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Minimality of group action on compact topological space implies that finite orbit of any non-empty open set covers the space
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I have that X is a compact topological space and G is a group that acts on X minimally in the sense that $forall$ x $in$X, $overlinecup_gin G (g.x)$ = X.
I want to show that given any non-empty open subset U of X, I can find finitely many elements g$_1$, g$_2$, ... , g$_n$ of G such that $cup_i=1^n$ (g.U) = X.
It seems to me that since orbit of any element comes arbitrarily close to any other element, somehow orbit of any open set should come arbitrarily close to any open set. If I could show that orbit of open set U actually forms an open set (may be around every point), then I could invoke compactness. However, that seems as if I am trying to draw a stronger conclusion, in which case, I have no idea how to approach this problem.
algebraic-topology dynamical-systems group-actions
asked Jul 14 at 18:41
HumbleStudent
643311
add a comment |Â
up vote
2
down vote
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I have that X is a compact topological space and G is a group that acts on X minimally in the sense that $forall$ x $in$X, $overlinecup_gin G (g.x)$ = X.
I want to show that given any non-empty open subset U of X, I can find finitely many elements g$_1$, g$_2$, ... , g$_n$ of G such that $cup_i=1^n$ (g.U) = X.
It seems to me that since orbit of any element comes arbitrarily close to any other element, somehow orbit of any open set should come arbitrarily close to any open set. If I could show that orbit of open set U actually forms an open set (may be around every point), then I could invoke compactness. However, that seems as if I am trying to draw a stronger conclusion, in which case, I have no idea how to approach this problem.
algebraic-topology dynamical-systems group-actions
asked Jul 14 at 18:41
HumbleStudent
643311
3
Can you see how $gcdot U$ is open ? Can you conclude if you know that $X=Gcdot U$ ? Can you prove that $X=Gcdot U$ ?
– Max
Jul 14 at 18:46
to conclude that g.U is open for every open U is like saying that group acts continuously i.e. for any g $in$ Gthe map from X $rightarrow$ X given by x mapping to g.x is continuous. However, I don't have stuff like that given.
– HumbleStudent
Jul 14 at 18:59
1
You probably do actually; either it's implicit, or it's written somewhere. Otherwise it's probably false
– Max
Jul 14 at 20:04
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have that X is a compact topological space and G is a group that acts on X minimally in the sense that $forall$ x $in$X, $overlinecup_gin G (g.x)$ = X.
I want to show that given any non-empty open subset U of X, I can find finitely many elements g$_1$, g$_2$, ... , g$_n$ of G such that $cup_i=1^n$ (g.U) = X.
It seems to me that since orbit of any element comes arbitrarily close to any other element, somehow orbit of any open set should come arbitrarily close to any open set. If I could show that orbit of open set U actually forms an open set (may be around every point), then I could invoke compactness. However, that seems as if I am trying to draw a stronger conclusion, in which case, I have no idea how to approach this problem.
algebraic-topology dynamical-systems group-actions
asked Jul 14 at 18:41
HumbleStudent
643311
I have that X is a compact topological space and G is a group that acts on X minimally in the sense that $forall$ x $in$X, $overlinecup_gin G (g.x)$ = X.
I want to show that given any non-empty open subset U of X, I can find finitely many elements g$_1$, g$_2$, ... , g$_n$ of G such that $cup_i=1^n$ (g.U) = X.
It seems to me that since orbit of any element comes arbitrarily close to any other element, somehow orbit of any open set should come arbitrarily close to any open set. If I could show that orbit of open set U actually forms an open set (may be around every point), then I could invoke compactness. However, that seems as if I am trying to draw a stronger conclusion, in which case, I have no idea how to approach this problem.
algebraic-topology dynamical-systems group-actions
asked Jul 14 at 18:41
HumbleStudent
643311
asked Jul 14 at 18:41
HumbleStudent
643311
asked Jul 14 at 18:41
HumbleStudent
643311
asked Jul 14 at 18:41
HumbleStudent
643311
643311
3
Can you see how $gcdot U$ is open ? Can you conclude if you know that $X=Gcdot U$ ? Can you prove that $X=Gcdot U$ ?
– Max
Jul 14 at 18:46
to conclude that g.U is open for every open U is like saying that group acts continuously i.e. for any g $in$ Gthe map from X $rightarrow$ X given by x mapping to g.x is continuous. However, I don't have stuff like that given.
– HumbleStudent
Jul 14 at 18:59
1
You probably do actually; either it's implicit, or it's written somewhere. Otherwise it's probably false
– Max
Jul 14 at 20:04
add a comment |Â
3
Can you see how $gcdot U$ is open ? Can you conclude if you know that $X=Gcdot U$ ? Can you prove that $X=Gcdot U$ ?
– Max
Jul 14 at 18:46
to conclude that g.U is open for every open U is like saying that group acts continuously i.e. for any g $in$ Gthe map from X $rightarrow$ X given by x mapping to g.x is continuous. However, I don't have stuff like that given.
– HumbleStudent
Jul 14 at 18:59
1
You probably do actually; either it's implicit, or it's written somewhere. Otherwise it's probably false
– Max
Jul 14 at 20:04
3
3
Can you see how $gcdot U$ is open ? Can you conclude if you know that $X=Gcdot U$ ? Can you prove that $X=Gcdot U$ ?
– Max
Jul 14 at 18:46
Can you see how $gcdot U$ is open ? Can you conclude if you know that $X=Gcdot U$ ? Can you prove that $X=Gcdot U$ ?
– Max
Jul 14 at 18:46
to conclude that g.U is open for every open U is like saying that group acts continuously i.e. for any g $in$ Gthe map from X $rightarrow$ X given by x mapping to g.x is continuous. However, I don't have stuff like that given.
– HumbleStudent
Jul 14 at 18:59
to conclude that g.U is open for every open U is like saying that group acts continuously i.e. for any g $in$ Gthe map from X $rightarrow$ X given by x mapping to g.x is continuous. However, I don't have stuff like that given.
– HumbleStudent
Jul 14 at 18:59
1
1
You probably do actually; either it's implicit, or it's written somewhere. Otherwise it's probably false
– Max
Jul 14 at 20:04
You probably do actually; either it's implicit, or it's written somewhere. Otherwise it's probably false
– Max
Jul 14 at 20:04
add a comment |Â
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3
Can you see how $gcdot U$ is open ? Can you conclude if you know that $X=Gcdot U$ ? Can you prove that $X=Gcdot U$ ?
– Max
Jul 14 at 18:46
to conclude that g.U is open for every open U is like saying that group acts continuously i.e. for any g $in$ Gthe map from X $rightarrow$ X given by x mapping to g.x is continuous. However, I don't have stuff like that given.
– HumbleStudent
Jul 14 at 18:59
1
You probably do actually; either it's implicit, or it's written somewhere. Otherwise it's probably false
– Max
Jul 14 at 20:04