Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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All constant functions form an equicontinuous family that is not pointwise bounded

Statement refuted

Refuted: equicontinuity alone implies pointwise boundedness.

Facts & Assumptions

Given: F={fc:c∈R}\mathcal F=\{f_c:c\in\mathbb R\} on a nonempty metric space, where fc(x)=cf_c(x)=c.

[L1]

Equicontinuity and pointwise boundedness have the meanings of Equicontinuity, pointwise boundedness, and uniform boundedness for families in C(K,R)C(K,\mathbb R).

Counterexample

Proof

technique · direct
1.1

For every cc and all x,yx,y, ∣fc(x)−fc(y)∣=0|f_c(x)-f_c(y)|=0, so any positive δ\delta works simultaneously for every member of F\mathcal F.

givenL1algebra
1.2

At each fixed point aa, {fc(a):c∈R}=R\{f_c(a):c\in\mathbb R\}=\mathbb R, which is unbounded.

givenL1algebra
2.1

Hence this family is equicontinuous but not pointwise bounded.

step 1.1step 1.2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 6 results over 3 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources