The identity on $(0,1)$ is bounded with no greatest value, and on $[0,\mathrm{\infty })$ it is continuous and unbounded

Refuted claim: for the conclusions of the extreme value theorem it is enough that the domain be bounded, or that it be closed; that is, a continuous real function on a bounded domain attains a greatest value, and a continuous real function on a closed domain is bounded (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, Lower bound, bounded below, bounded set, Maximum and minimum of a set).

Both halves are false, and one function refutes both:

• on $E_1:=(0,1)$, which is bounded and not closed, the identity is continuous and bounded, $\sup$ of its image exists and equals $1$, and no point of $E_1$ attains it;
• on $E_2:=[0,\mathrm{\infty })$, which is closed and not bounded, the identity is continuous and unbounded.

Neither $E_1$ nor $E_2$ is compact (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 to either, and both are instances 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: the first is its bounded-not-closed case, the second its unbounded case. Together they show that neither half of "closed and bounded" can be dropped.

This item is the worked witness for FALSE: a continuous real function on a bounded domain attains a greatest value, which refutes the first half alone.

• The two failures are of different kinds. On $(0,1)$ the function is bounded and the supremum exists as a real number; what is missing is a point at which it is attained, and adding the endpoint $1$ restores it. On $[0,\mathrm{\infty })$ there is no supremum at all, and no endpoint can be added. That is the distinction 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 draws between its bounded-not-closed and unbounded cases.

• The domains are exactly the two minimal ways to fail compactness. By A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded a subset of $\mathbb{R}$ fails compactness by failing closedness or by failing boundedness; $E_1$ fails only the first, $E_2$ only the second, and each already kills the theorem.

• The same two domains admit failures of uniform continuity, with different witnesses: on $(0,1)$ use $1\mathrm{/}x$ ($x\mapsto 1\mathrm{/}x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1\mathrm{/}(k+2)$ and $1\mathrm{/}(k+3)$ defeating every $\delta$), and on the displayed half-line $[0,\mathrm{\infty })$ use $x^2$. For any $\delta >0$, choose a positive integer $n>1\mathrm{/}\delta$ by [L4]. Both $n$ and $n+1\mathrm{/}n$ lie in the half-line, their distance is $1\mathrm{/}n<\delta$, and their squares differ by $2+1\mathrm{/}{n}^2>2$. Thus $\epsilon =2$ defeats every proposed radius, exactly as in $x\mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, the pairs $k+1$ and $k+1+1\mathrm{/}(k+1)$ defeating every $\delta$. This does not assert failure on every unbounded closed set: on $\mathbb{Z}$, $\delta =1$ forces two points at distance less than $\delta$ to coincide, so every function is uniformly continuous. The identity witnesses used above are themselves uniformly continuous on both displayed domains, since their output distance equals their input distance.