Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value

Let $A\subseteq \mathbb{R}$, let $f:A\to \mathbb{R}$ be continuous on $A$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point) and let $K\subseteq A$ be nonempty and compact (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset). Then $\sup f[K]$ and $\inf f[K]$ exist and are attained: there are $p,q\in K$ with

$f(q)=\inf f[K]\le f(x)\le \sup f[K]=f(p)\text{for every }x\in K\mathrm{.}$

Equivalently, the set $f[K]$ has a maximum and a minimum (Maximum and minimum of a set), namely $\max f[K]=f(p)$ and $\min f[K]=f(q)$.

Nonemptiness of $K$ is a hypothesis, not an oversight. For $K=\mathrm{\varnothing }$ the set $f[K]$ is empty, and neither a supremum nor a maximum of the empty set exists in this library (Complete ordered field (least-upper-bound property) supplies suprema of nonempty sets bounded above only).

This theorem is stated twice in this library, on purpose. Its metric-space twin is A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, proved from the cover machinery of metric spaces; the proof below is $\mathbb{R}$-native, running through Heine-Borel for $\mathbb{R}$ and the order-completeness of $\mathbb{R}$, and it uses no cover argument beyond the one already spent in The image of a compact subset of $\mathbb{R}$ under a continuous real function is compact. That the two statements are the same statement in two vocabularies is proved in Dictionary: for $A\subseteq \mathbb{R}$ with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$, continuity and uniform continuity of $f:A\to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace, later on this page.

• The two ingredients, kept apart. Compactness of $K$ enters only through the compactness of the image; order-completeness of $\mathbb{R}$ enters only in the existence of $\sup S$ and $\inf S$. The bridge between them is closedness of $S$: a closed set contains the adherent points of itself, and the supremum of a nonempty bounded set is always adherent to it, by Epsilon characterisation of the supremum. Neither ingredient can be dropped: over $\mathbb{Q}$ the supremum need not exist, and on a noncompact domain the supremum exists and is not attained (The identity on $(0,1)$ is bounded with no greatest value, and on $[0,\mathrm{\infty })$ it is continuous and unbounded ↗).

• Attainment is exactly what the epsilon characterisation cannot give on its own. Epsilon characterisation of the supremum produces points of $S$ arbitrarily close to $\sup S$ for any nonempty bounded $S$; nothing there says one of them equals $\sup S$. What closedness adds is that the limiting value is not lost.

• The converse. If every continuous real function on a set $E$ attains a greatest value, then $E$ is compact. That is the content of Rudin 4.20, the sharp converse: on a noncompact $E\subseteq \mathbb{R}$ there is an unbounded continuous function and a bounded continuous function with no greatest value, and if $E$ is bounded there is a continuous function on $E$ that is not uniformly continuous, which exhibits, for every noncompact $E$, a bounded continuous function on $E$ with no greatest value.