Proof of Thickenings, Neighbourhoods with Compact Closure, Urysohn Functions and Partitions of Unity in a Locally Compact Metric Space

Throughout, closed subsets of compact sets are compact, compact sets are closed, the closure of a set is closed and contains it, a closed set is its own closure, and $\operatorname{dist}(x,A)\le d(x,a)$ for $a\in A$, with $d(x,a)<r$ for some $a\in A$ whenever $r>\operatorname{dist}(x,A)$; these are background facts of the setting. If $A\subseteq C\subseteq X$, every open ball meeting $A$ meets $C$, so $\operatorname{cl}A\subseteq\operatorname{cl}C$ by Closure of a Subset of a Metric Space (ii); in particular $\operatorname{cl}A\subseteq C$ when $C$ is closed.

Compact thickening. Let $K\neq\emptyset$. Let $\mathcal{U}$ be the set of open balls $B(x,r)$ with $x\in K$ and real $r>0$ such that $\bar{B}(x,4r)$ is compact and $\bar{B}(x,4r)\subseteq V$.

First, $\mathcal{U}$ covers $K$. Let $x\in K$. By Locally Compact Metric Space (i) there is $r_{0}>0$ with $\bar{B}(x,r_{0})$ compact, and since $V$ is open there is $r_{1}>0$ with $B(x,r_{1})\subseteq V$. Put $r=\min\{r_{0},r_{1}\}/8$. The closed ball $\bar{B}(x,4r)$ is closed by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (iv) and lies in $\bar{B}(x,r_{0})$, so it is compact; and it lies in $B(x,r_{1})\subseteq V$ because $4r<r_{1}$. So $x\in B(x,r)\in\mathcal{U}$.

Open balls are open, so by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (i) some finite $\mathcal{V}\subseteq\mathcal{U}$ covers $K$, and $\mathcal{V}\neq\emptyset$ because $K\neq\emptyset$. For each $U\in\mathcal{V}$ the set of pairs $(x,r)$ as in the definition of $\mathcal{U}$ with $U=B(x,r)$ is nonempty, so by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (xi) we may fix such a pair $(x_{U},r_{U})$ for every $U\in\mathcal{V}$. Let $\delta>0$ be the least element of the nonempty finite set $\{r_{U}:U\in\mathcal{V}\}$, given by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vi) and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (viii), and let

$$L=\bigcup_{U\in\mathcal{V}}\bar{B}(x_{U},4r_{U}),$$

the union of a finite set of compact subsets of $V$ (finite, as the image of $\mathcal{V}$ under $U\mapsto\bar{B}(x_{U},4r_{U})$, by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vi)); $L$ is compact by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (ii), and $L\subseteq V$.

Let $N=\{y\in X:\operatorname{dist}(y,K)\le\delta\}$ and $y\in N$. As $\operatorname{dist}(y,K)<2\delta$, there is $a\in K$ with $d(y,a)<2\delta$, and $a\in U$ for some $U\in\mathcal{V}$, that is $d(x_{U},a)<r_{U}$. Hence

$$d(x_{U},y)\le d(x_{U},a)+d(a,y)<r_{U}+2\delta\le3r_{U}<4r_{U},$$

so $y\in L$. Thus $N\subseteq L\subseteq V$. As $N$ is closed by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (vi), it is compact.

Compact neighbourhood. If $K=\emptyset$, take $W=\emptyset$: no open ball meets $\emptyset$, so $\operatorname{cl}W=\emptyset$, which is compact by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (i) (the empty subset of any cover is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iii)) and contained in $V$. If $K\neq\emptyset$, let $\delta$ and $N$ be as in (i) and put $W=\{x\in X:\operatorname{dist}(x,K)<\delta\}$, which is open by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (v). For $x\in K$ we have $\operatorname{dist}(x,K)\le d(x,x)=0<\delta$, so $K\subseteq W$. Since $W\subseteq N$ and $N$ is closed, $\operatorname{cl}W\subseteq N\subseteq V$, and $\operatorname{cl}W$ is compact as a closed subset of the compact set $N$.

Urysohn function. If $K=\emptyset$, the constant function $0$ is continuous with empty, hence compact, support by Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (i), and $\emptyset\subseteq V$; so $f=0$ lies in $C_{c}(X)$ by The Space of Continuous Functions with Compact Support and the Uniform Norm (i) and has the required properties. If $K\neq\emptyset$, let $\delta$ and $N$ be as in (i) and define $f:X\to\mathbb{R}$ by

$$f(x)=\max\Bigl\{0,\ 1-\frac{\operatorname{dist}(x,K)}{\delta}\Bigr\}.$$

Since $\operatorname{dist}(x,K)\ge0$, we have $0\le f(x)\le1$, and $f(x)=1$ for $x\in K$, where $\operatorname{dist}(x,K)=0$. For real $s,t$ one has $|\max\{0,s\}-\max\{0,t\}|\le|s-t|$, so with the Lipschitz estimate for the distance,

$$|f(x)-f(y)|\le\frac{|\operatorname{dist}(x,K)-\operatorname{dist}(y,K)|}{\delta}\le\frac{d(x,y)}{\delta}\qquad(x,y\in X).$$

Thus $f$ is Lipschitz, hence continuous. If $f(x)\neq0$, then $\operatorname{dist}(x,K)<\delta$, so $x\in N$; as $N$ is closed, $\operatorname{supp}f=\operatorname{cl}\{x:f(x)\neq0\}\subseteq N\subseteq V$ by Support of a Real Function on a Metric Space (i), and $\operatorname{supp}f$ is compact as a closed subset of $N$. Hence $f\in C_{c}(X)$ by The Space of Continuous Functions with Compact Support and the Uniform Norm (i).

