Lorenz attractor path-connected?
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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
asked 8 hours ago
Douglas SirkDouglas Sirk
362
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add a comment |
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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
asked 8 hours ago
Douglas SirkDouglas Sirk
362
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.
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3
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You had two chances to spell Lorenz right and you have failed twice :P
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– Wojowu
8 hours ago
add a comment |
$begingroup$
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
asked 8 hours ago
Douglas SirkDouglas Sirk
362
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
asked 8 hours ago
Douglas SirkDouglas Sirk
362
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.
asked 8 hours ago
Douglas SirkDouglas Sirk
362
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.
edited 1 hour ago
Douglas Sirk
asked 8 hours ago
Douglas SirkDouglas Sirk
362
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.
asked 8 hours ago
Douglas SirkDouglas Sirk
362
asked 8 hours ago
Douglas SirkDouglas Sirk
362
362
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
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You had two chances to spell Lorenz right and you have failed twice :P
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– Wojowu
8 hours ago
add a comment |
3
$begingroup$
You had two chances to spell Lorenz right and you have failed twice :P
$endgroup$
– Wojowu
8 hours ago
3
3
$begingroup$
You had two chances to spell Lorenz right and you have failed twice :P
$endgroup$
– Wojowu
8 hours ago
$begingroup$
You had two chances to spell Lorenz right and you have failed twice :P
$endgroup$
– Wojowu
8 hours ago
add a comment |
1 Answer
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active
oldest
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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.
answered 8 hours ago
Piotr HajlaszPiotr Hajlasz
7,05642457
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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.
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– Douglas Sirk
6 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.
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– Douglas Sirk
1 hour ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 8 hours ago
Piotr HajlaszPiotr Hajlasz
7,05642457
$endgroup$
$begingroup$
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
6 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
1 hour ago
add a comment |
$begingroup$
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.
answered 8 hours ago
Piotr HajlaszPiotr Hajlasz
7,05642457
$endgroup$
$begingroup$
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
6 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
1 hour ago
add a comment |
$begingroup$
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.
answered 8 hours ago
Piotr HajlaszPiotr Hajlasz
7,05642457
$endgroup$
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.
answered 8 hours ago
Piotr HajlaszPiotr Hajlasz
7,05642457
answered 8 hours ago
Piotr HajlaszPiotr Hajlasz
7,05642457
answered 8 hours ago
Piotr HajlaszPiotr Hajlasz
7,05642457
answered 8 hours ago
Piotr HajlaszPiotr Hajlasz
7,05642457
7,05642457
$begingroup$
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
6 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
1 hour ago
add a comment |
$begingroup$
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
6 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
1 hour 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
6 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
6 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
1 hour 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
1 hour ago
add a comment |
Douglas Sirk is a new contributor. Be nice, and check out our Code of Conduct.
Douglas Sirk is a new contributor. Be nice, and check out our Code of Conduct.
Douglas Sirk is a new contributor. Be nice, and check out our Code of Conduct.
Douglas Sirk is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You had two chances to spell Lorenz right and you have failed twice :P
$endgroup$
– Wojowu
8 hours ago