Proof of Compactness of Intersections with Closed Sets and of Level Sets of Semicontinuous Functions

Throughout, $X\setminus S$ is the complement relative to $X$ of $S\subseteq X$; the finite-subcover criterion is condition 2 of Compact Subset Criterion via Open Covers in the Ambient Space, which characterises compactness by open covers in the ambient space; claims of Elementary Order Arithmetic in an Ordered Field (below, the order arithmetic lemma) are cited by number; and $\mathcal{T}_d$ contains the empty set and is closed under arbitrary unions, being a topology.

Claim 1. Since $C$ is closed in $(X,\mathcal{T}_d)$, the set $X\setminus C$ belongs to $\mathcal{T}_d$. Let $(U_i)_{i\in I}$ be an open cover of $K\cap C$ in $X$; we verify the finite-subcover criterion for $K\cap C$.

If $I$ is empty, then the union of the family is empty, so $K\cap C$ is empty and the subfamily indexed by the empty subset of $I$ is a subcover. Assume therefore that $I$ is nonempty.

For $i\in I$ set $W_i=U_i\cup(X\setminus C)$. Each $W_i$ is a union of members of $\mathcal{T}_d$ and hence belongs to $\mathcal{T}_d$. The family $(W_i)_{i\in I}$ is an open cover of $K$: let $x\in K$. If $x\in C$, then $x\in K\cap C$, so $x\in U_i$ for some $i\in I$ and therefore $x\in W_i$. If $x\notin C$, then $x\in X\setminus C$, and since $I$ is nonempty we may pick any $i\in I$ and get $x\in W_i$.

As $K$ is compact in $X$, the finite-subcover criterion gives a finite subset $J\subseteq I$ with $K\subseteq\bigcup_{j\in J}W_j$. Let $x\in K\cap C$. Then $x\in W_j$ for some $j\in J$; since $x\in C$ we have $x\notin X\setminus C$, and therefore $x\in U_j$. Hence $K\cap C\subseteq\bigcup_{j\in J}U_j$, so $(U_j)_{j\in J}$ is a subcover and $K\cap C$ is compact in $X$.

Claim 2. Write $S=\{x\in K: c\le w(x)\}$ and let $(U_i)_{i\in I}$ be an open cover of $S$ in $X$.

If $S$ is empty, the subfamily indexed by the empty subset of $I$ is a subcover and we are done. Assume therefore that $S$ is nonempty; then $I$ is nonempty, since a point of $S$ lies in some $U_i$.

Let $P$ be the set of all ordered pairs $(z,r)$ such that $z\in K$, $r\in\mathbb{R}$, $0<r$, and every $y\in K$ with $d(z,y)<r$ satisfies $w(y)<c$. For $(z,r)\in P$ let $B(z,r)$ be the open ball of centre $z$ and radius $r$, which belongs to $\mathcal{T}_d$ by Open Ball in a Metric Space is Open. Let

$$G=\bigcup_{(z,r)\in P}B(z,r),$$

which belongs to $\mathcal{T}_d$, being a union of members of $\mathcal{T}_d$ indexed by $P$ (and being the empty set, which belongs to $\mathcal{T}_d$, when $P$ is empty). No choice of a radius at each point is made here: $P$ is the set of all admissible pairs.

For $i\in I$ set $W_i=U_i\cup G$, which belongs to $\mathcal{T}_d$. The family $(W_i)_{i\in I}$ is an open cover of $K$: let $x\in K$. If $c\le w(x)$, then $x\in S$, so $x\in U_i\subseteq W_i$ for some $i\in I$. Otherwise $w(x)<c$; adding $-w(x)$ to both sides and using claim 1 of the order arithmetic lemma gives $0<c-w(x)$, so upper semicontinuity of $w$ at $x$ relative to $K$, applied with $\varepsilon=c-w(x)$, provides $\delta\in\mathbb{R}$ with $0<\delta$ such that every $y\in K$ with $d(x,y)<\delta$ satisfies

$$w(y)<w(x)+(c-w(x))=c .$$

Thus $(x,\delta)\in P$. Since $d(x,x)=0<\delta$ by the definition of a metric, $x\in B(x,\delta)\subseteq G$, and hence $x\in W_i$ for every $i\in I$; as $I$ is nonempty, $x$ is covered.

Since $K$ is compact in $X$, there is a finite subset $J\subseteq I$ with $K\subseteq\bigcup_{j\in J}W_j$. The sets $S$ and $G$ are disjoint: if $x\in S\cap G$, then $x\in B(z,r)$ for some $(z,r)\in P$, so $d(z,x)<r$, and $x\in K$ because $x\in S$; the defining property of $P$ then gives $w(x)<c$, contradicting $c\le w(x)$.

Now let $x\in S$. Then $x\in K$, so $x\in W_j=U_j\cup G$ for some $j\in J$; since $x\notin G$, we get $x\in U_j$. Hence $S\subseteq\bigcup_{j\in J}U_j$, so $(U_j)_{j\in J}$ is a subcover and $S$ is compact in $X$.

Claim 3. Let $-w:K\to\mathbb{R}$ be the function whose value at $x\in K$ is $-w(x)$. By claim 1 of Negation, Restriction, and Separated Differences of Semicontinuous Functions, applied at every point of $K$, the function $-w$ is upper semicontinuous on $K$. By claim 2 of the present lemma, applied to $-w$ and to the real number $-c$, the set

$$\{x\in K:\ -c\le -w(x)\}$$

is compact in $X$. Finally, for $x\in K$ the inequality $-c\le -w(x)$ holds if and only if $w(x)\le c$, by claim 4 of the order arithmetic lemma; hence this set is $\{x\in K: w(x)\le c\}$, which is therefore compact in $X$.