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PropositionStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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Topological-domain equicontinuity agrees with metric equicontinuity on a metric domain

Statement

Let (X,dX) and (Y,dY) be metric spaces, give X its metric topology, and let FC(X,Y). Then F is equicontinuous in the topological-domain sense if and only if it is equicontinuous in the published metric epsilon-delta sense.

Facts & Assumptions

Given: Metric spaces X,Y and a family FC(X,Y).

[L1]

Topological-domain equicontinuity requires, for each x and ε>0, one neighbourhood U of x on which every fF satisfies dY(f(y),f(x))<ε (Equicontinuity on a topological domain and pointwise relative compactness).

[L2]

Metric equicontinuity requires, for each x and ε>0, one δ>0 such that dX(x,y)<δ implies dY(f(x),f(y))<ε for all fF (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces).

[L3]

Every metric neighbourhood of x contains a positive-radius open ball about x (The balls B(x,1/n), n1, form a countable neighbourhood base at x, so every metric space is first countable).

Proof

technique · direct
1.1

Suppose [L1] holds, and fix xX and ε>0. Choose its common neighbourhood U; by [L3], some ball B(x,δ) lies in U. The same δ works for every fF, so [L2] holds.

L1L3
1.2

Conversely suppose [L2] holds. For fixed x and ε>0, let δ>0 be the common radius supplied there. The open neighbourhood B(x,δ) then satisfies [L1] for every fF.

L1L2
2.1

Steps 1.1 and 1.2 prove both directions without changing the order of the family quantifier.

step 1.1step 1.2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 84 results over 16 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