Proof of Extreme Value Theorem on a Compact Subset of a Metric Space

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 $n$, $[n]$ is the initial segment determined by $n$. We use the ordered field structure of $\mathbb{R}$, whose order $\le$ is in particular a total order, hence reflexive, antisymmetric, transitive and comparing any two elements, and we write $s<t$ to mean $s\le t$ and $s\ne t$. We also use the properties of the absolute value recorded in Properties of the Absolute Value in an Ordered Field, in particular $t\le|t|$ and $|-t|=|t|$ for every real $t$.

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

For each $x\in V_{c}$ we have $f(x)<c$, hence $0<c-f(x)$, so the continuity hypothesis provides a real $\delta_{x}>0$ such that every $y\in K$ with $d(x,y)<\delta_{x}$ satisfies $|f(y)-f(x)|<c-f(x)$; since $f(y)-f(x)\le|f(y)-f(x)|$, such a $y$ satisfies $f(y)-f(x)<c-f(x)$ and therefore $f(y)<c$, using compatibility of the order with addition (condition 1 of Ordered Field). Let

$$W_{c}=\bigcup_{x\in V_{c}}B_{d}(x,\delta_{x})$$

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 $r>0$ because $d(x,z)<\delta_{x}$. If $w\in B_{d}(z,r)$, then the triangle inequality of Metric Space gives $d(x,w)\le d(x,z)+d(z,w)<d(x,z)+r=\delta_{x}$, 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 $f(y)<c$ by the choice of $\delta_{x}$, that is $y\in V_{c}$. 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: $f$ attains a maximum.

Suppose not: suppose there is no $x_{\max}\in K$ with $f(x)\le f(x_{\max})$ for every $x\in K$. Then for every $x\in K$ there is $y\in K$ for which $f(y)\le f(x)$ fails. For such $x$ and $y$ we get $f(x)\le f(y)$ by comparability, and $f(x)\ne f(y)$ because $f(x)=f(y)$ would give $f(y)\le f(x)$ by reflexivity; that is, $f(x)<f(y)$.

Consider the family $(V_{f(y)})_{y\in K}$ 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 $f(x)<f(y)$, and then $x\in V_{f(y)}$. Since $(K,\mathcal{T}_{K})$ is compact, Compact Topological Space and Compact Subset provides a finite subset $J\subseteq K$ with

$$K\subseteq\bigcup_{y\in J}V_{f(y)},$$

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 $f(\beta(k))$: there is $j\in[n]$ with $f(\beta(k))\le f(\beta(j))$ for every $k\in[n]$. Put $y^{*}=\beta(j)$, an element of $J\subseteq K$. Then $y^{*}\in V_{f(y)}$ for some $y\in J$, that is $f(y^{*})<f(y)$, so $f(y^{*})\le f(y)$ and $f(y^{*})\ne f(y)$. Since $\beta$ maps $[n]$ onto $J$, we have $y=\beta(k)$ for some $k\in[n]$, whence $f(y)=f(\beta(k))\le f(\beta(j))=f(y^{*})$. Antisymmetry of the total order now gives $f(y^{*})=f(y)$, a contradiction.

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

Step 3: $f$ attains a minimum.

Let $g:K\to\mathbb{R}$ be defined by $g(x)=-f(x)$, the additive inverse in $\mathbb{R}$. For $x,y\in K$ we have $g(y)-g(x)=-(f(y)-f(x))$, hence $|g(y)-g(x)|=|f(y)-f(x)|$, so $g$ satisfies the same continuity hypothesis as $f$, with the same $\delta$ for each $x$ and $\varepsilon$. By Step 2 applied to $g$ there is $x_{\min}\in K$ with $g(x)\le g(x_{\min})$ for every $x\in K$, that is $-f(x)\le -f(x_{\min})$. Adding $f(x)+f(x_{\min})$ to both sides and using compatibility of the order with addition (condition 1 of Ordered Field) gives $f(x_{\min})\le f(x)$ for every $x\in K$.

Steps 2 and 3 together give the assertion.