Partitions of Unity and Paracompactness

Fix $x\in X$ and a neighbourhood $N$ of $x$ meeting only $A_{i_1},\ldots,A_{i_n}$. Choose an open neighbourhood $O$ of $x$ with $O\subseteq N$. If $O\cap\overline{A_j}\ne\varnothing$, choose $y\in O\cap\overline{A_j}$; the open neighbourhood $O$ of $y$ then meets $A_j$, so $N$ meets $A_j$ and $j\in\{i_1,\ldots,i_n\}$.

The inclusion $\bigcup_i\overline{A_i}\subseteq\overline{\bigcup_iA_i}$ holds because each $\overline{A_i}$ is contained in every closed set containing $A_i$, in particular in $\overline{\bigcup_iA_i}$.

Thus $O$ meets only $\overline{A_{i_1}},\ldots,\overline{A_{i_n}}$, so the closed family is locally finite.

Let $x\in\overline{\bigcup_iA_i}$ and take $N$ as in step 1.1; if $x\notin\bigcup_i\overline{A_i}$, then for each $i_k$ an open neighbourhood of $x$ misses $A_{i_k}$, and its finite intersection with an open neighbourhood inside $N$ misses every $A_i$, contradicting the closure criterion.

Hence $\overline{\bigcup_iA_i}=\bigcup_i\overline{A_i}$ by steps 1.2 and 2.2; if every $A_i$ is closed, the right-hand side is $\bigcup_iA_i$, so that union is closed.

By compactness, fix a finite subfamily $\mathcal V\subseteq\mathcal U$ covering $X$.

The family $\mathcal V$ is open, covers $X$, refines $\mathcal U$, and is locally finite because every point has the neighbourhood $X$, which meets only members of the finite family $\mathcal V$.

Thus $\mathcal V$ is the refinement required by [F2], and $X$ is paracompact.

For each member of $\mathcal U$, take all ambient open $O$ whose trace $O\cap A$ is that member; together with $X\setminus A$, these ambient open sets form an open cover $\mathcal W$ of $X$.

By [F2], fix a locally finite open cover $\mathcal V$ refining $\mathcal W$.

The nonempty traces $V\cap A$ for $V\in\mathcal V$ cover $A$, are open in $A$, and refine $\mathcal U$: a $V$ meeting $A$ cannot be contained in $X\setminus A$, so its containing member of $\mathcal W$ is an ambient representative of a member of $\mathcal U$.

These traces are locally finite in $A$, because the trace on $A$ of a neighbourhood in $X$ meeting only finitely many $V$ meets only the corresponding finitely many traces.

The family in step 2.1 is therefore the locally finite open refinement required for $A$, so $A$ is paracompact.

For every $x\in F$, Hausdorffness gives disjoint open sets $U,V$ with $x\in U$ and $p\in V$; hence $p\notin\overline U$, since $X\setminus V$ is closed and contains $U$. Thus the family of all open $U$ with $U\cap F\ne\varnothing$ and $p\notin\overline U$, together with $X\setminus F$, is an open cover $\mathcal U$ of $X$.

Take a locally finite open cover $\mathcal W$ refining $\mathcal U$, and put $H:=\bigcup\{W\in\mathcal W:W\cap F\ne\varnothing\}$.

The set $H$ is open and contains $F$: a member of $\mathcal W$ containing a point of $F$ cannot refine $X\setminus F$, so it occurs in the defining union.

Every $W$ occurring in $H$ lies in an eligible $U$ of step 1.1, so $p\notin\overline W$; local finiteness and [L1] give $\overline H=\bigcup\overline W$, whence $p\notin\overline H$.

The open sets $X\setminus\overline H$ and $H$ contain $p$ and $F$ respectively and are disjoint. By [F3], $X$ is regular.

For each $x\in E$, regularity supplies an open $U$ containing $x$ with $\overline U\cap F=\varnothing$; therefore the family of all such $U$, together with $X\setminus E$, is an open cover $\mathcal U$ of $X$.

Take a locally finite open cover $\mathcal W$ refining $\mathcal U$, and set $H:=\bigcup\{W\in\mathcal W:W\cap E\ne\varnothing\}$.

The open set $H$ contains $E$, because a member of $\mathcal W$ containing a point of $E$ cannot lie inside $X\setminus E$.

Every member $W$ used in $H$ lies in one of the eligible $U$ and so has $\overline W\cap F=\varnothing$; hence $\overline H=\bigcup\overline W$ is disjoint from $F$ by [L2].

The open sets $H$ and $X\setminus\overline H$ contain $E$ and $F$ respectively and are disjoint, so [F2] proves normality.

