# property of a topological group??

• Oct 8th 2011, 12:05 AM
student2011
property of a topological group??
Hello;

Show that if $G$ is a first countable topological group, then there is a sequence of symmetric neighborhoods $(U_{n})_{n}$ of the neutral element $e$ of $G$ such that

(1){ ${U_{n}:n\in Z }$} is a local base at $e$ in G

(2) $U_{n+1}^{3}\subset U_{n}$, for every $n$

I understand number 1 is due to first countability of $G$, but how can we get (2). Please guide me. Every comment or guidance is highly appreciated

Recall that a neighborhood of the identity element is said to be symmetric if it equal to its inverse.
• Oct 8th 2011, 01:40 AM
girdav
Re: property of a topological group??
Did you try to use the continuity of the map $f\colon G^3\to G$, $f(x,y,z)=xyz$?
• Oct 9th 2011, 10:28 PM
student2011
Re: property of a topological group??
Ok, thanks a lot my instructor. Now, I understand the concept.

Let $U$ be any neighborhood of the identical element $e$ in $G$. By continuity of $(x,y,z)\Rightarrow xyz$ there is a neighborhood $W_{1}$ of $e$ such that $W_{1}^{3}\subset U$. By continuity of $(x,y) \Rightarrow xy^{-1}$ there is a neighborhood $W_{2}$ of $e$ such that $W_{2}W_{2}^{-1} \subset W_{1}$. That's mean any neighborhood of $e$ contains a symmetric neighborhood. Now $V=W_{2}W{2}^{-1}$ is a symmetric neighborhood of $e$ and $V^{3} \subset W_{1}^{3} \subset U$. Because of that, we can construct such a sequence of symmetric neighborhood of $e$.

Thank you very much for the very helpful guidance.(Clapping)