Let $G$ be a topological group acting continuously on a Hausdorff space $X$. For $K\subset X $ compact, define \begin{equation} G_K=\{ g\in G : g K\cap K \neq \varnothing \}.\end{equation}
Is $G_K$ closed?
I suspect so, but I am not too sure of my proof, which is the following one. Denote by $\theta: G\times X \rightarrow X$ the action of $G$, and by $\theta_K: G\times K\rightarrow X $ its restriction to the subspace $G\times K$. Then $G_K =\pi_1( \theta_K^{-1}(K))$ where $\pi_1: G\times K \rightarrow G$ is the projection onto the first factor. Since $\theta_K^{-1}(K)$ is a closed subset of $G\times K$ and $\pi_1$ is a closed map, the result follows.
For context, I was wondering how to prove that the right action $\sigma$ used in the definition of an associated bundle $P\times_\rho F$, where $P$ is a principal $G$-bundle and $\rho$ an action of $G$ on $F$, is proper. I recall that $\sigma$ is given by $(p,f) \mapsto (pg, \rho(g ^{-1}) f)$ and that $P\times_\rho F$ is the orbit space of $\sigma$. Note that the action on the first factor is the usual right action of $G$ on a $G$-bundle.
It seems to me that the proof given by Jack Lee here that the right action of $G$ on a $G$-bundle is proper can be used to prove that $\sigma$ is proper too by just ignoring what is happening on $F$. This seemed strange at first. However, I then recalled that a continuous action of a topological group $G$ on a Hausdorff space $X$ is proper if and only if $G_K$ defined above is compact for any $K$ compact. Take then $X=P\times F$ and let $K$ be a compact subset of $P\times F$. For $\pi_P:P\times F\rightarrow P$, $\pi_F:P\times F\rightarrow F$ the Cartesian projections, write $K_P=\pi_P(K)$, $K_F=\pi_F(K)$. Then \begin{equation} G_K=G_{K_P}\cap G_{K_F}\end{equation} where $G_K$ is taken with respect to $\sigma$, $G_{K_P}$ with respect to the $G$ action on $P$ and $G_{K_F}$ with respect to the action $(g,f)\mapsto \rho(g^{-1}) f$ on $F$. The space $G_{K_P}$ is compact since the right action of $G$ on a principal $G$-bundle is proper. If $G_{K_F}$ is closed then $G_K$ is compact regardless of what $\rho$ is, which lead me to my question.