Addition is continuous at $(a,b)$ because $|s-a|,|t-b|<\varepsilon/2$ gives $|(s+t)-(a+b)|<\varepsilon$. Multiplication is continuous there: after requiring $|t-b|<1$, one has $|st-ab|\le |s-a||t|+|a||t-b| < |s-a|(|b|+1)+|a||t-b|,$ which is less than $\varepsilon$ when both coordinate errors are smaller than $\varepsilon/(2(|a|+|b|+1))$. These coordinate conditions describe product neighbourhoods, so both operations are continuous.

The reverse triangle inequality $\big||s|-|t|\big|\le |s-t|$ makes absolute value continuous. Consequently $\max\{s,t\}=\frac{s+t+|s-t|}{2},\qquad \min\{s,t\}=\frac{s+t-|s-t|}{2}$ are continuous by step 1.1 and composition.

Reciprocal is continuous at $b\ne0$: if $|t-b|<|b|/2$, then $|t|>|b|/2$ and $\left|\frac1t-\frac1b\right| =\frac{|t-b|}{|t||b|}<\frac{2|t-b|}{|b|^2}.$ Thus division $(s,t)\mapsto s/t$ is the product of $s$ and $1/t$ and is continuous on $\mathbb R\times(\mathbb R\setminus\{0\})$; moreover $\operatorname{coz}(g)$ is open by [F1].

The map $(f,g):X\to\mathbb R^2$ is continuous by [L1], so composing it with the operations of steps 1.1 and 2.1 gives continuity of $f+g$, $fg$, $\max(f,g)$, and $\min(f,g)$; composing $f$ with absolute value gives continuity of $|f|$.

Restricting $f$ and $g$ to $\operatorname{coz}(g)$ and composing their product map with division gives continuity of $f/g$ there.

Iterating the binary operations of step 3.1 proves the finite assertions.

Fix $x\in X$ and a neighbourhood $N$ meeting only $\operatorname{coz}(f_{s_1}),\ldots,\operatorname{coz}(f_{s_n})$; every $f_s$ with $s\notin\{s_1,\ldots,s_n\}$ vanishes on $N$.

Thus at every point of $N$ the displayed pointwise sum equals the finite sum $f_{s_1}+\cdots+f_{s_n}$, so it is well defined and agrees on $N$ with a continuous function.

Since every point has such a neighbourhood $N$, the pointwise sum is continuous on $X$ and is nonnegative.

By [L1] the function $f$ is continuous, and the positivity hypothesis makes $\operatorname{coz}(f)=X$.

Therefore each $\varphi_s=f_s/f$ is continuous by [L2] and nonnegative. Since $f_s(x)\le f(x)$, it takes values in $[0,1]$, and positivity of $f$ gives $\operatorname{coz}(\varphi_s)=\operatorname{coz}(f_s)$.

At every $x\in X$, local finiteness makes the sum finite and gives $\sum_s\varphi_s(x)=\sum_sf_s(x)/f(x)=f(x)/f(x)=1$.

The cozero family is unchanged, hence locally finite, and equality of cozero sets also gives equality of supports. Thus [F1] says that $\{\varphi_s\}$ is a partition of unity.

We first prove a one-shrink construction for any open cover $\mathcal C$. Let $\mathcal R$ be the family of all open $R$ for which $\overline R\subseteq C$ for some $C\in\mathcal C$. By [L1] and [L3], $\mathcal R$ covers $X$. Take a locally finite open refining cover $\mathcal A$ of $\mathcal R$ by [F1], discard its empty members, and use [A1] to assign to each $A\in\mathcal A$ sets $R(A)\in\mathcal R$ and $C(A)\in\mathcal C$ with $A\subseteq R(A)\subseteq\overline{R(A)}\subseteq C(A).$ Then $\overline A\subseteq\overline{R(A)}\subseteq C(A)$.

Apply step 1.1 to $\mathcal U$. This gives a locally finite open cover $\{W_s\}_{s\in S}$ and assigned $U_s\in\mathcal U$ such that $\overline{W_s}\subseteq U_s$.

Apply step 1.1 again, now to the cover $\{W_s:s\in S\}$. Obtain a locally finite open cover $\{A_t\}_{t\in T}$ and a map $t\mapsto s(t)$ such that $\overline{A_t}\subseteq W_{s(t)}$. For $s\in S$ put $V_s:=\bigcup\{A_t:s(t)=s\}.$ The family $\{V_s\}_{s\in S}$ is an open cover. It is locally finite because any neighbourhood meeting only finitely many $A_t$ meets only the corresponding finitely many grouped unions $V_s$.

