I know that this is a very easy exercise from Lee's SM book, but I don't know how to start. here is the question:
Q: Suppose a connected topological group $G$ acts continuously on a discrete space $K$. Show that the action is trivial.
If $g\in G$ s.t. $m(g,k)=gk\neq k$ then one can look at $m^{-1}(gk)$. How to proceed from here? please don't answer directly and a hint suffices.
If $m: G \times K \to K$ is the action, $O(k):=\{m(g,k): g \in G\} \subseteq K$, the orbit of $k$, is connected as the continuous image of $G \times \{k\} \simeq G$, and so must be a singleton, as the only connected non-empty subsets of a discrete space are singletons. Ergo $m(g,k)=k = m(1,k)$ for any $g,k$.