A continuous real function on a compact subset of $\mathbb{R}$ is bounded

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 $\varepsilon$-$\delta$ 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 compact (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset). Then $f$ is bounded on $K$: there is a real $M \ge 0$ with

$|f(x)| \le M \qquad \text{for every } x \in K .$

Equivalently, $f[K]$ is a bounded subset of $\mathbb{R}$ (Lower bound, bounded below, bounded set).

The hypothesis is compactness of $K$, not of $A$, and it cannot be relaxed to boundedness of $K$ or to closedness of $K$ alone: the identity is unbounded on the closed set $[0,\infty)$, and $x \mapsto 1/x$ is unbounded on the bounded set $(0,1)$. The general statement of that is 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, later on this page.

• Boundedness is the weak half of the extreme value theorem. What compactness gives in addition is that the two bounds are attained, which is Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value; the supremum of $f[K]$ exists as soon as $f[K]$ is nonempty and bounded above, and the work is entirely in showing that it belongs to $f[K]$.

• Boundedness of the domain is not what is used. The proof never looks at $K$ after the first line: the whole content is that the image is compact, hence bounded. That is why the same one-line argument gives boundedness of a continuous function on any compact set, however complicated.