Each subfamily $\{A_t:s(t)=s\}$ is locally finite, so [L2] gives $\overline{V_s} =\bigcup_{s(t)=s}\overline{A_t}\subseteq W_s.$ Together with step 2.1 this yields $\overline{V_s}\subseteq W_s\subseteq\overline{W_s}\subseteq U_s$ for every $s$, with both displayed families locally finite open covers.

Apply [L1] to obtain $V_s,W_s,U_s$ as stated.

By [L2], $X$ is normal. For each $s$, the closed sets $\overline{V_s}$ and $X\setminus W_s$ are disjoint, so [L3] gives a continuous $f_s:X\to[0,1]$ equal to $1$ on $\overline{V_s}$ and $0$ on $X\setminus W_s$.

The cozero set of $f_s$ lies in $W_s$, while its support lies in $\overline{W_s}\subseteq U_s$; since $\{W_s\}$ is locally finite, so is the cozero family.

Because $\{V_s\}$ covers $X$ and $f_s=1$ on $V_s$, the pointwise sum $\sum_sf_s$ is positive everywhere.

By [L4], the normalized functions $\varphi_s=f_s/(\sum_tf_t)$ form a locally finite partition of unity; their supports equal those of $f_s$, so step 3.1 makes the partition subordinate to $\mathcal U$.

If $X$ is paracompact, [L1] supplies the asserted partition for $\mathcal U$.

Conversely, suppose every open cover admits such a partition. For the partition subordinate to $\mathcal U$, the cozero sets cover $X$ because their functions sum to one.

Each cozero set is open, locally finite among the cozero family, and contained in its support and hence in a member of $\mathcal U$; it is therefore a locally finite open refinement of $\mathcal U$.

By [F2], step 1.3 proves that $X$ is paracompact, completing the equivalence.

Compactness gives a finite subcover $\mathcal U_0=\{U_1,\ldots,U_n\}$ of $\mathcal U$.

By [L1] and [L2], apply the partition theorem to the finite cover $\mathcal U_0$ and take a locally finite partition $\{\varphi_s\}_{s\in S}$ subordinate to it.

Assign each $\varphi_s$ to the first $U_j$ containing its support, and set $h_j$ equal to the corresponding sum. By [L3] the $h_j$ are continuous; by [L4] their supports are contained in $U_j$; and $h_1+\cdots+h_n=1$.

Discarding the zero $h_j$ leaves a finite subordinate partition of unity.

Well order $\{C_\alpha\}$ by [A1], and write $R(x,n)=B(x,2^{-n})$. A ball $R(z,n+1)$ is chosen for $C_\alpha$ when $n$ is the least natural number with $R(z,n)\subseteq C_\alpha$ and, in addition, $R(z,n)\subseteq C_\beta$ for some $\beta<\alpha$. Let $\mathcal G_\alpha$ be the union of all balls chosen for $C_\alpha$.

Put $C'_\alpha:=C_\alpha\setminus\overline{\mathcal G_\alpha}$. Each $C'_\alpha$ is open and refines $C_\alpha$.

The $C'_\alpha$ cover. Otherwise let $C_\alpha$ be the first original member containing an omitted point $x$. Then $x\in\overline{\mathcal G_\alpha}$. By [L1], choose $N$ with $B(x,3\cdot2^{-N})\subseteq C_\alpha$, and put $\delta=2^{-(N+2)}$. Some chosen ball $R(z,n_z+1)$ meets $B(x,\delta)$; write its radius as $r=2^{-(n_z+1)}$. If $r>\delta$, then $d(x,z)<r+\delta<2r$, so its expanded ball $R(z,n_z)$ contains $x$. If $r\le\delta$, then $d(x,z)<r+\delta\le2\delta<2^{-N}$, so $R(z,N)\subseteq C_\alpha$ and minimality gives $n_z\le N$; hence $r\ge2^{-(N+1)}=2\delta$, a contradiction. Thus in every case an expanded chosen ball contains $x$. That expanded ball lies in some $C_\beta$ with $\beta<\alpha$, contradicting the choice of $\alpha$.

If $x\in C'_\alpha$ and $n$ is least with $R(x,n)\subseteq C_\alpha$ (which exists by [L1]), then $C_\alpha$ is the first cover member containing $R(x,n)$: otherwise $R(x,n+1)$ would be chosen for $C_\alpha$ and would contain $x$, contrary to $x\notin\overline{\mathcal G_\alpha}$. For each $n$ there is at most one such first member, and as $n$ increases their ordinal indices are nonincreasing. Infinitely many distinct indices would therefore give an infinite strictly descending sequence of ordinals, impossible because its range has a least member. Thus only finitely many $C'_\alpha$ contain $x$.

Thus $\{C'_\alpha\}$ is the point-finite open refinement required by [F2].

