Can we tell if the Lorenz attractor is path-connected? By the attractor I do not mean only the line weaving around, but rather its closure.

enter image description here

EDIT: The answer below is unsatisfactory, and possibly incorrect. It does not account for the path-component of the main orbit being enlarged when the the closure is taken. In the forking regions there will be a Cantor set times "T", with one leg of "T" butting into the main orbit.

  • $\begingroup$ The answer is both satisfactory and correct. It is an elementary exercise that if $\Bbb R$ is a dense open subset of $X$, which is a Hausdorff space, then $X \setminus \Bbb R$ is either at most 2 points or $X$ fails to be path-connected. This uses Hausdorffness to show that arc-connectedness is the same as path-connectedness. $\endgroup$ – Mike Miller Jan 22 at 22:40
  • $\begingroup$ @MikeMiller I assume you mean a one-to-one continuous image of $\mathbb R$, and not $\mathbb R$ itself. So you claim $X\setminus "\mathbb R"$ is not path connected. Maybe true. (and if so, it answers this question). But the path component of "$\mathbb R$" can certainly be extended in certain $X$'s. There are simple examples. So it doesn not automatically follow that the orbit is a path component. $\endgroup$ – Douglas Sirk Jan 22 at 22:44
  • $\begingroup$ I mean, the orbit itself is not homeomorphic to $\mathbb R$, that is clear. $\endgroup$ – Douglas Sirk Jan 22 at 22:52
  • $\begingroup$ "homeomorphic to a dense open subset", then, if you're going to be pedantic. And no, I claimed that either $X = [0,1], [0,1), (0,1), S^1$, or that $X$ is not path-connected. This is an exercise. You should prove it. (If you are having trouble, post it on MSE.) Note the assumption that this is a dense open set. A flowline of a differential equation (which exists for all time) is always diffeomorphic to either $S^1$ or $\Bbb R$, depending on whether or not the flowline is periodic. It is not, in this case. Therefore the flowline that you take the closure of is diffeomorphic to $\Bbb R$. $\endgroup$ – Mike Miller Jan 22 at 22:59
  • $\begingroup$ @MikeMiller but I cannot see how the flowline in the Lorenz attractor is homeomorphic to $\mathbb R$. in fact it looks like the flow line is a first category set $\endgroup$ – Douglas Sirk Jan 23 at 1:05

The question has been answered here: https://mathoverflow.net/a/297836/121665. It is connected but not path connected.

The situation is somewhat similar to the topologist's sine curve: the graph of $$ f(x)=\sin\frac{1}{x}, \quad x\in (0,1] $$ is path connected, but its closure (as a subset of $\mathbb{R}^2$) is connected, but not path connected.

  • $\begingroup$ I do not think the answer in the link directly applies to the Lorenz attractor. We can see in the picture that in the closure of the weaving line, there will be new paths touching the weaving line. These will appear where the "forking" occurs. $\endgroup$ – Douglas Sirk Jan 19 at 23:24
  • $\begingroup$ It now appears to me that every irreducible subcontinuum of the Lorenz attractor is an arc. And therefore it is path-connected. $\endgroup$ – Douglas Sirk Jan 20 at 3:52
  • 4
    $\begingroup$ That answer does apply. The key words are that the Lorenz attractor has a transversal Cantor set. Also, when you think you see "touching" in pictures, what is actually happening is two different orbits of the attractor, forming two different path components, that are getting exponentially close as time increases to $+\infty$ (or decreases to $-\infty$). $\endgroup$ – Lee Mosher Jan 20 at 17:49
  • $\begingroup$ Lee Mosher: Yes I think you are seeing what I did. In the closure there will be "T"-shapes coming off of the main orbit, parts of which were not part of the original orbit. There may be enough of this repeated forking to actually make it path-connected. Or the path component may end up being a first category set... I still cannot decide but this is a deeper problem than compared, say, to the Plykin attractor. $\endgroup$ – Douglas Sirk Jan 20 at 19:13
  • $\begingroup$ Lee Mosher: In the closure there is clearly a Cantor set times T-shape... the path component of the orbit gets enlarged. $\endgroup$ – Douglas Sirk Jan 20 at 19:42

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.