Lorenz attractor path-connected?












6












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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.






gn.general-topology ds.dynamical-systems path-connected







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  • 3




    $begingroup$
    You had two chances to spell Lorenz right and you have failed twice :P
    $endgroup$
    – Wojowu
    7 hours ago
















6












$begingroup$


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.






gn.general-topology ds.dynamical-systems path-connected







share|cite|improve this question









New contributor




Douglas Sirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 3




    $begingroup$
    You had two chances to spell Lorenz right and you have failed twice :P
    $endgroup$
    – Wojowu
    7 hours ago














6












6








6





$begingroup$


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.






gn.general-topology ds.dynamical-systems path-connected







share|cite|improve this question









New contributor




Douglas Sirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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.






gn.general-topology ds.dynamical-systems path-connected




gn.general-topology ds.dynamical-systems path-connected






share|cite|improve this question









New contributor




Douglas Sirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Douglas Sirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 23 mins ago







Douglas Sirk













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Douglas Sirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 7 hours ago









Douglas SirkDouglas Sirk

312




312




New contributor




Douglas Sirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Douglas Sirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Douglas Sirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    $begingroup$
    You had two chances to spell Lorenz right and you have failed twice :P
    $endgroup$
    – Wojowu
    7 hours ago














  • 3




    $begingroup$
    You had two chances to spell Lorenz right and you have failed twice :P
    $endgroup$
    – Wojowu
    7 hours ago








3




3




$begingroup$
You had two chances to spell Lorenz right and you have failed twice :P
$endgroup$
– Wojowu
7 hours ago




$begingroup$
You had two chances to spell Lorenz right and you have failed twice :P
$endgroup$
– Wojowu
7 hours ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

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
$$
f(x)=sinfrac{1}{x},
quad xin (0,1]
$$
is path connected, but its closure (as a subset of $mathbb{R}^2$) is connected, but not path connected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    4 hours ago












  • $begingroup$
    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
    30 mins ago











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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
$$
f(x)=sinfrac{1}{x},
quad xin (0,1]
$$
is path connected, but its closure (as a subset of $mathbb{R}^2$) is connected, but not path connected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    4 hours ago












  • $begingroup$
    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
    30 mins ago
















7












$begingroup$

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
$$
f(x)=sinfrac{1}{x},
quad xin (0,1]
$$
is path connected, but its closure (as a subset of $mathbb{R}^2$) is connected, but not path connected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    4 hours ago












  • $begingroup$
    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
    30 mins ago














7












7








7





$begingroup$

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
$$
f(x)=sinfrac{1}{x},
quad xin (0,1]
$$
is path connected, but its closure (as a subset of $mathbb{R}^2$) is connected, but not path connected.






share|cite|improve this answer









$endgroup$



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
$$
f(x)=sinfrac{1}{x},
quad xin (0,1]
$$
is path connected, but its closure (as a subset of $mathbb{R}^2$) is connected, but not path connected.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 7 hours ago









Piotr HajlaszPiotr Hajlasz

7,04642457




7,04642457












  • $begingroup$
    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
    4 hours ago












  • $begingroup$
    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
    30 mins ago


















  • $begingroup$
    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
    4 hours ago












  • $begingroup$
    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
    30 mins ago
















$begingroup$
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
4 hours ago






$begingroup$
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
4 hours ago














$begingroup$
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
30 mins ago




$begingroup$
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
30 mins ago










Douglas Sirk is a new contributor. Be nice, and check out our Code of Conduct.










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