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CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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A lower semicontinuous function on [0,1] with no maximum

Statement refuted

Every lower semicontinuous real-valued function on a nonempty compact set attains both a minimum and a maximum.

Facts & Assumptions

Given: Define g:[0,1]R by g(0)=1 and g(x)=x for 0<x1, with the one-dimensional convention identified by Upper and lower semicontinuity on subsets of Rn, Euclidean semicontinuity agrees with the published real-line definition.

[L1]

Lower semicontinuity is equivalent to relative openness of every strict superlevel set and to relative closedness of every weak sublevel set (Semicontinuity on Rn is characterized by strict open level sets and weak closed level sets).

[L2]

Every lower semicontinuous real-valued function on a nonempty compact Euclidean set is bounded below and attains a minimum (Semicontinuous extreme value theorem on compact Euclidean sets).

Counterexample

technique · direct
1.1

The weak sublevel {gα} is empty for α<1, equals {0,1} at α=1, equals {0}[α,1] for 1<α<0, and is all of [0,1] for α0. Each is relatively closed, so [L1] makes g lower semicontinuous.

L1givenalgebra
2.1

Every value of g is negative, but g(1/N)=1/N approaches zero through positive natural N. Hence supg([0,1])=0 and the supremum is not attained, so g has no maximum.

step 1.1algebra
3.1

The function does attain its minimum 1 at zero, as [L2] requires. It is only the unsupported opposite extremum that fails.

step 2.1L2

Depends on

Used by

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Dependency tree · two levels

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Sources