cover, then there are continuous functions ${displaystyle f_{U}:Xto [0,1],}$ such that ${displaystyle operatorname {supp} ~f_{U}subseteq U,}$ and such that ${displaystyle f:=sum _{Uin {mathcal {O}}}f_{U},}$ is a continuous function which is always non-zero and finite.

Theorem: In a paracompact Hausdorff space ${displaystyle X,}$, if ${displaystyle {mathcal {O}},}$ is an open cover, then there exists a partition of unity subordinate to it.

Proof (Lemma 1):

Let ${displaystyle {mathcal {V}},}$ be the collection of open sets meeting only finitely many sets in ${displaystyle {mathcal {O}},}$, and whose closure is contained in a set in ${displaystyle {mathcal {O}}}$. One can check as an exercise that this provides an open refinement, since paracompact Hausdorff spaces are regular, and since ${displaystyle {mathcal {O}},}$ is locally finite. Now replace ${displaystyle {mathcal {V}},}$ 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 ${displaystyle W_{U}=bigcup {Ain {mathcal {V}}:{bar {A}}subseteq U},}$. The property of ${displaystyle {mathcal {V}},}$ guarantees that every ${displaystyle Ain {mathcal {V}}}$ is contained in some ${displaystyle W_{U}}$. Therefore ${displaystyle {W_{U}:Uin {mathcal {O}}},}$ is an open refinement of ${displaystyle {mathcal {O}},}$. Since we have ${displaystyle W_{U}subseteq U}$, this cover is immediately locally finite.

Now we want to show that each ${displaystyle {bar {W_{U}}}subseteq U,}$. For every ${displaystyle xnotin U}$, we will prove that ${displaystyle xnotin {bar {W_{U}}}}$. Since we chose ${displaystyle {mathcal {V}}}$ to be locally finite, there is a neighbourhood ${displaystyle V[x]}$ of ${displaystyle x}$ such that only finitely many sets in ${displaystyle {mathcal {V}}}$ have non-empty intersection with ${displaystyle V[x]}$, and we note ${displaystyle A_{1},...,A_{n},...in {mathcal {V}}}$ those in the definition of ${displaystyle W_{U}}$. Therefore we can decompose ${displaystyle W_{U}}$ in two parts: ${displaystyle A_{1},...,A_{n}in {mathcal {V}}}$ who intersect ${displaystyle V[x]}$, and the rest ${displaystyle Ain {mathcal {V}}}$ who don't, which means that they are contained in the closed set ${displaystyle C:=Xsetminus V[x]}$. We now have ${displaystyle {bar {W_{U}}}subseteq {bar {A_{1}}}cup...cup {bar {A_{n}}}cup C}$. Since ${displaystyle {bar {A_{i}}}subseteq U}$ and ${displaystyle xnotin U}$, we have ${displaystyle xnotin {bar {A_{i}}}}$ for every ${displaystyle i}$. And since ${displaystyle C}$ is the complement of a neighbourhood of ${displaystyle x}$, ${displaystyle x}$ is also not in ${displaystyle C}$. Therefore we have ${displaystyle xnotin {bar {W_{U}}}}$.

Proof (Lemma 2):

Applying Lemma 1, let ${displaystyle f_{U}:Xto [0,1],}$ be continuous maps with ${displaystyle f_{U}upharpoonright {bar {W}}_{U}=1,}$ and ${displaystyle operatorname {supp} ~f_{U}subseteq U,}$ (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 ${displaystyle f=sum _{Uin {mathcal {O}}}f_{U},}$ is always finite and non-zero, take ${displaystyle xin X,}$, and let ${displaystyle N,}$ a neighbourhood of ${displaystyle x,}$ meeting only finitely many sets in ${displaystyle {mathcal {O}},}$; thus ${displaystyle x,}$ belongs to only finitely many sets in ${displaystyle {mathcal {O}},}$; thus ${displaystyle f_{U}(x)=0,}$ for all but finitely many ${displaystyle U,}$; moreover ${displaystyle xin W_{U},}$ for some ${displaystyle U,}$, thus ${displaystyle f_{U}(x)=1,}$; so ${displaystyle f(x),}$ is finite and ${displaystyle geq 1,}$. To establish continuity, take ${displaystyle x,N,}$ as before, and let ${displaystyle S={Uin {mathcal {O}}:N{text{ meets }}U},}$, which is finite; then ${displaystyle fupharpoonright N=sum _{Uin S}f_{U}upharpoonright N,}$, which is a continuous function; hence the preimage under ${displaystyle f,}$ of a neighbourhood of ${displaystyle f(x),}$ will be a neighbourhood of ${displaystyle x,}$.

Proof (Theorem):

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