Did you try to use the continuity of the map , ?
Hello;
Show that if is a first countable topological group, then there is a sequence of symmetric neighborhoods of the neutral element of such that
(1){ } is a local base at in G
(2) , for every
I understand number 1 is due to first countability of , 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.
Ok, thanks a lot my instructor. Now, I understand the concept.
Let be any neighborhood of the identical element in . By continuity of there is a neighborhood of such that . By continuity of there is a neighborhood of such that . That's mean any neighborhood of contains a symmetric neighborhood. Now is a symmetric neighborhood of and . Because of that, we can construct such a sequence of symmetric neighborhood of .
Thank you very much for the very helpful guidance.