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My question follows from the solutions of a simple problem: on a pool table with no friction, no holes, using point-like billiard balls, and following the laws of reflection, can you hit a ball positioned in the point $P$ so that its trajectory will never pass at $P$ again?

I resolved that in 2 ways:

  1. you unfold the pool table and draw another one on the edge where the balls hit first and so on, so the ball's trajectory is a straight line: we know that if the slope of this line is an irrational number, than the balls will never pass on the point $P$ again, because if so, an irrational number could be expressed as a fraction of 2 integers (the number of the unfolding for each edge), so the subset of the irrational numbers is the solution.

  2. It's possible to demonstrate that the pool table satisfies the hypotheses of Poincarè recurrence theorem: "If $S$ is a space with measure $\mu$ and $T:S \to S$ is a measure-preserving transformation, then for any set with positive measure $B⊆S$, the subset $A⊆B$ of points that never recur to $B$ has measure zero."(https://digitalcommons.coastal.edu/honors-theses/23/), and so there exist a subset of points that will never recur, but this subset's measure is $0$. So we have solved the problem this way too.

But we know that the measure of the subset of the irrational numbers is not $0$. So how is that possible?

Arc
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user967210
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  • Fantastic! This question is similar to a question I asked a few days ago, but I didn't think of pool table, see Can y=αx be solved for α irrational if the integer parts of x and y are unknowns?. See the comments about the solution set of $0$ measure in The power of irrationality: sin(x)+sin(πx), which I asked in mathoverflow (now closed). – Arc Sep 16 '21 at 12:47
  • Also, I suppose you are talking about point-like billiard balls, if the balls are extended objects then its a whole different story. But for point-like billiard balls, if you find a trajectory which hits $P$, then it certainly never hits $P$ again, exactly because of irrational slope, this point can be hit at most once, and the vast majority of points on the pool cannot be hit. The problem I wished to solve is to find that trajectory, but it seems there's no answer (someone mentioned Baker's theorem, but I wasn't able to follow it to a solution). – Arc Sep 16 '21 at 12:59
  • I edited the question to specify point-like balls, and also I have linked this question on the question I mentioned above. – Arc Sep 16 '21 at 13:10
  • @Arc Thanks for your editing! Your question is really interesting too. I read the answer about the set of 0 measure, and I fully agree that the set of configuration that occur there has 0 measure. But I still don't get the tricky point in my question, because here you have that the set of not recurring orbits has 0 measure, but they are linked with the subset of irrational numbers, whose measure is not 0. – user967210 Sep 17 '21 at 06:27
  • @Arc I'just throwing some ideas related to your problem, that maybe heads nowhere: the number of orizontal and vertical reflections of the pool table represents the integers in your problem, and we know from Kac's lemma that the expected time to return to P starting from P is 1/µ(A). Now the time is linked to the number of reflections and so to the integers, so is it possible to set up a limit for your integers using that? I found a deep anaysis of the subject here https://mathoverflow.net/questions/46414/how-quickly-will-billiard-trajectories-cluster/46422#46422 – user967210 Sep 17 '21 at 07:25

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First of all you have to develop more your second affirmation. Indeed what is $S$ in your example? It should be the space $S := \{(x,r)\}$ where $x$ is a point on the boundary of your billiard and $r$ is a slope. Then your application $T$ give the point and the slope just after the next rebound. Then you can show that there is a invariant measure $\mu$ on it which is continuous with Lesbegue.

Finally your set $B$ you are considering is $B := \{ (x,r), x=P \}$ which have measure $0$. And there is no contradiction with the fact that the subset $A \subset B$ with no recur is also a $0$ set.

If you would have take $B := \{ (x,r), x \in I \}$ with $I$ an non empty interval of the boundary of the billard, then you would have $\mu(B) \ne 0$, and the set of non-recur would be different (certainly not the irrational slope) but sill with measure $0$.

EtienneBfx
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