Proof of Continuous Partition of Unity Subordinate to a Finite Open Cover of a Compact Set in a Metric Space

Step 0 (finite maxima). The order of the ordered field $\mathbb{R}$ is a total order, so the maximum of two elements is defined on $\mathbb{R}$. For a natural number $r$ and a map $s:[r]\to\mathbb{R}$ with values $s_k$, define $\max_{k\in[r]}s_k$ by recursion on $r$: put $\max_{k\in[1]}s_k=s_1$, and, whenever $m+1\le r$,

$$\max_{k\in[m+1]}s_k=\max\Bigl\{\max_{k\in[m]}s_k,\ s_{m+1}\Bigr\},$$

the inner maximum being formed from the restriction of $s$ to $[m]$. An induction on $r$, using that $\max\{a,b\}$ is $a$ or $b$ and is an upper bound for both, gives:

(a) $s_k\le\max_{k'\in[r]}s_{k'}$ for every $k\in[r]$;

(b) there is $j\in[r]$ with $\max_{k\in[r]}s_k=s_j$.

Step 1 (auxiliary functions). For a nonempty subset $A\subseteq X$ and $z\in X$ write $\operatorname{dist}_d(z,A)$ for the distance from $z$ to $A$ in $(X,d)$. For $i\in[n]$ define $\varphi_i:X\to\mathbb{R}$ as follows. If $U_i=X$, put $\varphi_i(x)=1$ for every $x\in X$. If $U_i\ne X$, then $X\setminus U_i$ is nonempty, and we put

$$\varphi_i(x)=\min\bigl\{1,\ \operatorname{dist}_d(x,X\setminus U_i)\bigr\},$$

using the minimum of two elements. We record four properties, valid for every $i\in[n]$.

(i) $0\le\varphi_i(x)\le 1$ for every $x\in X$. In the first case this is immediate. In the second case $0\le\operatorname{dist}_d(x,X\setminus U_i)$ by claim 1 of The Distance to a Set is Nonexpansive, and the minimum of $1$ and a nonnegative real number is nonnegative and at most $1$.

(ii) $\varphi_i$ is continuous on $X$ as a map from $(X,d)$ to $(\mathbb{R},d_{\mathbb{R}})$. By claim 1 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, applied at each point of $X$ with $A=X$, every constant real-valued map on $X$ is continuous on $X$; we use this repeatedly below. In the first case $\varphi_i$ is such a constant map. In the second case the map $z\mapsto\operatorname{dist}_d(z,X\setminus U_i)$ is continuous on $X$ by claim 5 of The Distance to a Set is Nonexpansive, the constant map with value $1$ is continuous on $X$, and the minimum of two real-valued maps continuous on $X$ is continuous on $X$ by claim 4 of Continuity of the Absolute Value, Maximum and Minimum of Real-Valued Functions on a Metric Space.

(iii) If $\varphi_i(x)\ne 0$ then $x\in U_i$. In the first case $U_i=X$ and there is nothing to prove. In the second case suppose $x\notin U_i$, so $x\in X\setminus U_i$; then $\operatorname{dist}_d(x,X\setminus U_i)\le d(x,x)$ by claim 2 of The Distance to a Set is Nonexpansive, and $d(x,x)=0$ by condition 2 of the definition of a metric, so $\varphi_i(x)\le 0$, which together with (i) gives $\varphi_i(x)=0$.

(iv) If $x\in U_i$ then $0<\varphi_i(x)$. In the first case $\varphi_i(x)=1$. In the second case $U_i\in\mathcal{T}_d$, so by the definition of an open subset of a metric space there is a real number $r>0$ with $B_d(x,r)\subseteq U_i$, where $B_d(x,r)$ is the open ball with center $x$ and radius $r$. Every $y\in X\setminus U_i$ then satisfies $y\notin B_d(x,r)$, that is $r\le d(x,y)$. Hence $r$ is a lower bound for the set of real numbers of the form $d(x,y)$ with $y\in X\setminus U_i$, whose greatest lower bound is $\operatorname{dist}_d(x,X\setminus U_i)$ by the definition of the distance to a set, so $r\le\operatorname{dist}_d(x,X\setminus U_i)$ and therefore $0<\min\{1,r\}\le\varphi_i(x)$.

