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I Dense orbit and fixed point question

  1. Jan 2, 2017 #1
    Let f be a continuous function of a metric space, M, to itself with a dense orbit and a fixed point.
    I.e. there exists z such that the set {f(n)(z)} for all n ∊ N (where f(n) is the nth iterate of f) is dense in M, and there exists p such that f(p) = p.

    Does this imply that f spreads?
    I.e. does there exists δ > 0 such that for x ≠ y there is an n0 (depending on x and y) with |f(n0)(x) - f(n0)(y)| > δ.

    I can prove this for M = R or S1, but I don't know in general, or even in the case where M = the orbit and p.
    Last edited: Jan 2, 2017
  2. jcsd
  3. Jan 3, 2017 #2


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    I don't think so. The function ##f:\mathbb R\to\mathbb R## given by ##f(x)=x/2## has a dense orbit and a fixed point at 0. But given ##\delta>0##, for any two points ##x,y## separated by a distance less than ##\delta## there is no ##n## such that ##|f^{(n)}(x)-f^{(n)}(y)|<\delta##.

    Also, the identity function on any metric space has a dense orbit, and every point is a fixed point. But again it does not satisfy the above definition of 'spreads'.
  4. Jan 3, 2017 #3
    Thanks for the reply.
    I suspect your last < should have been >. But that doesn't matter since f(x) = x/2 doesn't have a dense orbit in R. The orbit of z converges to 0 for any z. The orbit of z for the identity function is just z, hardly dense.
  5. Jan 3, 2017 #4


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    You seem to be using the concept of 'orbit' in a non-standard way. I have only ever seen the orbit of a function defined in relation to a particular point. See for instance here. Since you have not specified a point, guesswork is needed as to what you mean when you write the 'orbit of a function' rather than the 'orbit of a function at a point'.

    Hence, when you write that f has a dense orbit in M, I interpret that to mean that the set ##\{f^{(n)}(x)\ :\ x\in M,\ n\in\mathbb N\}## is dense in M. It sounds like you are interpreting it to mean that, for every ##x\in X## the orbit of ##f## at ##x## is dense in ##X##.

    The last < was intended as written but, given that your question is different from what it appeared to be, the point is moot.
  6. Jan 3, 2017 #5
    The definition you've given (x is fixed at one point and n runs over all members of N) is exactly the one I gave, except I gave a particular point z in M.
    For a fixed x the orbit orbit of x for f(x) = x/2 is {f(x), f(f(x)), f(f(f(x))), ...} = {x/2, x/4, x/8, ...} which converges to 0 in R, which is not dense in R.
  7. Jan 3, 2017 #6
    This thread seems to be using the idea of dense periodic orbits either wrong, or in a way different from what I've seen in dynamical systems theory. I've seen it said that a system has dense periodic orbits in its phase space, meaning that the set of points which generate a periodic orbit is dense in the phase space. The individual orbits aren't dense, the points which generate orbits are.

    Unless(as I implied may be the case?) I'm confused and the terminology is being used differently here.
  8. Jan 3, 2017 #7
    The notion of dense orbits (or topological transitivity) frequently shows up in chaotic dynamical systems. A periodic orbit is finite and can't be dense unless it is the entire space.
    For example the logistic function 4x(1-x) on [0,1] has a dense orbit and two fixed points and spreads (post #1). It also has a dense set of periodic points, but I don't care.
  9. Jan 3, 2017 #8
    I would be grateful and delighted if someone could decide whether f spreads when M (metric) = {x0, x1, x2, ...} U {p}, and has no isolated points, while f is continuous, and f(xn) = xn+1, f(p) = p.
  10. Jan 4, 2017 #9
    I think you might want to rework your definition of spread, since it is always trivially true for any function you pick if you iterate the function 0 times. You have the condition only that there need exist a single iteration where the distance between the iterates for [itex]x[/itex] and [itex]y[/itex] is nonzero. The only functions that never spread under your definition are the constant functions.
  11. Jan 5, 2017 #10
    The definition I use for the orbit of a point is the same as everyone's. The definition I use for being dense is the same as everyone's. the definition I use for for "spread" may be new (I don't know), but is almost the same as the one for "sensitivity to initial conditions". I believe you are having difficulty with the quantifiers. Here is the definition that f spreads:
    There exists δ > 0 such that for x ≠ y there is an n0 (depending on x and y) with |f(n0)(x) - f(n0)(y)| > δ.

    It means that no matter how close together x and y are there is some iterate of f when applied to x and y separates them to more than δ apart.
    So for example if f is the identity function notice that |x - y| = |f(n0)(x) - f(n0)(y)|. Hence if |x - y| < δ no iterate separates them to more than δ apart.
    If f(x) = x2 on {1,∞) then f spreads using δ = 1. It's irrelevant, but f has a fixed point at 1, and no dense orbit.
  12. Jan 5, 2017 #11
    For the identity function: If x ≠ y then there always exists δ less than |x - y| and greater than 0. Such a δ is going to be less than |x - y| for every iteration. Therefore the identity spreads according to your definition.
  13. Jan 5, 2017 #12
    The existence of δ precedes the x and y. This is what I meant when I said, "I believe you are having difficulty with the quantifiers." I suggest you look up the definition of "sensitivity to initial conditions".
  14. Jan 6, 2017 #13
    So, the maximal δ if it exists could then be a measurement of the function's spread according to your notion?
  15. Jan 6, 2017 #14
    Indeed. There will always be a sup. I don't think it is necessarily easy to find. What do you think it is for the logistic function, post #7?

    I don't know if my definition of spread is different than Devaney's definition of "sensitive to initial conditions", i.e. there exists δ > 0 such that for any x and any neighborhood U containing x there is y in U and no with |f(n0))(x) - f(n0))(y)| > δ. I would like to know.

    And what about post #8?
  16. Jan 6, 2017 #15
    Honestly, I have no idea what you meant by the metric equaling a countable set of points plus a fixed point.
  17. Jan 6, 2017 #16
    M is a countable dense in itself metric space as given in post #8. Thanks for your interest, but I think I'm done with this thread.
  18. Jan 6, 2017 #17
    Sorry, I guess when I studied nonlinear dynamical systems I jumped to continuous systems too quickly. My reaction to this has been affected by that, I suppose.
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