Undergrad Chaos like phenomena on a simple metric space?

[Zafa Pi]

Let M = {p, x₁, x₂, x₃, ...} be a metric space with no isolated points.
f: M → M is continuous with f(x_n) = x_n+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?

 

[andrewkirk]

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.

 

[Zafa Pi]

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, θ_n+1 = θ_n + 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.