Step 2 (the degenerate case $C=\varnothing$). Suppose $C$ is empty. For every $i\in[n]$ put $h_i(x)=0$ for all $x\in X$ and $D_i=\varnothing$. Then $h_i$ is a constant map, hence continuous on $X$ as noted in Step 1(ii), and its only value $0$ satisfies $0\le 0\le 1$, so property 1 holds. The set $\varnothing$ is closed in $(X,\mathcal{T}_d)$ because its complement $X$ lies in $\mathcal{T}_d$, and $\varnothing\subseteq U_i$, and $h_i$ vanishes identically; so property 2 holds. Every term of the family $(h_i(x))_{i\in[n]}$ equals $0$, so claim 7 of Properties of Finite Sums, applied with the index $1$, gives $\sum_{i=1}^{n}h_i(x)=h_1(x)=0\le 1$, and property 3 holds. Property 4 holds vacuously. From now on assume that $C$ is nonempty.

Step 3 (a uniform positive lower bound on $C$). Define $\Phi:X\to\mathbb{R}$ by $\Phi(x)=\max_{i\in[n]}\varphi_i(x)$, formed as in Step 0. By (ii), claim 4 of Continuity of the Absolute Value, Maximum and Minimum of Real-Valued Functions on a Metric Space and induction along the recursion of Step 0, $\Phi$ is continuous on $X$.

Let $x\in C$. Since $C\subseteq\bigcup_{i\in[n]}U_i$ there is $i\in[n]$ with $x\in U_i$, so $0<\varphi_i(x)\le\Phi(x)$ by (iv) and Step 0(a).

The restriction of $\Phi$ to $C$ satisfies the continuity hypothesis of Extreme Value Theorem on a Compact Subset of a Metric Space: given $x\in C$ and a real number $\varepsilon>0$, continuity of $\Phi$ at $x$ relative to $X$ furnishes a real number $\delta>0$ such that every $y\in X$ with $d(x,y)<\delta$ satisfies $|\Phi(y)-\Phi(x)|<\varepsilon$, and in particular every such $y$ lying in $C$ does. Since $C$ is nonempty and compact in $(X,\mathcal{T}_d)$, that theorem provides $x_{\min}\in C$ with $\Phi(x_{\min})\le\Phi(x)$ for every $x\in C$. Put $c=\Phi(x_{\min})$. Then $0<c$, because $x_{\min}\in C$, and $c\le\Phi(x)$ for every $x\in C$.

Write $2=1+1$. By claim 8 of Elementary Order Arithmetic in an Ordered Field the inverse $2^{-1}$ exists and, since $0<c$, the number $\gamma=c\cdot 2^{-1}$ satisfies $0<\gamma$, $\gamma<c$ and $\gamma+\gamma=c$.

Step 4 (the cutoffs and their supports). For $i\in[n]$ define $\psi_i:X\to\mathbb{R}$ and $D_i\subseteq X$ by

$$\psi_i(x)=\max\bigl\{0,\ \varphi_i(x)-\gamma\bigr\},\qquad D_i=\{x\in X:\gamma\le\varphi_i(x)\}.$$

The map $x\mapsto\varphi_i(x)-\gamma$ is the sum of $\varphi_i$ and the constant map with value $-\gamma$, hence continuous on $X$ by (ii), by the continuity of constant maps recorded in Step 1(ii) and by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space; so $\psi_i$ is continuous on $X$ by claim 4 of Continuity of the Absolute Value, Maximum and Minimum of Real-Valued Functions on a Metric Space. By (i) and $0<\gamma$ we have $\varphi_i(x)-\gamma<1$, so $0\le\psi_i(x)\le 1$.

By (ii) and claim 2 of Semicontinuity Under Negation and Characterization of Continuity, $\varphi_i$ is upper semicontinuous on $X$; hence, by claim 3 of Semicontinuity via Sublevel and Superlevel Sets applied with the subset $X$ of $X$ itself, the set $D_i$ is closed in the topological space $X$ carrying the topology $\mathcal{T}_d$.

