Results 1 to 3 of 3

Thread: property of a topological group??

  1. #1
    Member
    Joined
    Feb 2011
    Posts
    81

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    678
    Thanks
    32

    Re: property of a topological group??

    Did you try to use the continuity of the map $\displaystyle f\colon G^3\to G$, $\displaystyle f(x,y,z)=xyz$?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2011
    Posts
    81

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Topological group/locally compact subgroup
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Nov 6th 2010, 05:59 PM
  2. [SOLVED] A question in topological group
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Sep 16th 2010, 11:55 PM
  3. Proof of Topological Property
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Feb 3rd 2010, 02:18 PM
  4. Abelian Group Property
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Oct 8th 2009, 05:53 PM
  5. Topological property
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 30th 2007, 01:58 PM

Search Tags


/mathhelpforum @mathhelpforum