Action of a compact Lie group is a proper action

Question

I want to show that the usual action of $U(1)$ on $S^3$ given by $e^{i\theta}\cdot(z_1,z_2)=(e^{i\theta}z_1,e^{i\theta}z_2)$ is a proper action.

It seems that the following more general statement should be true

Any action of a compact Lie group is a proper action

Recall that the action of a topological group $G$ on $X$ is a map $G\times X\rightarrow X$ given by $(g,x)=g\cdot x$. The action is said to be proper if the map $G\times X \rightarrow X \times X$ given by $(g,x)\mapsto(g\cdot x,x)$ is a proper, that is the preimage of every compact set is compact.

I have having trouble proving this: Let $U\subset X\times X$ be compact.The projections $\pi_1,\pi_2:X\times X\rightarrow X$ onto the first and second coordinates are continuous. Therefore $\pi_1(U)$ and $\pi_2(U)$ are compact in $X$.

I am unsure what the preimage of $U$ is however. Obviously it is $\{(g,x)\in G \times X \text{ such that }(g\cdot x,x) \in U\}$ But I do not know how to continue.

Answer 1

Let $f:G\times X\rightarrow X\times Y$ defined by $f(g,x)=(gx,x)$ and $U$ be a compact subset of $X\times X$, $\pi_2(U)=V$ is compact since $\pi_2$ continuous. $f^{-1}(U)\subset G\times V$ and is closed thus it is compact since it is a closed subset of a compact set.

Answer 2

Knowing that the action is continuous then the orbit map $\mathcal{O}_x:G\to X$ is continuous. Therefore assuming $U\subseteq X$ is compact and $X$ is Hausdorff (this is generally assumed for most applications, but please specify if this is not allowed) then we find that $U$ is closed in $X$. Therefore $\mathcal{O}_x^{-1}(U) \subseteq G$ is closed, and thus compact since $G$ is compact. Therefore $\mathcal{O}_x^{-1}(U)\times U$ will be compact in $G\times X$.