Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-02
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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} on a nonempty metric space, where fc(x)=c.

[L1]

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

Counterexample

Proof

technique · direct
1.1

For every c and all x,y, ∣fc(x)−fc(y)∣=0, so any positive δ works simultaneously for every member of F.

givenL1algebra
1.2

At each fixed point a, {fc(a):c∈R}=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 · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources