property of a topological group??

Hello;

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

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

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

I understand number 1 is due to first countability of $\displaystyle G$, 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.

Re: property of a topological group??

Did you try to use the continuity of the map $\displaystyle f\colon G^3\to G$, $\displaystyle f(x,y,z)=xyz$?

Re: property of a topological group??

Ok, thanks a lot my instructor. Now, I understand the concept.

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

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