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Topological-domain equicontinuity agrees with metric equicontinuity on a metric domain
Statement
Let and be metric spaces, give its metric topology, and let . Then 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 and a family .
Topological-domain equicontinuity requires, for each and , one neighbourhood of on which every satisfies (Equicontinuity on a topological domain and pointwise relative compactness).
Metric equicontinuity requires, for each and , one such that implies for all (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces).
Every metric neighbourhood of contains a positive-radius open ball about (The balls , , form a countable neighbourhood base at , so every metric space is first countable).
Proof
Suppose [L1] holds, and fix and . Choose its common neighbourhood ; by [L3], some ball lies in . The same works for every , so [L2] holds.
Conversely suppose [L2] holds. For fixed and , let be the common radius supplied there. The open neighbourhood then satisfies [L1] for every .
Steps 1.1 and 1.2 prove both directions without changing the order of the family quantifier.
Depends on
- Equicontinuity on a topological domain and pointwise relative compactness
- Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces
- The balls $B(x, 1/n)$, $n \ge 1$, form a countable neighbourhood base at $x$, so every metric space is first countable
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
- Topology, second edition, Section 45 (standard reference, not scraped)