Partitions of Unity and Paracompactness

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

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

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

Let $x\in \overline{{\bigcup }_i{A}_i}$ and take $N$ as in step 1.1; if $x\mathrm{\notin }{\bigcup }_i\overline{A_i}$, 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 $\overline{{\bigcup }_i{A}_i}={\bigcup }_i\overline{A_i}$ by steps 1.2 and 2.2; if every $A_i$ is closed, the right-hand side is ${\bigcup }_i{A}_i$, 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 $p\mathrm{\notin }\bar{U}$, since $X\setminus V$ is closed and contains $U$. Thus the family of all open $U$ with $U\cap F\mathrm{\ne }\mathrm{\varnothing }$ and $p\mathrm{\notin }\bar{U}$, 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 $H:=\bigcup \{W\in \mathcal{W}:W\cap F\mathrm{\ne }\mathrm{\varnothing }\}$.

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 $p\mathrm{\notin }\bar{W}$; local finiteness and [L1] give $\bar{H}=\bigcup \bar{W}$, whence $p\mathrm{\notin }\bar{H}$.

The open sets $X\setminus \bar{H}$ 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 $\bar{U}\cap F=\mathrm{\varnothing }$; 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 $H:=\bigcup \{W\in \mathcal{W}:W\cap E\mathrm{\ne }\mathrm{\varnothing }\}$.

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 $\bar{W}\cap F=\mathrm{\varnothing }$; hence $\bar{H}=\bigcup \bar{W}$ is disjoint from $F$ by [L2].

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

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

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

Reciprocal is continuous at $b\mathrm{\ne }0$: if $\mathrm{\mid }t-b\mathrm{\mid }<\mathrm{\mid }b\mathrm{\mid }\mathrm{/}2$, then $\mathrm{\mid }t\mathrm{\mid }>\mathrm{\mid }b\mathrm{\mid }\mathrm{/}2$ and $\mid \frac{1}{t}-\frac{1}{b}\mid =\frac{\mathrm{\mid }t-b\mathrm{\mid }}{\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }}<\frac{2\mathrm{\mid }t-b\mathrm{\mid }}{\mathrm{\mid }b{\mathrm{\mid }}^2}\mathrm{.}$ Thus division $(s,t)\mapsto s\mathrm{/}t$ is the product of $s$ and $1\mathrm{/}t$ and is continuous on $\mathbb{R}\times (\mathbb{R}\setminus \{0\})$; moreover $\mathrm{coz}(g)$ 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 $\mathrm{\mid }f\mathrm{\mid }$.

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

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

Fix $x\in X$ and a neighbourhood $N$ meeting only $\mathrm{coz}(f_{s_1}),\dots ,\mathrm{coz}(f_{s_n})$; every $f_s$ with $s\mathrm{\notin }\{s_1,\dots ,s_n\}$ 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 $\mathrm{coz}(f)=X$.

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

At every $x\in X$, local finiteness makes the sum finite and gives ${\sum }_s{\phi }_s(x)={\sum }_s{f}_s(x)\mathrm{/}f(x)=f(x)\mathrm{/}f(x)=1$.

The cozero family is unchanged, hence locally finite, and equality of cozero sets also gives equality of supports. Thus [F1] says that $\{{\phi }_s\}$ 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 $\bar{R}\subseteq C$ 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 $A\subseteq R(A)\subseteq \overline{R(A)}\subseteq C(A)\mathrm{.}$ Then $\bar{A}\subseteq \overline{R(A)}\subseteq C(A)$.

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 $V_s:=\bigcup \{A_t:s(t)=s\}\mathrm{.}$ 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 $\overline{V_s}={\bigcup }_{s(t)=s}\overline{A_t}\subseteq W_s\mathrm{.}$ 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 ${\sum }_s{f}_s$ is positive everywhere.

By [L4], the normalized functions ${\phi }_s=f_s\mathrm{/}({\sum }_t{f}_t)$ 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 ${\mathcal{U}}_0=\{U_1,\dots ,U_n\}$ of $\mathcal{U}$.

By [L1] and [L2], apply the partition theorem to the finite cover ${\mathcal{U}}_0$ and take a locally finite partition $\{{\phi }_s{\}}_{s\in S}$ subordinate to it.

