Open subsets in topological groups I'm starting to study topological groups, and I noticed that every single theorem in topological groups I have to use the following statement:
Let $G$ be a topological group and U an open subset of G, if $g\in G$, then $gU$ is an open subset of G.
I can't prove it, please anyone can help me please.
Thanks
 A: In fact the map $\varphi_g:G\to G:x\mapsto gx$ is a homeomorphism. It’s clearly a bijection, since $\varphi_g^{-1}=\varphi_{g^{-1}}$. To see that it’s continuous, let $U\subseteq G$ be open. The group operation is continuous, so $V=\{\langle x,y\rangle\in G\times G:xy\in U\}$ is open in $G\times G$. Let $\pi:G\times G\to G:\langle x,y\rangle\mapsto y$, and let $G_g=\{g\}\times G$; $\pi\upharpoonright G_g:G_g\to G$ is a homeomorphism, and $V\cap G_g$ is open in $G_g$, so $\pi[V\cap G_g]$ is open in $G$. But 
$$\pi[V\cap G_g]=\{x\in G:\langle g,x\rangle\in V\}=\{x\in G:gx\in U\}=\varphi_g^{-1}[U]\;,$$
so $\varphi_g^{-1}[U]$ is open in $G$, and $\varphi_g$ is continuous. Since $g$ was arbitrary, it follows immediately that $\varphi_g^{-1}=\varphi_{g^{-1}}$ is also continuous and hence that $\varphi_g$ is a homeomorphism. In particular, then, $gU=\varphi_g[U]$ is open for every $g\in G$ and open $U\subseteq G$.
A: Multiplication is continuous, so the map 
$g^{-1}:G\to G, u\mapsto g^{-1}u$ is continuos. Thus the preimage of an open set $U\subset G$ is open, and the preimage equals $gU$.
Hope I got it right.
A: It is enough to show that $\varphi_g: G \rightarrow G, x \mapsto gx$ is a homeomorphism. Then obviously an open set maps to an open set.
$\varphi_g$ is invertible with inverse $\varphi_{g^{-1}}$. Moreover, it is enough to show $\varphi_g$ is continuous for all $g\in G$, because then $\varphi_{g^{-1}}$ is also continuous.
Now $\psi_g: G \rightarrow G \times G, x \mapsto (g,x)$ is continuous w.r.t. the product topology (which is obvious from the universal property of the product topology, since the two projections $\pi_1: G\times G \rightarrow G, (g,x) \mapsto g$ and $\pi_2=\operatorname{id}_G$ are continuous). Then $\varphi_g$ is continuous as composition of the continuous map $\psi_g$ with the continuous group multiplication.
Edit: I cannot comment yet, so I have to make this an answer although it is just a remark to the solution of Brian. On request, I repeated the content of Brian's answer to make this a complete answer.
