# Uniform convergence over compacts subsets implies existence of a uiform convergente subsequence?

Let $$H$$ the group of all homeomorphisms of a locally compact second countable and totally bounded metric space $$X$$ onto itself, under the compact-open topology ($$X$$ is totally bounded if every sequence have Cauchy subsequence). Let $$G$$ a compact subgroup of $$H$$. Then, $$G$$ is a topological group and the natural function $$X\times G\rightarrow X$$ given by $$(g,x)\rightarrow g(x)$$ is continuos and give us a continuos action of $$G$$ on $$X$$. Since that $$G$$ is a compact topological group, given $$g\in G$$ exists a sequence of positive integers $$n_k\rightarrow \infty$$ such that $$g^{n_k}\rightarrow e$$. in particular, the sequence of function $$\{g^{n_k}\}$$ converge to indentity map of $$X$$ uniformly over compact subsets of $$X$$.

The question:

Does Exists a subsequence of $$\{g^{n_k}\}$$ with converge uniformly?

• Of course the sequence $(g^n)_n$ accumulates somewhere, but why at the identity? – LSpice Nov 27 '19 at 17:45
• The sequence $\{g^n,n\geq 1\}$ have a convergente subsequence, say $g^{n_r}\rightarrow h\in G$. as $G$ is a topological group we have that $g^{-n_r}\rightarrow h^{-1}$. If $U$ is arbitrary neighborhood of the identity $e\in G$, exists $m$ such that for $i,j>m$ implies $g^{n_i}g^{-n_j}$ is in $U$ (this is possible since that the multiplicative operation on $G$ is continuos). From this, we can contruct a sequence $g^{n_k}$ with converge to $e$. – Elizeu França Nov 27 '19 at 17:57
• I am not a specialist, so forgive me if I am wrong. I would say, consider X being the disjoint sum of a countable number of circles. Then let H the set of homeomorphisms of X which keep each circle invariant, acting on each circle by a rotation. H is compact for the compact-open topology, I believe. But for g acting by a rotation of angle 1/n on the n-th circle, any iterate of g has distance 1 to identity for the uniform topology distance, so you have no uniform convergence. – Rémi Peyre Nov 27 '19 at 18:40
• In this example, with a suitable metric the function $g$ is isometry. Thus, the set $G=\langle g\rangle$ uniformly equicontinuous. In general, if $X$ is totally bounded metric space and $G$ act on $X$ such that each $g\in G$ is uniformly continuos function and $G$ is uniformly equicontinuos, then $G$ is almost periodic (with implies the existence for each $g\in G$, of a sequence $g^{n_k}\rightarrow I_X$ uniformly; Theorem 4.38 Topological Dynamics of the Gottschalk and Hedlund). – Elizeu França Nov 28 '19 at 17:17
• In this example, $X$ is not totally bounded. The case of my interest $X$ is totally bounded metric space. Sorry for omitting this hypothesis, I will edit the post. Thank you. – Elizeu França Nov 28 '19 at 17:18