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I'm having some trouble proving the following proposition:~

Let $(G,\cdot)$ be a topological group, $x,y\in G$ and $W$ an open neighbourhood of $x\cdot y$. Then, there are open neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that$$U\cdot V\subseteq W$$Where $U\cdot V\subseteq W:=\{u\cdot v:u\in U,v\in V\}$

I tried the following:

Let $H=\cdot^{-1}(W)\subseteq G\times G$, then $(x,y)\in H$. Because the projection map is open in the product topology, $p_1(H)$ and $p_2(H)$ are open neighbourhoods of $x$ and $y$. After this I wasn't able to conclude anything.

How can this be done?

Eduardo Magalhães
  • 4,878

1 Answers1

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The map $f:G\times G\to G$ given by multiplication $f(g,h)=gh$ is continuous. So $f^{-1}(W)$ is an open set in the product topology on $G\times G$, and it contains the point $(x,y)$. Hence by definition of the product topology there are open sets $U,V\subseteq G$ such that $(x,y)\in U\times V\subseteq f^{-1}(W)$. In particular, if $u\in U, v\in V$ then $(u,v)\in U\times V\subseteq f^{-1}(W)$, and so:

$uv=f(u,v)\in W$

Mark
  • 39,605