Proof of Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls

Throughout, the open subsets of $X$ are the members of the topology $\mathcal{T}_{d}$, a set $\mathcal{U}$ of sets covers $K$ when $K\subseteq\bigcup\mathcal{U}$, and finite sets are as in Finite Sets (i); singletons are finite, unions of two finite sets are finite, unions of finite families of finite sets are finite, and images of finite sets under maps are 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), 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), 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 (x) 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 (vi), and segments of $\mathbb{N}$ are 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 (vii); set-builder conditions are predicative by the formula trees of Formula Trees with Values: Syntax, Two Readings, and Definedness.

Covers by open sets. Let $K\subseteq X$, and let $\mathcal{T}_{K}=\{K\cap U:U\in\mathcal{T}_{d}\}$ be the subspace topology on $K$. By Compact Topological Space and Compact Subset (ii) and Compact Topological Space and Compact Subset (i), $K$ is compact exactly when every subset of $\mathcal{T}_{K}$ covering $K$ has a finite subset covering $K$.

Suppose first that $K$ is compact, and let $\mathcal{U}$ be a set of open subsets of $X$ covering $K$. The image $\mathcal{W}$ of $\mathcal{U}$ under $U\mapsto K\cap U$ is a subset of $\mathcal{T}_{K}$ covering $K$, so some finite $\mathcal{W}_{0}\subseteq\mathcal{W}$ covers $K$. For each $W\in\mathcal{W}_{0}$ the set $\{U\in\mathcal{U}:W=K\cap U\}$ is nonempty, so 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) gives a map $c$ on $\mathcal{W}_{0}$ with $c(W)\in\mathcal{U}$ and $W=K\cap c(W)$ for every $W\in\mathcal{W}_{0}$. The image $\mathcal{V}=c(\mathcal{W}_{0})$ is a finite subset of $\mathcal{U}$, and it covers $K$: each $x\in K$ lies in some $W\in\mathcal{W}_{0}$, and $W\subseteq c(W)$.

Conversely, suppose that every set of open subsets of $X$ covering $K$ has a finite subset covering $K$, and let $\mathcal{W}\subseteq\mathcal{T}_{K}$ cover $K$. Put $\mathcal{U}=\{U\in\mathcal{T}_{d}:K\cap U\in\mathcal{W}\}$. Every $W\in\mathcal{W}$ has the form $K\cap U$ with $U\in\mathcal{T}_{d}$, and such a $U$ lies in $\mathcal{U}$; hence $\mathcal{U}$ covers $K$, and some finite $\mathcal{V}\subseteq\mathcal{U}$ covers $K$. The image of $\mathcal{V}$ under $U\mapsto K\cap U$ is then a finite subset of $\mathcal{W}$ covering $K$. So $K$ is compact.

Finite unions. Let $\mathcal{U}$ be a set of open subsets of $X$ covering $\bigcup\mathcal{K}$. For every $L\in\mathcal{K}$, $\mathcal{U}$ covers the compact set $L$, so by (i) the set $S_{L}$ of finite subsets of $\mathcal{U}$ covering $L$ is nonempty. 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 is a map $c$ on $\mathcal{K}$ with $c(L)\in S_{L}$ for every $L\in\mathcal{K}$. The set $\mathcal{V}=\bigcup_{L\in\mathcal{K}}c(L)$ is finite, it is contained in $\mathcal{U}$, and it covers every $L\in\mathcal{K}$, hence $\bigcup\mathcal{K}$. By (i), $\bigcup\mathcal{K}$ is compact.

Increasing covers. We have $U_{m}\subseteq U_{m'}$ whenever $m\le m'$ (by induction on $m'$). The image $\mathcal{U}=\{U_{m}:m\in\mathbb{N}\}$ of $\mathbb{N}$ covers $K$, so by (i) some finite $\mathcal{V}\subseteq\mathcal{U}$ covers $K$. For $V\in\mathcal{V}$ let $\mu(V)$ be the least $m\in\mathbb{N}$ with $V=U_{m}$, which exists by Arithmetic and Order of the Natural Numbers (xv). The set $\mu(\mathcal{V})\cup\{1\}$ is a nonempty finite subset of $\mathbb{N}$; the element $1$ is added because $\mathcal{V}$, and with it $\mu(\mathcal{V})$, may be empty. So it has a greatest element $M$ 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 (viii). Every $V\in\mathcal{V}$ equals $U_{\mu(V)}\subseteq U_{M}$, and therefore

$$K\subseteq\bigcup\mathcal{V}\subseteq U_{M}.$$

Sublevel sets of the distance. Put $O=\{x\in X:\operatorname{dist}(x,A)<\delta\}$ and $N=\{x\in X:\operatorname{dist}(x,A)\le\delta\}$. We use the Lipschitz estimate $|\operatorname{dist}(x,A)-\operatorname{dist}(y,A)|\le d(x,y)$ for $x,y\in X$.

Let $x\in O$ and $r=\delta-\operatorname{dist}(x,A)>0$. For $y$ in the open ball $B(x,r)$ we have $\operatorname{dist}(y,A)\le\operatorname{dist}(x,A)+d(x,y)<\delta$, so $B(x,r)\subseteq O$. Hence $O$ is open, which is (v).

Let $x\in X\setminus N$ and $r=\operatorname{dist}(x,A)-\delta>0$. For $y\in B(x,r)$ we have $\operatorname{dist}(y,A)\ge\operatorname{dist}(x,A)-d(x,y)>\delta$, so $B(x,r)\subseteq X\setminus N$. Hence $X\setminus N$ is open and $N$ is closed, which is (vi).

Closed balls. Let $x\in X$, $r\ge0$ and $y\in X\setminus\bar{B}(x,r)$, so that $s=d(x,y)-r>0$. For $z\in B(y,s)$ the triangle inequality gives $d(x,z)\ge d(x,y)-d(y,z)>r$, so $z\notin\bar{B}(x,r)$. Thus $X\setminus\bar{B}(x,r)$ is open, and $\bar{B}(x,r)$ is closed. This proves (iv).