Proof of The Space of Continuous Functions with Compact Support on a Locally Compact Separable Metric Space is Separable

Throughout, for $f,g\in C_{c}(X)$ the difference $f-g=f+(-1)g$ is the pointwise difference, by The Continuous Functions with Compact Support Form a Real Normed Space under the Uniform Norm (i) and The Space of Continuous Functions with Compact Support and the Uniform Norm (ii); the norm metric of the real normed space $(C_{c}(X),\|\cdot\|_{u})$ of The Continuous Functions with Compact Support Form a Real Normed Space under the Uniform Norm (ii) is $d_{u}(f,g)=\|f-g\|_{u}$ by The Norm Metric of a Real Normed Space (i); and a function $h\in C_{c}(X)$ with $|h(x)|\le\varepsilon$ for every $x\in X$ satisfies $\|h\|_{u}\le\varepsilon$, by The Continuous Functions with Compact Support Form a Real Normed Space under the Uniform Norm (iii). Sums and products of continuous real functions are continuous by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (v), and Lipschitz functions are continuous by A Lipschitz Map is Uniformly Continuous and Continuous (ii). Countable sets are as in Countable Set (i).

If $X=\emptyset$, the only function $X\to\mathbb{R}$ is the empty function, so $C_{c}(X)$ has the single element $0$; the set $\{0\}$ 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), hence countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (iii), and it meets every open ball of $C_{c}(X)$, so $C_{c}(X)$ is separable. Assume from now on that $X\neq\emptyset$.

Step 1: cut-offs, a dense sequence and tents. Let $K_{m}$ and $\varphi_{m}$, $m\in\mathbb{N}$, be as in Exhaustion of a Locally Compact Separable Metric Space by Compact Sets, with Cut-Off Functions. By Separable Metric Space and Separable Real Normed Space (i) there is a countable dense set $D\subseteq X$; it is nonempty, since $B(x,1)$ meets $D$ for any $x\in X$, so there is a surjective map $e:\mathbb{N}\to D$. Put $x_{k}=e(k)$; every open ball of $X$ contains some $x_{k}$.

For $q,k\in\mathbb{N}$ define the tent $t_{q,k}:X\to\mathbb{R}$ by $t_{q,k}(x)=\max\{0,\,1-q\,d(x,x_{k})\}$. Then $0\le t_{q,k}\le1$; $t_{q,k}(x)>0$ if and only if $d(x,x_{k})<1/q$; $t_{q,k}(x)>1/2$ if $d(x,x_{k})<1/(2q)$; and $t_{q,k}$ is continuous, being Lipschitz because $|\max\{0,a\}-\max\{0,b\}|\le|a-b|$ and $|d(x,x_{k})-d(y,x_{k})|\le d(x,y)$. For $q,n\in\mathbb{N}$ put

$$s_{q,n}=\sum_{k=1}^{n}t_{q,k},\qquad w_{q,n}(x)=\frac{1}{\max\{s_{q,n}(x),1/2\}}\quad(x\in X).$$

Then $s_{q,n}\ge0$ is continuous as a finite sum of continuous functions, and $0<w_{q,n}\le2$. Since $|1/a-1/b|\le4|a-b|$ for $a,b\ge1/2$ and $|\max\{a,1/2\}-\max\{b,1/2\}|\le|a-b|$, we have $|w_{q,n}(x)-w_{q,n}(y)|\le4|s_{q,n}(x)-s_{q,n}(y)|$, so $w_{q,n}$ is continuous because $s_{q,n}$ is.

Step 2: an enumeration of the rational combinations of tents. The set $\mathbb{Q}$ is countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (ix) and contains $0$, so there is a surjective map $\sigma:\mathbb{N}\to\mathbb{Q}$; and by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (ii) there is a bijection $\tau:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$, written $\tau(j)=(\tau_{1}(j),\tau_{2}(j))$. Let $\mathcal{S}$ be the set of maps from $\mathbb{N}\times\mathbb{N}$ to the set of functions $X\to\mathbb{R}$. By Recursion on the Natural Numbers Starting at One (i) there is a map $n\mapsto E_{n}$ from $\mathbb{N}$ to $\mathcal{S}$ with

$$E_{1}(q,j)=\sigma(j)\,t_{q,1},\qquad E_{n+1}(q,j)=E_{n}\bigl(q,\tau_{1}(j)\bigr)+\sigma\bigl(\tau_{2}(j)\bigr)\,t_{q,n+1}\qquad(n,q,j\in\mathbb{N}).$$

By induction on $n$, every $E_{n}(q,j)$ is continuous, as a sum of real multiples of tents. We claim: for all $n,q\in\mathbb{N}$ and all rational numbers $c_{1},\dots,c_{n}$ there is $j\in\mathbb{N}$ with

$$E_{n}(q,j)=\sum_{k=1}^{n}c_{k}\,t_{q,k}.$$

For $n=1$ take $j$ with $\sigma(j)=c_{1}$. If the claim holds for $n$ and $c_{1},\dots,c_{n+1}$ are rational, take $j'$ with $E_{n}(q,j')=\sum_{k=1}^{n}c_{k}t_{q,k}$, $j''$ with $\sigma(j'')=c_{n+1}$, and $j$ with $\tau(j)=(j',j'')$; then $E_{n+1}(q,j)=\sum_{k=1}^{n+1}c_{k}t_{q,k}$.

