connected topological group is generated by any neighborhood of identity

Theorem - Let $G$ be a connected topological group and $e$ its identity element. If $U$ is any open neighborhood of $e$ , then $G$ is generated by $U$ .

$\\,$

Proof: Let $U$ be an open neighborhood of $e$ . For each $n\\in\\mathbb{N}$ we denote by $U^{n}$ the set of elements of the form $u_{1}\\dots u_{n}$ , where each $u_{i}\\in U$ . Let $W:=\\bigcup_{n\\in\\mathbb{N}}U^{n}$ .

Since each $U^{n}$ is open (see this entry (http://planetmath.org/BasicResultsInTopologicalGroups) - 3), we have that $W$ is an open set. We now see that it is also closed.

Let $g\\in\\overline{W}$ , the closure of $W$ . Since $gU^{-1}$ is an open neighborhood of $g$ , it must intersect $W$ . Thus, let $h\\in W\\cap gU^{-1}$ .

•
Since $h\\in gU^{-1}$ , then $h=gu^{-1}$ for some element $u\\in U$ .
•
Since $h\\in W$ , then $h\\in U^{n}$ for some $n\\in\\mathbb{N}$ , i.e. $h=u_{1}\\dots u_{n}$ with each $u_{i}\\in U$ .

We then have $g=u_{1}\\dots u_{n}u$ , i.e. $g\\in U^{n+1}\\subseteq W$ . Hence, $W$ is closed.

Since $G$ is connected and $W$ is open and closed, we must have $W=G$ . This means that $G$ is generated by $U$ . $\\square$

单词	ConnectedTopologicalGroupIsGeneratedByAnyNeighborhoodOfIdentity
释义	connected topological group is generated by any neighborhood of identity Theorem - Let $G$ be a connected topological group and $e$ its identity element. If $U$ is any open neighborhood of $e$ , then $G$ is generated by $U$ . $\\,$ Proof: Let $U$ be an open neighborhood of $e$ . For each $n\\in\\mathbb{N}$ we denote by $U^{n}$ the set of elements of the form $u_{1}\\dots u_{n}$ , where each $u_{i}\\in U$ . Let $W:=\\bigcup_{n\\in\\mathbb{N}}U^{n}$ . Since each $U^{n}$ is open (see this entry (http://planetmath.org/BasicResultsInTopologicalGroups) - 3), we have that $W$ is an open set. We now see that it is also closed. Let $g\\in\\overline{W}$ , the closure of $W$ . Since $gU^{-1}$ is an open neighborhood of $g$ , it must intersect $W$ . Thus, let $h\\in W\\cap gU^{-1}$ . • Since $h\\in gU^{-1}$ , then $h=gu^{-1}$ for some element $u\\in U$ . • Since $h\\in W$ , then $h\\in U^{n}$ for some $n\\in\\mathbb{N}$ , i.e. $h=u_{1}\\dots u_{n}$ with each $u_{i}\\in U$ . We then have $g=u_{1}\\dots u_{n}u$ , i.e. $g\\in U^{n+1}\\subseteq W$ . Hence, $W$ is closed. Since $G$ is connected and $W$ is open and closed, we must have $W=G$ . This means that $G$ is generated by $U$ . $\\square$
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