On the group Γ of all homeomorphisms of a locally compact space X, the topology Tβ is the coarsest for which the mappings u→u and u→u−1 of Γ into Cc(X;X) are continuous.
Let us denote the latter topology for the moment by C′. Since u→u−1 is continuous with respect to Tβ and since Tβ is finer than the topology of compact convergence, it is clear that Tβ is finer than C′. To prove the converse, endow X′ with its unique uniformity; let u0∈Γ and let V be an entourage of X′; then we have to prove that there is a compact subset K of X and a symmetric entourage W of X′ such that the relations
u∈Γ, (u0(x),u(x))∈W and (u0−1(x),u−1(x))∈W for all x∈K
imply
(u0(x),u(x))∈V for all x∈X.
Let V1 be a symmetric open entourage of X′ such that V21⊂V; then K1=X′−V1(ω) is a compact subset of X. Choose a symmetric open entourage W of X′ such that W⊂V and W(ω)∩W(u0−1(K1))=∅; this is possible by Proposition 4 of Chapter II, § 4, no. 3. Let K2=X′−W(ω), which is a compact subset of X. We shall see that W and the compact set K=K1∪K2 do what is required. Since W⊂V, it is enough to show that the relation
(u0−1(x),u−1(x))∈W for all x∈K1(u∈Γ)
implies that
(u(y),ω)∈V1 for all y∈W(ω);
for we shall then also have (u0(y),ω)∈V1 and thence
(u0(y),u(y))∈V21⊂V
for all y∈W(ω)=X′−K2. Now if we had y∈W(ω) and
u(y)∈X′−V1(ω)=K1,
it would follow that y∈u−1(K1)⊂W(u0−1(K1)), contrary to the choice of W; the proof is therefore complete.
In general the group Γ, endowed with Tβ, is not locally compact; but we have the following criterion:
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.