Let G be a locally compact group, and let U be an open set in G. Show that Ug is an open set in G for all g in G.

Question

Let $G$ be a locally compact group. Let $U \subseteq G$ be open. Show that $Ug \subseteq G$ is open for all $g \in G$.

I'm really not sure how to prove this. I think it will involve the condition on a topology that any union of open sets is open, and any finite intersection of open sets is open, but I don't know how to express $Ug$ as a union/intersection of open sets in $G$. I would appreciate a hint!

Answer 1

Hint: In a topological group operations are continuous. In particular, $t_g:G\rightarrow G$ given by $t_g(x) = xg$ is continuous. Similarly with $t_{g^{-1}}$. Next $$t_g\cdot t_{g^{-1}}=?,\,t_{g^{-1}}\cdot t_g = ?$$

Remark. You don't need local compactness.

Answer 2

The map $f_g: G \to G$ defined by $f_g(x)=xg$ is a homeomorphism:

• continuous as multiplication $G \times G \to G$ is continuous and we just restrict to second element to $g$.
• Its inverse is of the same form $f_{g'}$ with $g'=g^{-1}$ by simple group laws.

So $f_g$ is an open map and so $f_g[U]=Ug$ is open for any open $U \subseteq G$. This is all true in any topological group.