Page $37$ of this book, chapter focus on topological dynamics. They here reference a fact from topology that I am unfamiliar with:
Since the projection $\pi:H\to K$ is continuous, and $H$ is compact, $\pi(\overline{A})=\overline{\pi(A)}$
The set $A$ in question is the forward orbit of a point $(x_0,1)\in H$, under a continuous dynamic. $H$ is a product set $K\times G$, where $G$ is a compact group and $K$ is a compact topological space.
My only thoughts:
The closure of $A$ is of course a closed set, and closed subsets of compact spaces are themselves compact. Perhaps if we cover the forward orbit by open neighbourhoods of every point in the orbit, which is somehow meaningful. I know that $\pi(\overline{A})\subseteq\overline{\pi(A)}$ since $\pi$ is continuous, but where compactness is relevant, and how to show the reverse direction, is beyond me.
Best Answer
The set $\overline{A}$ is compact as the closed subset of a compact space. Continuous functions map compact sets to compact sets. So $\pi(\overline{A})$ is compact. At least if $Y$ is a Hausdorff space, $\pi(\overline{A})$ is then also closed. So $\pi(\overline{A})$ is a closed set containing $\pi(A)$. Since $\overline{\pi(A)}$ is the smallest closed set containing $\pi(A)$, we have $\overline{\pi(A)}\subseteq \pi(\overline{A})$.