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I'm not sure exactly what you mean by "space of subsets". All I was trying to point out was that the argument that proves the recurrence theorem itself uses a

I'm not sure exactly what you mean by "space of subsets". All I was trying to point out was that the argument that proves the recurrence theorem itself uses a volume of space around the initial state, and how "preimages" of that volume work. So it applies to volumes of states, too. The Lorenz attractor generally avoids the recurrence because its dynamics are dissipative: nothing drives points near the attractor to points far from the attractor. But once you are on the attractor, you can't just stay on the attractor forever getting "smeared out" without recurrence: you can only get at most smeared out over the finite area of the attractor, and eventually the smearing reaches your initial location on the attractor again.

rssoconnor 3y ago
You are correct about the Lorenz attractor being dissipative. I just assumed the chaos of the Lorenz attractor and, say a frictionless double pendulum, were the same phenomenon. However it seems they are, in fact, quite different. Even still > So it applies to volumes of states, too. I'm pretty sure this is false. As I mention in another comment, http://philsci-archive.pitt.edu/9838/1/recurrence.pdf http://philsci-archive.pitt.edu/9838/1/recurrence.pdf says "In [the case of the space of probability distributions over phase space], the reason that classical dynamics fails to abide by the linear recurrence theorem is that the distribution space is infinite-dimensional, even if phase space has finite volume: distributions can have structure on arbitrarily short scales. Yes the proof of the recurrence theorem uses volumes; but the result still only applies to individual points within the volume (or maybe sets of measure 0 at best).