$VV^{-1} \subseteq U$

Question

If $U$ is a neighbourhood of $e$ (identity) in a topological group,show there is a neighbourhood $V$ of $e$ for which $VV^{-1} \subseteq U$.

It's exercise from Basic topology,Armstrong,Promblem 18 of Chapter 4,and my approach get a little problem,so can you help me???

score 2 · Accepted Answer · answered Oct 02 '18 at 14:05

2

HINT

Use the fact that the map $G\times G\to G$ given by $(g,h)\mapsto gh^{-1}$ is continuous.

answered Oct 02 '18 at 14:05

Aweygan

:but it's not $V \times V$. – Daniel Xu Oct 02 '18 at 14:12
@DanielXu Write out what continuity at $(e,e)$ means for that function! – Henno Brandsma Oct 02 '18 at 14:23
@Henno Brandsma There is a neighbourhood of (e,e) whose image is U? – Daniel Xu Oct 02 '18 at 14:27
@DanielXu there is a neighbourhood $O_1 \times O_2$ of $(e,e)$ whose image lies inside $U$. Now take $V= O_1 \cap O_2$. – Henno Brandsma Oct 02 '18 at 14:28
@Henno Brandsma oh,i get it,thank you. – Daniel Xu Oct 02 '18 at 14:30

1 Answers1