Let be a sequence in converging to a point , where , and . Since is compact, there is a subsequence of that converges to some . Since also converges (its a subsequence of a convergent sequence), it follows that also converges, say to ( is closed). Hence converges to , which implies that (every subsequence of a convergent sequence converges to the same limit). Since , it follows that , which means that is closed.
Formally it's 3-tuple such that is a topological space and a group, with the condition that the map g,h)\mapsto gh^{-1}" alt="\alpha:G\times G\to Gg,h)\mapsto gh^{-1}" /> is continuous.
The result. That's why I was talking nonsense about hard lemmas and such. You need that for the general theorem.You mean the proof is also valid generally? Or just the result?