Common open neighborhood as preimage of open neighborhood under continuous group action Let G be a separable, metrisable, Hausdorff topological group acting on a compact Hausdorff space X.
Let $f: X \rightarrow \mathbb{C}$ be a continuous complex valued function on X. I am trying to reason why there is a neighhborhood of the identity $V\ni 1_G$ such that $\forall g \in V, \ |f(g\cdot x)- f(x)|<\epsilon \ \forall x \in X$.
It is obvious from definition that we have such an open neighbourhood $V_x \forall x\in X$. However I am not at all sure why we have one that works $\forall x$.
I think it has to do with compactness. But we would here have to be taking the intersection of open neighborhoods rather than closed so I don't see how this would work.
 A: Suppose to the contrary, there exists $\epsilon>0$, a sequence $\{g_n\}$ of elements of $G$ convergent to $1_G$, and a sequence 
$\{x_n\}$ of points of $X$ such that $|f(g_n\cdot x_n)-f(x_n)|\ge 2\epsilon$ for each $n$. Since the space $X$ is compact, the sequence $\{x_n\}$ has a cluster point $x^*$. Since the function $f$ is continuous, there exists a neighborhood $U$ of the point $x^*$ such that $|f(x’)-f(x^*)|<\epsilon$ for each $x’\in U$. Now we assume that the action $G\times X:\to X$, $(g,x)\mapsto g(x)$ is a continuous map. Therefore there exist a neighborhood $V$ of $1_G$ and a neighborhood $W\subset U$ of $x$ such that $\{g\cdot x: g\in V, x\in W\}\subset U$. Since the sequence $\{g_n\}$ converges to $1_G$, there exists $N$ such that $g_n\in V$ for each $n>N$. Since $x^*$ is a cluster point of the sequence $\{x_n\}$, there is $n>N$ such that $x_n\in W$. Then $$|f(g_n\cdot x_n)-f(x_n)|\le |f(g_n\cdot x_n)-f(x^*)|+|f(x^*)-f(x_n)|\le\epsilon+\epsilon=2\epsilon,$$ a contradiction.  
