Show that limit set of dynamical system is closed

saxen
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Homework Statement



Define the w-limit set (omega) of a point. Show that w(x) is closed.

Homework Equations



The Attempt at a Solution



The definition of a limit set is the set of points to which there exists a sequence t_n→∞ such that \phi(t_n,x) → y

The second question. I was first thinking that I could try to find a sequence to show that y is in the closure of w(x) but my real analysis/topology skills are bad and I could really use some help!

All help is greatly appreciated!
 
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saxen said:

Homework Statement



Define the w-limit set (omega) of a point. Show that w(x) is closed.

Homework Equations



The Attempt at a Solution



The definition of a limit set is the set of points to which there exists a sequence t_n→∞ such that \phi(t_n,x) → y

The second question. I was first thinking that I could try to find a sequence to show that y is in the closure of w(x) but my real analysis/topology skills are bad and I could really use some help!

All help is greatly appreciated!

You want to show that if ##y## is a limit of ω-limit points, i.e. there is a sequence of ω-limit points ##y_m## such that ##y_m \rightarrow y##, then ##y## is a ω-limit point. Your thinking about how to do it is exactly correct. Take it step by step. Can you show there is a ##t_1## such that |\phi(t_1,x)-y|<1? Now can you show there is a ##t_2>t_1## such that |\phi(t_2,x)-y|<\frac{1}{2}? Etc, etc. It's not hard. Use the definitions and the triangle inequality.
 
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this is almost a tautology. a closed set is one which contains its limit points. so if a set of limit points exist, that set will be closed automatically. they don't have to be the unique limit of a sequence, just that there is an indexed infinite set which comes arbitrarily close to it. (but maybe I'm approaching this backwards...)
 
xaos said:
this is almost a tautology. a closed set is one which contains its limit points. so if a set of limit points exist, that set will be closed automatically. they don't have to be the unique limit of a sequence, just that there is an indexed infinite set which comes arbitrarily close to it. (but maybe I'm approaching this backwards...)

You are oversimplifying. This is a special notion of limit points. The ω-limit points aren't even all limit points. They are only the 'future' limit points of the dynamical trajectory. It's really best (and not even hard) to take this from first principles.
 
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if i understand you, you're looking at showing the intersection set of all the tails of a given trajectory is closed?
 
xaos said:
if i understand you, you're looking at showing the intersection set of all the tails of a given trajectory is closed?

I'm just saying approach it directly. Don't use any abstract properties of limit points. Construct a sequence of ##t_n## that show ##y## is an ω-limit point.
 
I forgot about this thread. I had some problem with the last part of this question but found a proof that helped me:

phi denotes the flow.

Let y be in the closure of w(x). Then there exists a sequence y_n in w(x) such that |y-y_n| < 1/2n. Moreover chosoe a sequence s.t t_n --> inf |phi (t_n,x)-y_n| <1/2n. Then |phi (t_n,x)-y| < 1/n and y is in w(x).

edit: sorry for format, TeX is not working for me.
 
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