# Lorenz attractor path-connected?

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.

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.

• 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. – Mike Miller Jan 22 at 22:40
• @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. – Douglas Sirk Jan 22 at 22:44
• I mean, the orbit itself is not homeomorphic to $\mathbb R$, that is clear. – Douglas Sirk Jan 22 at 22:52
• "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 $\Bbb R$. – Mike Miller Jan 22 at 22:59
• @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 – Douglas Sirk Jan 23 at 1:05

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.
• 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 $-\infty$). – Lee Mosher Jan 20 at 17:49