Proof of Exhaustion of a Locally Compact Separable Metric Space by Compact Sets, with Cut-Off Functions

Throughout, closed subsets of compact sets are compact, the interior of a set is open and contained in it, and $\operatorname{dist}(x,A)\le d(x,a)$ for $a\in A$; these are background facts of the setting. A set $O$ that is open and contained in a set $A$ lies in $\operatorname{int}A$, since each $x\in O$ has an open ball about it inside $O\subseteq A$ (Open Subset of a Metric Space (i), Interior of a Subset of a Metric Space (ii)).

If $X=\emptyset$, the sets $K_{m}=\emptyset$ and the functions $\varphi_{m}=0$ have all the properties: 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), $0$ is continuous with compact support, and its support $\emptyset$ is compact. Assume from now on that $X\neq\emptyset$.

Step 1: a dense sequence. By Separable Metric Space and Separable Real Normed Space (i) there is a countable set $D\subseteq X$ that is dense. Picking any $x\in X$, the ball $B(x,1)$ meets $D$, so $D\neq\emptyset$ and Countable Set (i) gives a surjective map $e:\mathbb{N}\to D$. Put $x_{k}=e(k)$ for $k\in\mathbb{N}$. Every open ball of $X$ contains some $x_{k}$.

Step 2: thickening radii. Let $\mathcal{K}$ be the set of nonempty compact subsets of $X$. For $L\in\mathcal{K}$ and real $\delta$ put $N_{\delta}(L)=\{y\in X:\operatorname{dist}(y,L)\le\delta\}$, a closed set 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), and let

$$A(L)=\{\delta\in\mathbb{R}:0<\delta\le1\text{ and }N_{\delta}(L)\text{ is compact}\}.$$

By Thickenings, Neighbourhoods with Compact Closure, Urysohn Functions and Partitions of Unity in a Locally Compact Metric Space (i), applied to $L$ and the open set $X$, there is $\delta>0$ with $N_{\delta}(L)$ compact; then $N_{\min\{\delta,1\}}(L)$ is a closed subset of $N_{\delta}(L)$, so $\min\{\delta,1\}\in A(L)$. Thus $A(L)$ is nonempty and bounded above by $1$, and by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vii) we may put

$$\tau(L)=\tfrac12\sup A(L),\qquad0<\tau(L)\le\tfrac12 .$$

If $0<s<2\tau(L)$, some $\delta\in A(L)$ exceeds $s$, and $N_{s}(L)$ is compact as a closed subset of $N_{\delta}(L)$; in particular $N_{\tau(L)}(L)$ is compact. Moreover $L\subseteq O(L)\subseteq N_{\tau(L)}(L)$, where $O(L)=\{y\in X:\operatorname{dist}(y,L)<\tau(L)\}$ 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), because $\operatorname{dist}(y,L)\le d(y,y)=0$ for $y\in L$.

For $x\in X$ the set $\{x\}$ lies in $\mathcal{K}$ (it 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), since one member $U$ of a cover containing $x$ suffices and $\{U\}$ 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 $\operatorname{dist}(y,\{x\})=d(y,x)$, so $N_{s}(\{x\})=\bar{B}(x,s)$ for real $s\ge0$.

Step 3: balls about the dense sequence. Every $x\in X$ lies in $\bar{B}(x_{k},\tau(\{x_{k}\}))$ for some $k\in\mathbb{N}$. Indeed, put $t=\tau(\{x\})$; by Step 1 there is $k$ with $d(x,x_{k})<t/2$. For $y\in\bar{B}(x_{k},t)$ we have $d(x,y)\le d(x,x_{k})+d(x_{k},y)<3t/2$, so $\bar{B}(x_{k},t)$ is a closed subset of $\bar{B}(x,3t/2)$, which is compact by Step 2 as $3t/2<2t$. Since $0<t\le1$, this gives $t\in A(\{x_{k}\})$, so $\tau(\{x_{k}\})\ge t/2$. Hence $d(x_{k},x)<t/2\le\tau(\{x_{k}\})$.