Well order the point-finite cover. For $x\in X$ let $\rho_x=\sup\{r>0:B(x,r)\subseteq C_\alpha \text{ for some }\alpha\}.$ Put $m_x=\min\{1,\rho_x/4\}$ when $\rho_x<\infty$, and $m_x=1$ otherwise. Then $0<m_x\le1$ and $B(x,2m_x)$ lies in some cover member: its radius is strictly below $\rho_x$. Assign $x$ to the first $C_\alpha$ containing $B(x,2m_x)$, and let $C'_\alpha$ be the union of all $B(x,m_x)$ assigned to $\alpha$.

The selected smaller balls cover $X$ and each lies in its assigned $C_\alpha$, so $\{C'_\alpha\}$ is an open refining cover.

Fix $x$. If $C'_\alpha$ meets $B(x,m_x/8)$, choose a ball $B(y,m_y)\subseteq C'_\alpha$ meeting it. We claim $x\in C_\alpha$. Otherwise $x\notin B(y,2m_y)$, while intersection gives $d(x,y)<m_y+m_x/8$; hence $m_y<m_x/8\le1/8$. Thus the truncation in step 1.1 is inactive at $y$ and $\rho_y=4m_y$. But $B(y,5m_y)\subseteq B(x,2m_x),$ because $d(x,y)+5m_y<6m_y+m_x/8<7m_x/8$; the right-hand ball lies in some cover member by step 1.1. This contradicts the definition of $\rho_y$. Thus $x\in C_\alpha$.

The input cover is point-finite, so $x$ belongs to only finitely many $C_\alpha$. Step 2.2 shows that $B(x,m_x/8)$ meets only the corresponding finitely many $C'_\alpha$; hence the new cover is locally finite.

Hence $\{C'_\alpha\}$ is a locally finite open refinement of the point-finite cover. For an arbitrary metric open cover, first apply [L1] and then this construction; refinement is transitive, so the result refines the original cover.

Apply [L1] to the arbitrary cover $\mathcal U$.

The resulting locally finite open refinement is exactly the condition in [F1], so $X$ is paracompact.

By [L1] and [L2], $X$ is paracompact and Hausdorff.

Applying [L3] to the given cover yields the required locally finite subordinate partition of unity.

Let $\mathcal U$ be an arbitrary open cover, and take a locally finite open refining cover $\mathcal V$ by [F2].

Suppose $\mathcal V$ were infinite. For every $n$ the family of $(n+1)$-element subsets of $\mathcal V$ is nonempty; [A1] chooses one $E_n$. Then $E=\bigcup_nE_n$ is at most countable by [L2] and infinite because it has finite subsets of arbitrarily large size. Hence, by the definition of at-most-countable, $E$ is countably infinite; enumerate its distinct members as $(V_n)_{n\in\mathbb N}$.

By [A1] choose $d_n\in V_n$ for every $n$. The range $D=\{d_n:n\in\mathbb N\}$ is infinite: if it were finite, local finiteness would make only finitely many distinct $V_n$ meet $D$, but every $V_n$ contains $d_n\in D$. Moreover the singleton family $\{\{d\}:d\in D\}$ is locally finite, since a neighbourhood meeting only finitely many $V_n$ can contain only finitely many points $d\in D$.

Hausdorffness makes points closed, so [L1] makes $D$ closed. For each $d\in D$, local finiteness gives a neighbourhood meeting only finitely many points of $D$; pass to an open subneighbourhood and remove those finitely many other closed points. Using [A1] along an enumeration of $D$ yields open sets $O_d$ with $O_d\cap D=\{d\}$.

The open set $X\setminus D$, together with the at most countable family $\{O_d:d\in D\}$, is an open cover with no finite subcover, contradicting countable compactness in [F1].

Hence $\mathcal V$ is finite. By [L3], select for each member of this finite refining family one containing member of $\mathcal U$; the selected members form a finite subcover of $\mathcal U$.

Since $\mathcal U$ was arbitrary, [F1] proves that $X$ is compact.

The family of all open $V$ for which $\overline V\subseteq U$ for some $U\in\mathcal U$ covers $X$ by [L1]; by Lindelöfness take a sequence $V_0,V_1,\ldots$ covering $X$.

By [A1], choose $U_n\in\mathcal U$ with $\overline{V_n}\subseteq U_n$ for each $n$.

Put $W_n:=U_n\setminus\bigcup_{i<n}\overline{V_i}$. Each $W_n$ is open and lies in $U_n$.

The $W_n$ cover $X$: if $n$ is the least index with $x\in U_n$, then $x\notin\overline{V_i}$ for $i<n$, since $\overline{V_i}\subseteq U_i$, and hence $x\in W_n$.