FALSE: a continuous real function on a bounded domain attains a greatest value

False claim: if $E \subseteq \mathbb{R}$ is nonempty and bounded (Lower bound, bounded below, bounded set) and $f : E \to \mathbb{R}$ is continuous on $E$ (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), then $f$ attains a greatest value on $E$: there is $p \in E$ with $f(x) \le f(p)$ for every $x \in E$ (Maximum and minimum of a set).

Why it is tempting. The extreme value theorem is often remembered as "a continuous function on a bounded interval attains its bounds", and on $[a,b]$ that is true. The hypothesis that actually does the work is compactness, which for a subset of $\mathbb{R}$ is closed and bounded (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded); dropping closedness loses the theorem even though the function may stay bounded.

What is true. Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value gives the conclusion on a nonempty compact domain, and the hypothesis cannot be weakened: for every noncompact $E$ there is a bounded continuous function on $E$ whose supremum is not attained, which 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. The witness below is that theorem's simplest instance.

• The domain is bounded and not closed, and that is exactly the gap. The set $(0,1)$ is not closed and hence is not compact by A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded, so Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value does not apply. Adding the two endpoints repairs everything: on $[0,1]$ the same $f$ attains the value $1$ at the point $1$.

• Boundedness of the function is not the issue either. The witness above is bounded, so the failure is not a blow-up; it is the loss of the point at which the supremum would be attained. A function on the same domain that is unbounded, such as $x \mapsto 1/x$, fails the conclusion for the cruder reason that no upper bound exists at all, and both failures are catalogued together in The identity on $(0,1)$ is bounded with no greatest value, and on $[0,\infty)$ it is continuous and unbounded ↗ on the companion page.

• The least value fails in the same way, by symmetry: $f$ has no least value on $(0,1)$ either, and $\inf f[E] = 0 \notin f[E]$. The statement is written for the greatest value only because that is the form the false claim usually takes.