Bourbaki

Proposition 5

Tag 0CCF, permanent. Cite it as [Bourbaki, Tag 0CCF].

General Topology, chapter X, § 2. Equicontinuous sets

Let G,G′G, G' be two topological groups endowed with their left uniformities, and let HH be a set of homomorphisms of GG into G′G'. Then the following conditions are equivalent: a) HH is equicontinuous at the identity element ee of GG, b) HH is equicontinuous, c) HH is uniformly equicontinuous.

It is enough to show that a) implies c). Let V′V' be a neighbourhood of the identity element e′e' of G′G'; then, by hypothesis, there is a neighbourhood VV of ee in GG such that u(V)⊂V′u(V) \subset V' for all u∈Hu \in H; since the elements of HH are homomorphisms, the relation x−1y∈Vx^{-1} y \in V implies that we have

(u(x))−1u(y)=u(x−1y)∈V′. (u(x))^{-1} u(y) = u(x^{-1} y) \in V'.
In view of the definition of the entourages of the left uniformities of GG and G′G' (Chapter III, § 3, no. 1), the result follows.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.