Assign each ${\phi }_s$ 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 $C_{\alpha }^{\mathrm{\prime }}:=C_{\alpha }\setminus \overline{{\mathcal{G}}_{\alpha }}$. Each $C_{\alpha }^{\mathrm{\prime }}$ is open and refines $C_{\alpha }$.

The $C_{\alpha }^{\mathrm{\prime }}$ 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 $B(x,3\cdot 2^{-N})\subseteq C_{\alpha }$, 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 $d(x,z)<r+\delta \le 2\delta <2^{-N}$, so $R(z,N)\subseteq C_{\alpha }$ and minimality gives $n_z\le N$; hence $r\ge 2^{-(N+1)}=2\delta$, 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 $x\in C_{\alpha }^{\mathrm{\prime }}$ 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 $x\mathrm{\notin }\overline{{\mathcal{G}}_{\alpha }}$. 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 $C_{\alpha }^{\mathrm{\prime }}$ contain $x$.

Thus $\{C_{\alpha }^{\mathrm{\prime }}\}$ is the point-finite open refinement required by [F2].

Well order the point-finite cover. For $x\in X$ let ${\rho }_x=\sup \{r>0:B(x,r)\subseteq C_{\alpha }\text{ for some }\alpha \}\mathrm{.}$ Put $m_x=\min \{1,{\rho }_x\mathrm{/}4\}$ when ${\rho }_x<\mathrm{\infty }$, and $m_x=1$ otherwise. Then $0<m_x\le 1$ 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 $C_{\alpha }^{\mathrm{\prime }}$ 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 $\{C_{\alpha }^{\mathrm{\prime }}\}$ is an open refining cover.

Fix $x$. If $C_{\alpha }^{\mathrm{\prime }}$ meets $B(x,m_x\mathrm{/}8)$, choose a ball $B(y,m_y)\subseteq C_{\alpha }^{\mathrm{\prime }}$ meeting it. We claim $x\in C_{\alpha }$. Otherwise $x\mathrm{\notin }B(y,2m_y)$, while intersection gives $d(x,y)<m_y+m_x\mathrm{/}8$; hence $m_y<m_x\mathrm{/}8\le 1\mathrm{/}8$. 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 $d(x,y)+5m_y<6m_y+m_x\mathrm{/}8<7m_x\mathrm{/}8$; 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 $B(x,m_x\mathrm{/}8)$ meets only the corresponding finitely many $C_{\alpha }^{\mathrm{\prime }}$; hence the new cover is locally finite.

Hence $\{C_{\alpha }^{\mathrm{\prime }}\}$ 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 $E={\bigcup }_n{E}_n$ 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 $\bar{V}\subseteq U$ for some $U\in \mathcal{U}$ covers $X$ by [L1]; by Lindelöfness take a sequence $V_0,V_1,\dots$ 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 $x\mathrm{\notin }\overline{V_i}$ for $i<n$, since $\overline{V_i}\subseteq U_i$, and hence $x\in W_n$.

The cover is locally finite: for $x\in V_k$, the neighbourhood $V_k$ is disjoint from $W_n$ for every $n>k$, while it can meet only $W_0,\dots ,W_k$.

Thus $\{W_n\}$ is a locally finite open refinement of $\mathcal{U}$, and [F1] proves paracompactness.

The displayed family is a topology, and $X$ is compact because every open cover of this finite set already has a finite subcover.

Its closed sets include $\{a\}$ and $\{b\}$, while every open set containing $a$ contains $c$ and every open set containing $b$ contains $c$.

Thus the disjoint closed sets $\{a\}$ and $\{b\}$ have no disjoint open neighbourhoods, so $X$ is not normal by [F2].

By [F1] the compact space $X$ is paracompact, and step 3.1 refutes the displayed assertion.

By [A1] and [L1], ${\omega }_1+1$ is compact, hence paracompact by [L3], and its initial segment ${\omega }_1$ is countably compact but noncompact.

The initial segment ${\omega }_1$ is open in ${\omega }_1+1$, since its complement is the closed singleton consisting of the top endpoint.

If ${\omega }_1$ were paracompact, its Hausdorffness from [L4] would let [L2] make it compact, contradicting step 1.1.

Thus a paracompact space has the nonparacompact subspace ${\omega }_1$, which refutes the displayed hereditary assertion.