Proof of Heine-Cantor Theorem: Continuity on a Compact Metric Space or a Compact Subset Implies Uniform Continuity

Proof of (ii). Let $\mathcal{T}_{K}$ be the subspace topology on $K$ induced by $\mathcal{T}_{d_X}$. By Compact Topological Space and Compact Subset (ii), the hypothesis on $K$ says that $(K,\mathcal{T}_{K})$ is compact. By Uniformly Continuous Map Between Metric Spaces (i) we must show that for every real $\varepsilon>0$ there is a real $\delta>0$ such that all $x,y\in K$ with $d_X(x,y)<\delta$ satisfy $d_Y(f(x),f(y))<\varepsilon$. The data are fixed in this order: first $\varepsilon$; then $\eta$, determined by $\varepsilon$; then a finite subcover $\mathcal{V}$ of a cover $\mathcal{U}$ determined by $\eta$; then a choice map $\varphi$ on $\mathcal{V}$; then $\delta$, determined by $\varphi$; and only then the points $x,y\in K$, on which nothing fixed before depends.

Let $\varepsilon$ be a real number with $\varepsilon>0$. If $K=\emptyset$, then $\delta=1$ serves, since $K$ has no points. Assume from now on that $K\neq\emptyset$.

Step 1: a cover of $K$ defined by a condition. Put $\eta=\varepsilon/2$. Then $0<\eta$ and $\eta+\eta=\varepsilon$, by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vii) applied in the ordered field $\mathbb{R}$ to $0<\varepsilon$. Let $P$ be the set of all $(x,r)\in K\times\mathbb{R}$ with $0<r$ such that every $z\in K$ with $d_X(x,z)<r+r$ satisfies $d_Y(f(z),f(x))<\eta$. For $(x,r)\in P$, the open ball $B_{d_X}(x,r)$ is a subset of $X$ that is open in $(X,d_X)$, so it lies in $\mathcal{T}_{d_X}$ by Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (iii), and hence $K\cap B_{d_X}(x,r)\in\mathcal{T}_{K}$ by Subspace Topology (i). Let $\mathcal{U}$ be the set of all $W\in\mathcal{T}_{K}$ such that $W=K\cap B_{d_X}(x,r)$ for some $(x,r)\in P$. Thus $\mathcal{U}\subseteq\mathcal{T}_{K}$.

We show $K\subseteq\bigcup\mathcal{U}$. Let $x\in K$. By Continuous Map Between Metric Spaces (ii), $f$ is continuous at $x$ relative to $K$, so Continuous Map Between Metric Spaces (i), applied with $\eta$ in place of $\varepsilon$, gives a real $\delta_{0}>0$ such that every $z\in K$ with $d_X(x,z)<\delta_{0}$ satisfies $d_Y(f(z),f(x))<\eta$. Put $r=\delta_{0}/2$; then $0<r$ and $r+r=\delta_{0}$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vii), so $(x,r)\in P$. As $d_X(x,x)=0<r$ by Metric Space (iii), $x\in B_{d_X}(x,r)$ by Open Ball in a Metric Space (ii), so $x\in K\cap B_{d_X}(x,r)$, which belongs to $\mathcal{U}$. The numbers $\delta_{0}$ and $r$ serve only to show that $x$ is covered: no radius is selected for each point, since $P$ and $\mathcal{U}$ are defined by conditions.

Step 2: finitely many radii and their least element. As $(K,\mathcal{T}_{K})$ is compact and $\mathcal{U}$ is a subset of $\mathcal{T}_{K}$ with $K\subseteq\bigcup\mathcal{U}$, Compact Topological Space and Compact Subset (i) gives a finite subset $\mathcal{V}$ of $\mathcal{U}$ with $K\subseteq\bigcup\mathcal{V}$; and $\mathcal{V}\neq\emptyset$, because $K\neq\emptyset$. For $W\in\mathcal{V}$ let $C_{W}$ be the set of all radii $r\in\mathbb{R}$ such that there is $x$ with $(x,r)\in P$ and $W=K\cap B_{d_X}(x,r)$. Each $C_{W}$ is a set by Sets and Maps: Ordinary Notation (iii) and a subset of $\mathbb{R}$, so $C_{W}\in\mathcal{P}(\mathbb{R})$; hence $W\mapsto C_{W}$ is a map from $\mathcal{V}$ to $\mathcal{P}(\mathbb{R})$, by Maps and Relations Given by Formulas (i) as in Sets and Maps: Ordinary Notation (v), that is, a family $(C_{W})_{W\in\mathcal{V}}$ of sets. Moreover $C_{W}\neq\emptyset$ for every $W\in\mathcal{V}$: as $W\in\mathcal{V}\subseteq\mathcal{U}$, there is $(x,r)\in P$ with $W=K\cap B_{d_X}(x,r)$, and then $r\in C_{W}$. 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), applied to the finite set $\mathcal{V}$ and this family, gives a map $\varphi:\mathcal{V}\to\bigcup_{W\in\mathcal{V}}C_{W}$ with $\varphi(W)\in C_{W}$ for every $W\in\mathcal{V}$. Only the radius $\varphi(W)$ is chosen; no centre is selected for $W$.

