Chaos like phenomena on a simple metric space?

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  • Thread starter Zafa Pi
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Let M = {p, x1, x2, x3, ...} be a metric space with no isolated points.
f: M → M is continuous with f(xn) = xn+1, and f(p) = p.
We say f separates if ∃ δ > 0, ∋ for any y and z there is some n with |fn(y) - fn(z)| > δ, where fn+1(y) = f(fn(y)).
QUESTION: Does f separate?
 

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  • #2
andrewkirk
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I find it very hard to think about abstract problems like this without a concrete example.

Here's a concrete example I thought of:

Let M be a subspace of the unit circle ##S^1## in ##\mathbb R^2##, with the metric inherited from ##\mathbb R^2##. Since all points in M have radius 1 we can fully specify a point by its polar angle ##\theta##. We set the angles as ##\theta_p=0,\theta_1=1,\theta_{n+1}=2\theta_n##. Note that under this specification two points are identical if their angles differ by a multiple of ##2\pi##, so we deduct whatever multiple of ##2\pi## is necessary to ensure that a point's angle lies in the range ##[0,2\pi)##.

I am pretty confident that with this definition M is dense in the unit circle ##S^1## so that there are no isolated points, and that the map ##f## that doubles the angle of a point is continuous. So this space satisfies the premises of the problem.

I claim that this space separates. For a small ##\delta##, eg say ##\delta=0.01##, given any ##y,z\in M## we can find ##n## such that ##|f^n(y)-f^n(z)>\delta##. I have not proven this but I can see a way one would go about doing so.

Have a go at seeing if you can prove that this space and the function ##f## satisfy the premises and that ##f## separates.

Assuming that works, the next step would be to see if we can generalise that result to all space-function combinations that satisfy the premises or, alternatively, whether we can find a counterexample - a space and function that satisfy the premises but the function does not separate.
 
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We set the angles as θp=0,θ1=1,θn+1=2θnθp=0,θ1=1,θn+1=2θn\theta_p=0,\theta_1=1,\theta_{n+1}=2\theta_n.
I agree with what you say about this space and function. However, the same is true if 0 is eliminated, i.e. you don't need the fixed point.
Again, if we are on the circle group and this time θ1 = 1, θn+1 = θn + 1, we get density, f is continuous, no fixed point, and no separation.

There are many examples where the result is true (everyone I've tried), but I've been unable to show it is always true. I have spent a lot of time trying.
I think the result would have cool applications.
 

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