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In my first question Lorenz attractor path-connected?, some are saying the Lorenz attractor $\mathscr L$ is not path-connected.

But suppose $x$ and $y$ are two points in different path components of the attractor. By Zorn's Lemma there is a subcontinuum of $\mathscr L$ which is irreducible between $x$ and $y$ (no proper subcontinuum of it contains these two points). I am having hard time imagining any irreducible subcontinua in $\mathscr L$ other than paths...

Are the "butterfly's wings" connected, individually?

EDIT: It seems a lot of people are not understanding my question. I want you to demonstrate an irreducible subcontinuum of $\mathscr L$ which is not an arc. If it is not path-connected, then there must be such an example. And the example cannot be $\mathscr L$ itself, because $\mathscr L$ is the union of two proper subcontinua, and these sub continua extend to points in their complements by small connecting arcs.

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    $\begingroup$ Look at this image commons.wikimedia.org/wiki/File:ContinuBJK.svg $\endgroup$
    – Anton Petrunin
    Commented Jan 22, 2019 at 20:01
  • $\begingroup$ @AntonPetrunin so could the butterfly wings be of that type? $\endgroup$
    – Douglas Sirk
    Commented Jan 22, 2019 at 20:03
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    $\begingroup$ The principle is the same. $\endgroup$
    – Anton Petrunin
    Commented Jan 22, 2019 at 20:05
  • $\begingroup$ @AntonPetrunin right, it seems possible, but it's not obvious. The wing could be Cantor set times circle, for instance, which is not connected. $\endgroup$
    – Douglas Sirk
    Commented Jan 22, 2019 at 20:07
  • $\begingroup$ @AntonPetrunin but what are you applying this principle towards? It does NOT apply, for instance, to the entire $\mathscr L$, because $\mathscr L$ is not indecomposable like the BJK continuum. You can perhaps apply the principle to each wing... but I'd need supporting arguments. It is not enough to just say "it's like the BJK continuum". $\endgroup$
    – Douglas Sirk
    Commented Jan 22, 2019 at 22:22

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Here is an example of an irreducible continuum between two points that is not a path:

https://www.encyclopediaofmath.org/index.php/Irreducible_continuum

The situation with the Lorenz attractor is similar: the orbit $E$ connected, in fact path connected, but not closed. The closure of a connected space is connected. However, you cannot connect points in $cl(E)\setminus E$ to $E$ by a curve.

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  • $\begingroup$ but does Lorenz contain a sin(1/x) continuum? $\endgroup$
    – Douglas Sirk
    Commented Jan 22, 2019 at 20:02
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    $\begingroup$ @DouglasSirk No, it does not, but my answer to your statement: I am having hard time imagining any irreducible subcontinua in L other than paths... $\endgroup$
    – Piotr Hajlasz
    Commented Jan 22, 2019 at 20:04
  • $\begingroup$ but what irreducible continuum does it contain which is not an arc? that is my question $\endgroup$
    – Douglas Sirk
    Commented Jan 22, 2019 at 20:05
  • $\begingroup$ @DouglasSirk I do not understand the question. $\endgroup$
    – Piotr Hajlasz
    Commented Jan 22, 2019 at 20:07
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    $\begingroup$ @DouglasSirk Reffering to your previous post: I do not mean only the line weaving around, but rather its closure. The `weaving line' $L$ is connected, in fact path connected, but not closed. The closure of a connected space is connected. However, you cannot connect points in $cl{L}\setminus L$ to $L$ by a curve. This is a very similar situation to that described in the link provided in my answer. $\endgroup$
    – Piotr Hajlasz
    Commented Jan 22, 2019 at 20:14

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