- #1
Lie
- 15
- 0
Hello!
Could anyone help me to resolve the impasse below?
Th: Let G be a topological group and H subgroup of G. If H and G/H (quotient space of G by H) are compact, then G itself is compact.
Proof: Since H is compact, the the natural mapping g of G onto G/H is a closed mapping. Therefore if a family S of closed subsets of G has the finite intersection property, then so does {g(F): F in S}. So that G/H is compact, then [tex]\bigcap_{F \in S} g(F) \neq \varnothing .[/tex] But as I conclude that [tex]\bigcap_{F \in S} F \neq \varnothing \; ?[/tex]
If necessary we also know that G/H is Haudorff space.
Thankful! :)
Could anyone help me to resolve the impasse below?
Th: Let G be a topological group and H subgroup of G. If H and G/H (quotient space of G by H) are compact, then G itself is compact.
Proof: Since H is compact, the the natural mapping g of G onto G/H is a closed mapping. Therefore if a family S of closed subsets of G has the finite intersection property, then so does {g(F): F in S}. So that G/H is compact, then [tex]\bigcap_{F \in S} g(F) \neq \varnothing .[/tex] But as I conclude that [tex]\bigcap_{F \in S} F \neq \varnothing \; ?[/tex]
If necessary we also know that G/H is Haudorff space.
Thankful! :)