I am stuck on how to prove the statement
"A continuous action of a compact group $G$ on a uniform space $X$ is equicontinuous."
So, essentially we want to show that for every entourage $\alpha$ of $X$, and every $y\in X$, there exists a neighbourhood $U$ of $y$ such that for each $t\in G$, and $x\in U$, $$(tx,ty) \in \alpha$$
I am confused on how to prove this and on where the compactness comes in. So far, I have only managed to show the existence of open sets $V\subset G$, and a neighbourhood $U$ of $y$ such that for all $t\in V$, $x\in U$, $$(tx, ty)\in \alpha$$
Any help is much appreciated!
Thanks.
Let $C_u(X,X)$ be the uniform space of continuous functions on $X$ with respect to uniform limit. In other words, entourages contain sets of the form $$\mathcal{U}(\alpha) = \{(f,g): \; (f(x),g(x)) \in \alpha \; \mathrm{for} \; \mathrm{all} \; x \in X\}.$$ The set $\Lambda$ of multiplications $x \mapsto t \cdot x$ with $t \in G$ is a compact subset.
Fix an entourage $\alpha$ and choose another entourage $\beta$ such that $$(w,x),(x,y),(y,z) \in \beta \; \Longrightarrow \; (w,z) \in \alpha.$$ In other words $\beta^{(3)} \subseteq \alpha.$ For any $t \in G,$ define $$W_t := \mathcal{U}(\beta)[t] = \{f \in C_u(X,X): \; (f(x),t \cdot x) \in \beta \; \mathrm{for} \; \mathrm{all} \; x \in X\}.$$ Sets of this form always have nonempty interior (if $\gamma^{(2)} \subseteq \beta$ then $\mathcal{U}(\gamma)[t] \subseteq \mathrm{Int}(W_t)$), and it is clear that they cover $\Lambda.$
By compactness there is a finite cover $$\Lambda \subseteq \bigcup_{i=1}^n W_{t_i}.$$
Fix an arbitrary $x$. Then the set $$U := \bigcap_{i=1}^n t_i^{-1} \cdot \beta[t_i \cdot x] = \{y \in X: \, (t_i \cdot x,t_i \cdot y) \in \beta \; \mathrm{for} \; \mathrm{all} \; i\}$$ is an open neighborhood of $x$ in $X$. For any $y \in U$ and $t \in G$, choose an index $i$ with $t \in W_{t_i};$ then $$(t \cdot x, t_i \cdot x), (t_i \cdot x, t_i \cdot y), (t_i \cdot y, t \cdot y) \in \beta$$ and therefore $(t \cdot x, t \cdot y) \in \alpha.$