CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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An upper semicontinuous function on $[0,1]$ that is bounded below and attains no minimum, so the semicontinuous extreme value theorem is genuinely one-sided

Statement refuted

Refuted claim: an upper semicontinuous function on a nonempty compact subset of $\mathbb{R}$ that is bounded below attains a minimum (Upper and lower semicontinuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ , Maximum and minimum of a set).

What Semicontinuous extreme value theorem: an upper semicontinuous function on a nonempty compact $K \subseteq \mathbb{R}$ is bounded above and attains a maximum, and a lower semicontinuous one is bounded below and attains a minimum proves is the one-sided statement: an upper semicontinuous function on a nonempty compact set attains a maximum, and a lower semicontinuous one attains a minimum. The refuted claim mixes the two, and it is false.

Counterexample

Define $f : [0,1] \to \mathbb{R}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) by

$f(0) := 1, \qquad f(x) := x \quad \text{for } 0 < x \le 1 .$

Then $f$ is upper semicontinuous on $[0,1]$ , bounded below by $0$ , with $\inf f[\,[0,1]\,] = 0$ (Greatest lower bound (infimum)), and $f(x) > 0$ for every $x \in [0,1]$ : the infimum is not attained, so $f$ has no minimum. It does attain a maximum, namely $1$ at $x = 0$ , as Semicontinuous extreme value theorem: an upper semicontinuous function on a nonempty compact $K \subseteq \mathbb{R}$ is bounded above and attains a maximum, and a lower semicontinuous one is bounded below and attains a minimum requires.

Facts & Assumptions

Given: The function $f : [0,1] \to \mathbb{R}$ with $f(0) = 1$ and $f(x) = x$ for $0 < x \le 1$ .

[L1]

$f$ is upper semicontinuous at $c$ when for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $f(x) < f(c) + \varepsilon$ for every $x \in [0,1] \cap N_\delta(c)$ (Upper and lower semicontinuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L2]

The identity is continuous, so for $c \ne 0$ the $\varepsilon$ - $\delta$ condition for $f$ on a neighbourhood of $c$ avoiding $0$ is that of the identity (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, 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).

[L3]

A nonempty set of reals bounded below has a greatest lower bound (Every nonempty set bounded below has an infimum, Greatest lower bound (infimum), Lower bound, bounded below, bounded set); $m$ is a minimum of $S$ when $m \in S$ and $m \le s$ for every $s \in S$ (Maximum and minimum of a set).

[L4]

For every real $\eta > 0$ there is a natural $n \ge 1$ with $1/n < \eta$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[L5]

An upper semicontinuous function on a nonempty compact set attains a maximum (Semicontinuous extreme value theorem: an upper semicontinuous function on a nonempty compact $K \subseteq \mathbb{R}$ is bounded above and attains a maximum, and a lower semicontinuous one is bounded below and attains a minimum); upper semicontinuity is equivalent to the strict sublevel sets being relatively open ( $f$ is upper semicontinuous on $A$ if and only if $\{x \in A : f(x) < \alpha\}$ is relatively open in $A$ for every real $\alpha$ , lower semicontinuous if and only if $\{x \in A : f(x) > \alpha\}$ is, and continuous if and only if it is both).

Verification

technique · direct

1.1

$f$ is upper semicontinuous at $0$ : $f(0) = 1$ and $f(x) \le 1$ for every $x \in [0,1]$ , so $f(x) < f(0) + \varepsilon = 1 + \varepsilon$ for every $x \in [0,1]$ and every real $\varepsilon > 0$ ; any $\delta$ works.

1.2

$f$ is upper semicontinuous at every $c \in (0,1]$ : taking $\delta := \min\{c, \varepsilon\} > 0$ , every $x \in [0,1]$ with $|x - c| < \delta$ satisfies $x > c - \delta \ge 0$ , hence $x \ne 0$ and $f(x) = x < c + \varepsilon = f(c) + \varepsilon$ .

L1L2

1.3

$f$ is bounded below by $0$ and $f(x) > 0$ for every $x \in [0,1]$ : for $x = 0$ the value is $1 > 0$ , and for $0 < x \le 1$ the value is $x > 0$ .

2.1

$\inf f[\,[0,1]\,] = 0$ : the set $f[\,[0,1]\,]$ is nonempty and bounded below by $0$ by step 1.3, so its infimum $\ell$ exists and $\ell \ge 0$ ; and for every real $\eta > 0$ there is a natural $n \ge 1$ with $1/n < \eta$ , and then $f(1/n) = 1/n < \eta$ , so no positive real is a lower bound and $\ell = 0$ .

step 1.3L3L4

3.1

$f$ has no minimum: a minimum would be a value $f(x_{0})$ that is a lower bound of $f[\,[0,1]\,]$ , hence at most the infimum $0$ ; but every value of $f$ is strictly positive.

step 1.3step 2.1L3

4.1

So $f$ is upper semicontinuous on the nonempty compact set $[0,1]$ , is bounded below, and attains no minimum, which refutes the claim. It does attain a maximum, $f(0) = 1 \ge f(x)$ for every $x \in [0,1]$ , in agreement with the semicontinuous extreme value theorem.

step 1.1step 1.2step 3.1L5∎

Remarks

The failing hypothesis, named exactly. $f$ is not lower semicontinuous at $0$ : taking $\varepsilon := 1/2$ , every neighbourhood of $0$ contains points $x$ with $f(x) = x < 1/2 = f(0) - \varepsilon$ . Lower semicontinuity is precisely what Semicontinuous extreme value theorem: an upper semicontinuous function on a nonempty compact $K \subseteq \mathbb{R}$ is bounded above and attains a maximum, and a lower semicontinuous one is bounded below and attains a minimum requires for a minimum, and it is precisely what is missing.
Reflecting the example gives the dual failure. The function $-f$ is lower semicontinuous on $[0,1]$ , bounded above, and attains no maximum, by Upper and lower semicontinuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ ; so neither half of the theorem can be strengthened to the other extremum.

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Sources

Semi-continuity (Wikipedia) (standard reference, not scraped)

An upper semicontinuous function on [0,1][0,1][0,1] that is bounded below and attains no minimum, so the semicontinuous extreme value theorem is genuinely one-sided