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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

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    $\begingroup$ Both the multiplication and the inversion in the group are continuous. This means that the map that multiplies by a fixed element from the group is in fact a homeomorphism, so restricting it to any open subset gives a homeomorphism onto the image. $\endgroup$ – Tobias Kildetoft Oct 12 '12 at 9:34
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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 $U\subseteq G$.

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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.

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    $\begingroup$ It’s right, but you probably should point out that there’s quite a bit of work hidden in the jump from multiplication is continuous to $u\mapsto g^{-1}u$ is continuous. Multiplication, after all, is a function from $G\times G$ to $G$, while you’re looking at a function from $G$ to $G$. $\endgroup$ – Brian M. Scott Oct 12 '12 at 9:45
  • $\begingroup$ @BrianM.Scott Thanks, I am sorry I forgot about that. $\endgroup$ – lka Oct 12 '12 at 9:48

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