Partition of unity. Step 1: compact pieces. For $i\in[n]$ define $e_{i}:X\to\mathbb{R}$ by $e_{i}(x)=\operatorname{dist}(x,X\setminus V_{i})$ if $V_{i}\neq X$, and $e_{i}(x)=1$ if $V_{i}=X$. Then $e_{i}(x)>0$ if and only if $x\in V_{i}$: if $V_{i}\neq X$ and $x\in V_{i}$, then $x$ lies outside the closed set $X\setminus V_{i}$, so $\operatorname{dist}(x,X\setminus V_{i})>0$ (positivity of the distance off a closed set); and if $x\notin V_{i}$, then $\operatorname{dist}(x,X\setminus V_{i})\le d(x,x)=0$. For real $c$, the set $\{x:e_{i}(x)>c\}$ is open and $\{x:e_{i}(x)\ge c\}$ is closed: for $V_{i}\neq X$ they are the complements of the closed set $\{x:e_{i}(x)\le c\}$ and of the open set $\{x:e_{i}(x)<c\}$ of Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (vi) and Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (v), and for $V_{i}=X$ each is $X$ or $\emptyset$.

For $M\in\mathbb{N}$ let $U_{M}=\bigcup_{i=1}^{n}\{x\in X:e_{i}(x)>1/M\}$, an open set as a union of open sets, with $U_{M}\subseteq U_{M+1}$ because $1/(M+1)<1/M$. Every $x\in K$ lies in some $V_{i}$, so $e_{i}(x)>0$ and, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (ix), $e_{i}(x)>1/M$ for some $M$; thus the sets $U_{M}$ cover $K$. By Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (iii) there is $M\in\mathbb{N}$ with $K\subseteq U_{M}$. For $i\in[n]$ put

$$K_{i}=\{x\in K:e_{i}(x)\ge1/M\}=K\cap\{x\in X:e_{i}(x)\ge1/M\}.$$

Each $K_{i}$ is closed, as the intersection of two closed sets, and lies in $K$, so it is compact; $K_{i}\subseteq V_{i}$ since $e_{i}>0$ on $K_{i}$; and $K\subseteq K_{1}\cup\dots\cup K_{n}$ since $K\subseteq U_{M}$.

Step 2: Urysohn functions. For $i\in[n]$, the set of $g\in C_{c}(X)$ with $0\le g\le1$, $g=1$ on $K_{i}$ and $\operatorname{supp}g\subseteq V_{i}$ is nonempty by (iii), applied to the compact set $K_{i}$ and the open set $V_{i}\supseteq K_{i}$. By Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (xi) there are such functions $g_{1},\dots,g_{n}$, $g_{i}$ belonging to the $i$-th set.

Step 3: the partition. For $i\in[n+1]$ and $x\in X$ put

$$P_{i}(x)=\prod_{j=1}^{i-1}\bigl(1-g_{j}(x)\bigr),\qquad\text{so }P_{1}=1\text{ and }P_{i+1}=P_{i}\,(1-g_{i}),$$

and for $i\in[n]$ put $h_{i}=P_{i}\,g_{i}$. Since $0\le1-g_{j}\le1$, induction on $i$ gives $0\le P_{i}\le1$, hence $0\le h_{i}\le1$; and since constants, sums, real multiples and products of continuous real functions are continuous, induction on $i$ gives that $1-g_{i}$, $P_{i}$ and $h_{i}$ are continuous. If $h_{i}(x)\neq0$ then $g_{i}(x)\neq0$, so $\operatorname{supp}h_{i}\subseteq\operatorname{supp}g_{i}\subseteq V_{i}$, and $\operatorname{supp}h_{i}$ is compact as a closed subset of the compact set $\operatorname{supp}g_{i}$. Hence $h_{i}\in C_{c}(X)$.

From $P_{i+1}=P_{i}-P_{i}g_{i}$ we get $h_{i}=P_{i}-P_{i+1}$, and the sum telescopes:

$$\sum_{i=1}^{n}h_{i}(x)=P_{1}(x)-P_{n+1}(x)=1-\prod_{j=1}^{n}\bigl(1-g_{j}(x)\bigr)\qquad(x\in X).$$

As $P_{n+1}\ge0$, the sum is at most $1$ everywhere. If $x\in K$, then $x\in K_{j}$ for some $j\in[n]$, so $g_{j}(x)=1$, the factor $1-g_{j}(x)$ vanishes, $P_{n+1}(x)=0$, and the sum equals $1$.