Proof of Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded

Each result cited is universally quantified over the data in its own statement. No choice principle is used: every object below is either given by a hypothesis, obtained from an existential statement, or defined by a formula.

Throughout, for a set $\mathcal{S}$ of sets, $\bigcup\mathcal{S}$ is the union of its members, and a subset of a set $\mathcal{U}$ is called finite in the sense of Finite Sets §finite. We use two consequences of 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 §image, 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 §subset, 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 §small 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 §union: the image of a finite set under a map is finite, and a subset of a finite set, a singleton, and the union of two finite sets are finite.

Clause closed-subset. Let $(X,\mathcal{T})$, $K$ and $C$ be as in Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded §closed-subset, so $C\subseteq K\subseteq X$. Let $\mathcal{T}_{K}$ and $\mathcal{T}_{C}$ be the subspace topologies on $K$ and on $C$. By Compact Topological Space and Compact Subset §compact-subset and Compact Topological Space and Compact Subset §compact, the hypothesis says: for every subset $\mathcal{W}$ of $\mathcal{T}_{K}$ with $K\subseteq\bigcup\mathcal{W}$ there is a finite subset $\mathcal{V}$ of $\mathcal{W}$ with $K\subseteq\bigcup\mathcal{V}$; and we must show the same for $C$ and $\mathcal{T}_{C}$.

Let $\mathcal{U}$ be a subset of $\mathcal{T}_{C}$ with $C\subseteq\bigcup\mathcal{U}$. Let

$$\mathcal{U}'=\{U\in\mathcal{T}:C\cap U\in\mathcal{U}\},$$

a set by Sets and Maps: Ordinary Notation §set-builder. Since $C$ is closed in $(X,\mathcal{T})$, $X\setminus C\in\mathcal{T}$. Let $\mathcal{W}$ be the union of the image of $\mathcal{U}'$ under the map $U\mapsto K\cap U$ from $\mathcal{U}'$ to $\mathcal{P}(K)$, given by Sets and Maps: Ordinary Notation §maps, with the singleton $\{K\cap(X\setminus C)\}$. Every member of $\mathcal{W}$ has the form $K\cap U$ with $U\in\mathcal{T}$ (namely $U\in\mathcal{U}'$ or $U=X\setminus C$) and is a subset of $K$, so $\mathcal{W}\subseteq\mathcal{T}_{K}$ by Subspace Topology §subspace-topology.

$\mathcal{W}$ covers $K$: let $x\in K$. If $x\notin C$, then $x\in K\cap(X\setminus C)\in\mathcal{W}$. If $x\in C$, then $x\in W$ for some $W\in\mathcal{U}$; as $W\in\mathcal{T}_{C}$, $W=C\cap U$ for some $U\in\mathcal{T}$ by Subspace Topology §subspace-topology; then $C\cap U=W\in\mathcal{U}$, so $U\in\mathcal{U}'$, and $x\in K\cap U\in\mathcal{W}$.

By compactness of $K$ there is a finite subset $\mathcal{V}$ of $\mathcal{W}$ with $K\subseteq\bigcup\mathcal{V}$. Let $\mathcal{E}$ be the image of $\mathcal{V}$ under the map $V\mapsto C\cap V$ from $\mathcal{V}$ to $\mathcal{P}(C)$, and let $\mathcal{V}^{*}=\mathcal{E}\cap\mathcal{U}$. Then $\mathcal{E}$ is finite, hence so is its subset $\mathcal{V}^{*}$, and $\mathcal{V}^{*}\subseteq\mathcal{U}$.

$\mathcal{V}^{*}$ covers $C$: let $x\in C$. Then $x\in K$, so $x\in V$ for some $V\in\mathcal{V}$. Since $x\in C$, $x\notin K\cap(X\setminus C)$, so $V\neq K\cap(X\setminus C)$ and hence $V=K\cap U$ for some $U\in\mathcal{U}'$. As $C\subseteq K$, $C\cap V=C\cap U$, which lies in $\mathcal{U}$ by the definition of $\mathcal{U}'$; so $C\cap V\in\mathcal{E}\cap\mathcal{U}=\mathcal{V}^{*}$, and $x\in C\cap V$.

Thus every subset $\mathcal{U}$ of $\mathcal{T}_{C}$ covering $C$ has a finite subset $\mathcal{V}^{*}$ covering $C$: $(C,\mathcal{T}_{C})$ is compact by Compact Topological Space and Compact Subset §compact, so $C$ is compact in $(X,\mathcal{T})$ by Compact Topological Space and Compact Subset §compact-subset.

Clause closed-bounded. Let $(X,d)$ and $K$ be as in Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded §closed-bounded, with $\mathcal{T}_{d}$ the topology of Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces §topology, so that $(X,\mathcal{T}_{d})$ is a topological space, and $\mathcal{T}_{K}$ the subspace topology on $K$. The metric $d$ satisfies Metric Space §nonnegative, Metric Space §zero, Metric Space §symmetric and Metric Space §triangle.