Step 3: the countable family. For $(m,q,n,j)\in\mathbb{N}^{4}$ let

$$\Phi(m,q,n,j)=\varphi_{m}\,w_{q,n}\,E_{n}(q,j),$$

a pointwise product of continuous functions, hence continuous. If $\Phi(m,q,n,j)(x)\neq0$ then $\varphi_{m}(x)\neq0$; so by Support of a Real Function on a Metric Space (i) and Closure of a Subset of a Metric Space (ii) the support of $\Phi(m,q,n,j)$ lies in $\operatorname{supp}\varphi_{m}$, and it is compact, as a closed subset of the compact set $\operatorname{supp}\varphi_{m}$. Hence $\Phi(m,q,n,j)\in C_{c}(X)$ by The Space of Continuous Functions with Compact Support and the Uniform Norm (i). The set $\mathbb{N}^{4}=((\mathbb{N}\times\mathbb{N})\times\mathbb{N})\times\mathbb{N}$ is countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (i) and Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (vii), so the image $\mathcal{D}=\Phi(\mathbb{N}^{4})\subseteq C_{c}(X)$ is countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (vi).

Step 4: approximation. Let $f\in C_{c}(X)$ and $\varepsilon>0$, and choose in this order: by Exhaustion of a Locally Compact Separable Metric Space by Compact Sets, with Cut-Off Functions (ii), $m\in\mathbb{N}$ with $\operatorname{supp}f\subseteq\operatorname{int}K_{m}\subseteq K_{m}$; by Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (v), $\eta>0$ with $|f(x)-f(y)|<\varepsilon/2$ for all $x,y\in X$ with $d(x,y)<\eta$; by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (ix), $q\in\mathbb{N}$ with $1/q<\eta$; $n\in\mathbb{N}$ with $K_{m}\subseteq\bigcup_{k=1}^{n}B(x_{k},1/(2q))$, which exists 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), since these unions are open, increase with $n$, and cover $X$ (each $x\in X$ has some $x_{k}$ in $B(x,1/(2q))$); rational numbers $c_{1},\dots,c_{n}$ with $|c_{k}-f(x_{k})|<\varepsilon/2$, which exist by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (x) and are taken together 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); and, by Step 2, $j\in\mathbb{N}$ with $E_{n}(q,j)=\sum_{k=1}^{n}c_{k}t_{q,k}$. Put $g=\Phi(m,q,n,j)\in\mathcal{D}$, and abbreviate $t_{k}=t_{q,k}$, $s=s_{q,n}$, $w=w_{q,n}$.

Let $x\in X$. If $k\in[n]$ and $t_{k}(x)>0$, then $d(x,x_{k})<1/q<\eta$, so

$$|c_{k}-f(x)|\le|c_{k}-f(x_{k})|+|f(x_{k})-f(x)|<\varepsilon;$$

as the terms with $t_{k}(x)=0$ vanish, this gives

$$\Bigl|\sum_{k=1}^{n}\bigl(c_{k}-f(x)\bigr)t_{k}(x)\Bigr|\le\varepsilon\,s(x).\tag{$*$}$$

If $x\in K_{m}$, some $k\in[n]$ has $d(x,x_{k})<1/(2q)$, so $s(x)\ge t_{k}(x)>1/2$ and $w(x)=1/s(x)$, while $\varphi_{m}(x)=1$ by Exhaustion of a Locally Compact Separable Metric Space by Compact Sets, with Cut-Off Functions (iii); hence by $(*)$

$$|g(x)-f(x)|=\Bigl|\frac{\sum_{k=1}^{n}c_{k}t_{k}(x)}{s(x)}-f(x)\Bigr|=\frac{\bigl|\sum_{k=1}^{n}(c_{k}-f(x))t_{k}(x)\bigr|}{s(x)}\le\varepsilon .$$

If $x\notin K_{m}$, then $x\notin\operatorname{supp}f$, so $f(x)=0$ by Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (iii), and $(*)$ reads $|\sum_{k=1}^{n}c_{k}t_{k}(x)|\le\varepsilon s(x)$; since $0\le\varphi_{m}(x)\le1$,

$$|g(x)-f(x)|=\varphi_{m}(x)\,\frac{\bigl|\sum_{k=1}^{n}c_{k}t_{k}(x)\bigr|}{\max\{s(x),1/2\}}\le\frac{\varepsilon\,s(x)}{\max\{s(x),1/2\}}\le\varepsilon .$$

Thus $|(f-g)(x)|\le\varepsilon$ for every $x\in X$, and $d_{u}(f,g)=\|f-g\|_{u}\le\varepsilon$.

Step 5: conclusion. Let $f\in C_{c}(X)$ and $r>0$. Step 4 with $\varepsilon=r/2$ gives $g\in\mathcal{D}$ with $d_{u}(f,g)\le r/2<r$, so $\mathcal{D}$ meets the open ball of radius $r$ about $f$. Hence $\mathcal{D}$ is a countable dense subset of $(C_{c}(X),d_{u})$ by Dense Subset of a Metric Space (i), this metric space is separable by Separable Metric Space and Separable Real Normed Space (i), and the normed space $(C_{c}(X),\|\cdot\|_{u})$ is separable by Separable Metric Space and Separable Real Normed Space (ii).