Proof of Semicontinuous Functions Attain Their Extrema on a Compact Set

Steps 1 and 2 below follow the published TheoremBase proof of the extreme value theorem on a compact subset of a metric space, adapted: the continuity hypothesis used there is replaced by upper semicontinuity, which is all that the covering argument consumes, and the conclusion about minima is obtained from the maximum case by negation rather than proved separately. The source version is recorded in the citation attached to this proof.

Write $\mathcal{T}_{d}$ for the collection of subsets of $X$ that are open in $(X,d)$, a topology by Metric Open Sets Form a Topology, and $\mathcal{T}_{K}=\{K\cap W: W\in\mathcal{T}_{d}\}$ for the subspace topology on $K$, which is a topology by The Subspace Topology is a Topology. By the definition of a compact subset, the hypothesis on $K$ says that the topological space $(K,\mathcal{T}_{K})$ is compact. For a natural number $n$, write $[n]$ for the initial segment determined by $n$. We use the elementary order arithmetic of the ordered field $\mathbb{R}$, and the fact that $\le$ is a total order, hence reflexive, antisymmetric, transitive and comparing any two elements.

Step 1: for every $c\in\mathbb{R}$ the set $V_{c}=\{y\in K: u(y)<c\}$ belongs to $\mathcal{T}_{K}$.

Let $x\in V_{c}$, so that $u(x)<c$. Claim 1 of Elementary Order Arithmetic in an Ordered Field, adding $-u(x)$, gives $0<c-u(x)$. Since $u$ is upper semicontinuous at $x$ relative to $K$, applying the defining condition with $\varepsilon=c-u(x)$ provides $\delta_{x}\in\mathbb{R}$ with $0<\delta_{x}$ such that every $y\in K$ with $d(x,y)<\delta_{x}$ satisfies

that is $y\in V_{c}$. Let

be the union of the family of open balls $B_{d}(x,\delta_{x})$ indexed by $x\in V_{c}$.

The set $W_{c}$ is open in $(X,d)$. Indeed, let $z\in W_{c}$, say $z\in B_{d}(x,\delta_{x})$ with $x\in V_{c}$, and put $r=\delta_{x}-d(x,z)$, which satisfies $0<r$ by claim 1 of Elementary Order Arithmetic in an Ordered Field because $d(x,z)<\delta_{x}$. If $w\in B_{d}(z,r)$, then the triangle inequality in the definition of a metric gives

using claims 1 and 2 of Elementary Order Arithmetic in an Ordered Field, so $B_{d}(z,r)\subseteq B_{d}(x,\delta_{x})\subseteq W_{c}$.

Moreover $K\cap W_{c}=V_{c}$. If $y\in K\cap W_{c}$, then $y\in B_{d}(x,\delta_{x})$ for some $x\in V_{c}$, so $d(x,y)<\delta_{x}$ and hence $y\in V_{c}$ by the choice of $\delta_{x}$. Conversely, if $x\in V_{c}$, then $x\in K$ and $d(x,x)=0<\delta_{x}$, so $x\in B_{d}(x,\delta_{x})\subseteq W_{c}$. Hence $V_{c}=K\cap W_{c}\in\mathcal{T}_{K}$.

Step 2: proof of claim 1.

Suppose, for contradiction, that there is no $x_{\max}\in K$ with $u(x)\le u(x_{\max})$ for every $x\in K$. Then for every $x\in K$ there is $y\in K$ for which $u(y)\le u(x)$ fails. For such $x$ and $y$, comparability gives $u(x)\le u(y)$, and $u(x)\ne u(y)$, since $u(x)=u(y)$ would give $u(y)\le u(x)$ by reflexivity. So for every $x\in K$ there is $y\in K$ with $u(x)<u(y)$.

Consider the family $(V_{u(y)})_{y\in K}$ of subsets of $K$ indexed by the set $K$. Each member belongs to $\mathcal{T}_{K}$ by Step 1, and the family covers $K$: given $x\in K$, there is $y\in K$ with $u(x)<u(y)$, and then $x\in V_{u(y)}$. Since $(K,\mathcal{T}_{K})$ is compact, Compact Topological Space and Compact Subset provides a finite subset $J\subseteq K$ with

where a union indexed by the empty set is empty. Since $K$ is nonempty, $J$ is nonempty. Being finite and nonempty, $J$ has $n$ elements for some natural number $n$, so there is a bijection $\beta:[n]\to J$.

Apply Greatest Element of a Finite Family in a Totally Ordered Set to the set $\mathbb{R}$ with its total order and to the $n$-tuple in $\mathbb{R}$ whose $k$-th component is $u(\beta(k))$: there is $j\in[n]$ with $u(\beta(k))\le u(\beta(j))$ for every $k\in[n]$. Put $y^{*}=\beta(j)$, an element of $J\subseteq K$. Then $y^{*}\in V_{u(y)}$ for some $y\in J$, that is $u(y^{*})<u(y)$, so $u(y^{*})\le u(y)$ and $u(y^{*})\ne u(y)$. Since $\beta$ maps $[n]$ onto $J$, we have $y=\beta(k)$ for some $k\in[n]$, whence $u(y)=u(\beta(k))\le u(\beta(j))=u(y^{*})$. Antisymmetry now gives $u(y^{*})=u(y)$, a contradiction.

Hence the supposition was false, and there exists $x_{\max}\in K$ such that $u(x)\le u(x_{\max})$ for every $x\in K$.

Step 3: proof of claim 2.

Let $w:K\to\mathbb{R}$ be lower semicontinuous on $K$, and let $-w:K\to\mathbb{R}$ be the function whose value at $y\in K$ is the additive inverse of $w(y)$. By claim 1 of Semicontinuity Under Negation and Characterization of Continuity, applied at each point of $K$, the function $-w$ is upper semicontinuous on $K$. By claim 1 of the present theorem, proved in Step 2 and applied to $-w$, there is $x_{\min}\in K$ such that

for every $x\in K$. Claim 4 of Elementary Order Arithmetic in an Ordered Field converts this into $w(x_{\min})\le w(x)$ for every $x\in K$.