An extraction step. Let $(U_{m})_{m\in\mathbb{N}}$ be a family of members of $\mathcal{T}_{d}$ with $U_{m}\subseteq U_{m'}$ whenever $m\le m'$, and with $K\subseteq\bigcup_{m\in\mathbb{N}}U_{m}$. We show that $K\subseteq U_{M}$ for some $M\in\mathbb{N}$. Let $\mathcal{U}$ be the image of $\mathbb{N}$ under the map $m\mapsto K\cap U_{m}$ from $\mathbb{N}$ to $\mathcal{P}(K)$; then $\mathcal{U}\subseteq\mathcal{T}_{K}$ by Subspace Topology §subspace-topology, and $K\subseteq\bigcup\mathcal{U}$ since each $x\in K$ lies in some $U_{m}$, hence in $K\cap U_{m}$. By the hypothesis, Compact Topological Space and Compact Subset §compact-subset and Compact Topological Space and Compact Subset §compact, there is a finite subset $\mathcal{V}$ of $\mathcal{U}$ with $K\subseteq\bigcup\mathcal{V}$. For $V\in\mathcal{V}$ the set $\{m\in\mathbb{N}:V=K\cap U_{m}\}$ is nonempty, so it has a least element by Arithmetic and Order of the Natural Numbers §well-order, unique by antisymmetry of $\le$; let $\mu:\mathcal{V}\to\mathbb{N}$ be the map sending $V$ to this least element, given by Sets and Maps: Ordinary Notation §maps. The set $T=\mu[\mathcal{V}]\cup\{1\}$ is a finite nonempty subset of $\mathbb{N}$, 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 §extremes, $\le$ being a total order on $\mathbb{N}$ by Arithmetic and Order of the Natural Numbers §partial-order and Arithmetic and Order of the Natural Numbers §trichotomy. For $V\in\mathcal{V}$, $V=K\cap U_{\mu(V)}\subseteq U_{\mu(V)}\subseteq U_{M}$ as $\mu(V)\le M$. Hence $K\subseteq\bigcup\mathcal{V}\subseteq U_{M}$.

$K$ is closed. By Closed Subset of a Topological Space §closed and Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces §topology we must show that $X\setminus K$ is open in $(X,d)$. Let $x\in X\setminus K$. For $m\in\mathbb{N}$ let $U_{m}=\{y\in X:1/m<d(y,x)\}$, where $1/m\in\mathbb{R}$ by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §quotient; $U_{m}$ is a set by Sets and Maps: Ordinary Notation §set-builder, its condition being predicative by Predicative Substitution of Values of Functions and Binary Operations, and Predicative Formulas for Negatives, Differences, Absolute Values, Distances, Reciprocals and Quotients in an Ordered Field §value-substitution and Predicative Substitution of Values of Functions and Binary Operations, and Predicative Formulas for Negatives, Differences, Absolute Values, Distances, Reciprocals and Quotients in an Ordered Field §quotient-formula.

Each $U_{m}$ is open in $(X,d)$: let $y\in U_{m}$ and $r=d(y,x)-1/m>0$, a real number by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §negative. For $z$ in the open ball $B_{d}(y,r)$ we have $d(y,z)<r$ by Open Ball in a Metric Space §membership, and $d(y,x)\le d(y,z)+d(z,x)$, so $d(z,x)\ge d(y,x)-d(y,z)>d(y,x)-r=1/m$, that is $z\in U_{m}$. So $B_{d}(y,r)\subseteq U_{m}$, and $U_{m}$ is open by Open Subset of a Metric Space §open; thus $U_{m}\in\mathcal{T}_{d}$ by Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces §topology.

If $m\le m'$ then $1/m'\le1/m$, so $U_{m}\subseteq U_{m'}$. The $U_{m}$ cover $K$: for $y\in K$ we have $y\neq x$, as $x\notin K$, so $d(y,x)>0$ and $1/d(y,x)\in\mathbb{R}$ by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §quotient; by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §archimedean there is $m\in\mathbb{N}$ with $1/d(y,x)<m$, whence $1/m<d(y,x)$ and $y\in U_{m}$. By the extraction step, $K\subseteq U_{M}$ for some $M\in\mathbb{N}$. Now if $y\in B_{d}(x,1/M)$, then $d(y,x)=d(x,y)<1/M$, so $y\notin U_{M}$ and hence $y\notin K$. Thus $B_{d}(x,1/M)\subseteq X\setminus K$. As $x\in X\setminus K$ was arbitrary, $X\setminus K$ is open in $(X,d)$, so $X\setminus K\in\mathcal{T}_{d}$ and $K$ is closed in $(X,\mathcal{T}_{d})$.

$K$ is bounded. If $K=\emptyset$, the condition of Bounded Subset of a Metric Space §bounded holds with $M=0$ vacuously. Otherwise let $x_{0}\in K$, and for $m\in\mathbb{N}$ let $U_{m}=B_{d}(x_{0},m)$, which is open in $(X,d)$ by Open Ball in a Metric Space is Open §open, so $U_{m}\in\mathcal{T}_{d}$. If $m\le m'$ then $U_{m}\subseteq U_{m'}$ by Open Ball in a Metric Space §membership. For $y\in K$, The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §archimedean gives $m\in\mathbb{N}$ with $d(x_{0},y)<m$, so $y\in U_{m}$. By the extraction step, $K\subseteq B_{d}(x_{0},M)$ for some $M\in\mathbb{N}$. For $u,v\in K$,

$$d(u,v)\le d(u,x_{0})+d(x_{0},v)=d(x_{0},u)+d(x_{0},v)<2M,$$

so $K$ is bounded in $(X,d)$, with the real number $2M$ (here $2=1+1$ in $\mathbb{R}$ by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §two, and $M\in\mathbb{R}$ by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §naturals).