Step 4: the recursion. Let

$$K_{1}=\bar{B}\bigl(x_{1},\tau(\{x_{1}\})\bigr),\qquad K_{m+1}=N_{\tau(K_{m})}(K_{m})\cup\bar{B}\bigl(x_{m+1},\tau(\{x_{m+1}\})\bigr)\quad(m\in\mathbb{N})$$

(by Step 2 and Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (ii), applied to the two-element family $\{N_{\tau(L)}(L),\bar{B}(x_{m+1},\tau(\{x_{m+1}\}))\}$, 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 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 (v), the rule $g(m,L)=N_{\tau(L)}(L)\cup\bar{B}\bigl(x_{m+1},\tau(\{x_{m+1}\})\bigr)$ defines a map $g:\mathbb{N}\times\mathcal{K}\to\mathcal{K}$; as $K_{1}\in\mathcal{K}$ by Step 2, Recursion on the Natural Numbers Starting at One (i) with $a=\mathcal{K}$ and $c=K_{1}$ gives the sequence, and each $K_{m}$ lies in $\mathcal{K}$). For $m\in\mathbb{N}$ write $\tau_{m}=\tau(K_{m})$ and define $\varphi_{m}:X\to\mathbb{R}$ by

$$\varphi_{m}(x)=\max\Bigl\{0,\ 1-\frac{\operatorname{dist}(x,K_{m})}{\tau_{m}}\Bigr\}.$$

Exhaustion. By Step 2, $K_{m}\subseteq O(K_{m})\subseteq N_{\tau_{m}}(K_{m})\subseteq K_{m+1}$ with $O(K_{m})$ open, so $K_{m}\subseteq O(K_{m})\subseteq\operatorname{int}K_{m+1}\subseteq K_{m+1}$. By the recursion, $\bar{B}(x_{k},\tau(\{x_{k}\}))\subseteq K_{k}$ for every $k\in\mathbb{N}$, so by Step 3 every $x\in X$ lies in some $K_{k}$, and $\bigcup_{m\in\mathbb{N}}K_{m}=X$.

Compact sets. The sets $\operatorname{int}K_{m}$, $m\in\mathbb{N}$, are open; they increase, since $\operatorname{int}K_{m}\subseteq K_{m}\subseteq\operatorname{int}K_{m+1}$; and they cover $X$, since each $x\in X$ lies in some $K_{m}\subseteq\operatorname{int}K_{m+1}$. For a compact $C\subseteq X$, Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (iii) gives $m\in\mathbb{N}$ with $C\subseteq\operatorname{int}K_{m}$.

Cut-off functions. Let $m\in\mathbb{N}$. As $\operatorname{dist}(x,K_{m})\ge0$, we have $0\le\varphi_{m}(x)\le1$, and $\varphi_{m}(x)=1$ for $x\in K_{m}$, where the distance vanishes. Since $|\max\{0,s\}-\max\{0,t\}|\le|s-t|$ for real $s,t$, the Lipschitz estimate for the distance gives $|\varphi_{m}(x)-\varphi_{m}(y)|\le d(x,y)/\tau_{m}$, so $\varphi_{m}$ is Lipschitz, hence continuous. If $\varphi_{m}(x)\neq0$, then $\operatorname{dist}(x,K_{m})<\tau_{m}$, so $x\in N_{\tau_{m}}(K_{m})$; this set is closed, so by Support of a Real Function on a Metric Space (i) and Closure of a Subset of a Metric Space (ii)

$$\operatorname{supp}\varphi_{m}=\operatorname{cl}\{x:\varphi_{m}(x)\neq0\}\subseteq N_{\tau_{m}}(K_{m})\subseteq K_{m+1},$$

where we used that the closure of a subset of a closed set $N$ lies in $\operatorname{cl}N=N$. Being closed in the compact set $K_{m+1}$, $\operatorname{supp}\varphi_{m}$ is compact, and $\varphi_{m}\in C_{c}(X)$ by The Space of Continuous Functions with Compact Support and the Uniform Norm (i).