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All I have studied is forward orbits. So I was wondering whether there can be backward orbits. If that were to be the case, then shouldn't the transformation, T be invertible? This is not guaranteed though, right?

Sorry if this is too basic, but just a thought. Any constructive input is welcomed.

user43901
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1 Answers1

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Yes: if $\langle X,f\rangle$ is a discrete dynamical system, and $f$ is invertible, the backward orbit of $x\in X$ is exactly what you’d expect, namely, $\left\{f^{-n}(x):n\in\Bbb N\right\}$ (where $\Bbb N$ includes $0$). And as you imply, $f$ need not be invertible, in which case a point need not have a backward orbit.

Brian M. Scott
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  • Thanks Brian! That helps a lot. – user43901 Mar 30 '13 at 03:22
  • @user43901: You’re welcome! – Brian M. Scott Mar 30 '13 at 03:23
  • On a related note, Brian, I know about contraction maps. But what a contraction map is. But is there something called as an expansion map? Will it be something like $||fx -fy|| \leq c ||x-y||$ for all $x,y \in V$ such that $c>1$? I am essentially taking the Lipschitz and making $c>1$. – user43901 Mar 30 '13 at 03:45
  • @user43901: In general I don’t think that you get much interesting when $c>1$. I wouldn’t call such a map expansive: it doesn’t actually force any expansion. For that you’d have to require something like $|fx-fy|\ge c|x-y|$ for some $c>1$. You might be interested in the notion of an expansive homeomorphism, though. – Brian M. Scott Mar 30 '13 at 03:53
  • Awesome. Thanks Brian! – user43901 Mar 30 '13 at 04:13