If $x\in D_i$ then $0<\gamma\le\varphi_i(x)$, so $\varphi_i(x)\ne 0$ and $x\in U_i$ by (iii); thus $D_i\subseteq U_i$. If $x\notin D_i$ then $\varphi_i(x)<\gamma$, so $\varphi_i(x)-\gamma<0$ and $\psi_i(x)=0$.

Step 5 (normalization). Define $\Psi:X\to\mathbb{R}$ and $\Theta:X\to\mathbb{R}$ by

$$\Psi(x)=\sum_{i=1}^{n}\psi_i(x),\qquad \Theta(x)=\max\{\Psi(x),\gamma\}.$$

By claim 1 of Properties of Finite Sums, $\Psi$ is obtained from the maps $\psi_i$ by iterated addition, so $\Psi$ is continuous on $X$ by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space and induction on $n$; and $\Theta$ is then continuous on $X$ by the continuity of constant maps recorded in Step 1(ii) and claim 4 of Continuity of the Absolute Value, Maximum and Minimum of Real-Valued Functions on a Metric Space.

Since $0<\gamma\le\Theta(x)$ for every $x\in X$, we have $\Theta(x)\ne 0$, so the map $\Theta^{-1}:X\to\mathbb{R}$ with $\Theta^{-1}(x)=(\Theta(x))^{-1}$ is defined and is continuous on $X$ by claim 2 of Continuity of the Reciprocal of a Nonvanishing Real-Valued Function on a Metric Space; moreover $0<\Theta^{-1}(x)$ by claim 7 of Elementary Order Arithmetic in an Ordered Field. We shall also use that multiplying a non-strict inequality between real numbers by a positive real number preserves it, which follows from claim 10 of Elementary Order Arithmetic in an Ordered Field when the inequality is strict and is trivial when it is an equality. Define

$$h_i(x)=\psi_i(x)\,\Theta^{-1}(x)\qquad(i\in[n],\ x\in X),$$

which is continuous on $X$ by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space.

Property 1. $h_i(x)$ is a product of two nonnegative numbers, so $0\le h_i(x)$. By claims 5 and 6 of Properties of Finite Sums applied to the nonnegative family $(\psi_i(x))_{i\in[n]}$ we get $0\le\Psi(x)$ and $\psi_i(x)\le\Psi(x)$; and $\Psi(x)\le\Theta(x)$ by Step 0(a). Multiplying $\psi_i(x)\le\Theta(x)$ by $\Theta^{-1}(x)>0$ gives $h_i(x)\le\Theta(x)\Theta^{-1}(x)=1$.

Property 2. This was proved in Step 4, together with $h_i(x)=\psi_i(x)\Theta^{-1}(x)=0$ whenever $x\notin D_i$.

Property 3. By claim 3 of Properties of Finite Sums with $\lambda=\Theta^{-1}(x)$,

$$\sum_{i=1}^{n}h_i(x)=\Theta^{-1}(x)\sum_{i=1}^{n}\psi_i(x)=\Theta^{-1}(x)\,\Psi(x),$$

and multiplying $\Psi(x)\le\Theta(x)$ by $\Theta^{-1}(x)>0$ gives $\Theta^{-1}(x)\Psi(x)\le 1$.

Property 4. Let $x\in C$. By Step 0(b) there is $j\in[n]$ with $\Phi(x)=\varphi_j(x)$, and $c\le\Phi(x)$ by Step 3, so $\varphi_j(x)-\gamma\ge c-\gamma=\gamma>0$ and therefore $\psi_j(x)=\varphi_j(x)-\gamma\ge\gamma$. By claim 6 of Properties of Finite Sums, $\gamma\le\psi_j(x)\le\Psi(x)$, so $\Theta(x)=\max\{\Psi(x),\gamma\}=\Psi(x)$ and, by the display in Property 3,

$$\sum_{i=1}^{n}h_i(x)=\Psi(x)^{-1}\Psi(x)=1 .$$

This completes the proof.