Let $\Delta=\varphi(\mathcal{V})$ be the image of $\mathcal{V}$ under $\varphi$. By The Image and the Preimage of a Class under a Class (i), its elements are the sets $v$ with $(W,v)\in\varphi$ for some $W\in\mathcal{V}$, that is, the values $\varphi(W)$ for $W\in\mathcal{V}$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (i)); as $\varphi(W)\in C_{W}\subseteq\mathbb{R}$, $\Delta$ is a subset of $\mathbb{R}$. It 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 (vi), and nonempty because $\mathcal{V}$ is nonempty. As the order of $\mathbb{R}$ is a total order, $\Delta$ has a least element 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), unique by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iii); let $\delta=\min\Delta$. Then $\delta\in\Delta$, so $\delta=\varphi(W_{0})\in C_{W_{0}}$ for some $W_{0}\in\mathcal{V}$; thus $(x,\delta)\in P$ for some $x$, and hence $0<\delta$. And $\delta\le\varphi(W)$ for every $W\in\mathcal{V}$, as $\varphi(W)\in\Delta$.

Step 3: the estimate. Let $x,y\in K$ with $d_X(x,y)<\delta$. As $x\in K\subseteq\bigcup\mathcal{V}$, there is $W\in\mathcal{V}$ with $x\in W$. Put $r=\varphi(W)$. As $r\in C_{W}$, there is a point $c$ with $(c,r)\in P$ and $W=K\cap B_{d_X}(c,r)$; it is used only in this paragraph. Then $c\in K$, $0<r$ and $\delta\le r$. As $x\in B_{d_X}(c,r)$, $d_X(c,x)<r$ by Open Ball in a Metric Space (ii), and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ii), applied to $d_X(c,x)<r$ and $0\le r$, gives $d_X(c,x)<r+r$; as $x\in K$ and $(c,r)\in P$, it follows that $d_Y(f(x),f(c))<\eta$. Next, $d_X(x,y)\le\delta\le r$, so by Metric Space (v) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ii), applied to $d_X(c,x)<r$ and $d_X(x,y)\le r$,

$$d_X(c,y)\le d_X(c,x)+d_X(x,y)<r+r,$$

whence $d_X(c,y)<r+r$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i); as $y\in K$ and $(c,r)\in P$, it follows that $d_Y(f(y),f(c))<\eta$. Finally, by Metric Space (v), Metric Space (iv) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ii), applied to $d_Y(f(x),f(c))<\eta$ and $d_Y(f(y),f(c))\le\eta$,

$$d_Y(f(x),f(y))\le d_Y(f(x),f(c))+d_Y(f(y),f(c))<\eta+\eta=\varepsilon ,$$

so $d_Y(f(x),f(y))<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i), with $\eta+\eta=\varepsilon$ from Step 1. As $\varepsilon>0$ was arbitrary, $f$ is uniformly continuous on $K$.

Proof of (i). Let $\mathcal{T}_{X}$ be the subspace topology on $X$ induced by $\mathcal{T}_{d_X}$. Every $U\in\mathcal{T}_{d_X}$ is a subset of $X$, by Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (iii), so $X\cap U=U$; hence, by Subspace Topology (i), $\mathcal{T}_{X}=\mathcal{T}_{d_X}$, and $(X,\mathcal{T}_{X})$ is compact. By Compact Topological Space and Compact Subset (ii), $X$ is compact in $(X,\mathcal{T}_{d_X})$. As $f$ is continuous, it is continuous on $X$, by Continuous Map Between Metric Spaces (ii). So (ii), applied with $K=X$, shows that $f$ is uniformly continuous on $X$, that is, uniformly continuous, by Uniformly Continuous Map Between Metric Spaces (i).