Hello;
Show that ifis a first countable topological group, then there is a sequence of symmetric neighborhoods
of the neutral element
of
(1){} is a local base at
(2), for every
I understand number 1 is due to first countability of
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.
Letbe any neighborhood of the identical element
there is a neighborhood
of
. By continuity of
there is a neighborhood
of
. That's mean any neighborhood of
is a symmetric neighborhood of
. Because of that, we can construct such a sequence of symmetric neighborhood of
Thank you very much for the very helpful guidance.
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