Mixing
Is this map chain-transitive
Let $f:mathbb R^2→mathbb R^2$ be a continuous self-map and let $δ$ be a positive real number. A (finite or infinite) sequence $(x_{n})_{n≥0}$ is a $δ$-chain if $$d(f(x_{n}),x_{n+1})<δ$$ for all $n$.
The map $f$ is called chain-transitive if for every $x,y∈mathbb R^2$ and every $δ>0$ there is a finite $δ$-chain $x₀,...,x_{n}$ such that $x₀=x$ and $x_{n}=y$.
Assuming a map g verify: $g^{k}(x)$ tends to infinity as $k→∞$ for all $x in mathbb R^2$, i.e., the successive iterations of $x$ by $g$ goes to infinity, or the orbit of the points $x in mathbb R^2$ diverge.
My question is: is this map $g$ chain-transitive?
geometry analysis differential-geometry dynamical-systems ergodic-theory
asked Dec 31 '18 at 16:53
ChinaChina
1,4271029
add a comment |
Let $f:mathbb R^2→mathbb R^2$ be a continuous self-map and let $δ$ be a positive real number. A (finite or infinite) sequence $(x_{n})_{n≥0}$ is a $δ$-chain if $$d(f(x_{n}),x_{n+1})<δ$$ for all $n$.
The map $f$ is called chain-transitive if for every $x,y∈mathbb R^2$ and every $δ>0$ there is a finite $δ$-chain $x₀,...,x_{n}$ such that $x₀=x$ and $x_{n}=y$.
Assuming a map g verify: $g^{k}(x)$ tends to infinity as $k→∞$ for all $x in mathbb R^2$, i.e., the successive iterations of $x$ by $g$ goes to infinity, or the orbit of the points $x in mathbb R^2$ diverge.
My question is: is this map $g$ chain-transitive?
geometry analysis differential-geometry dynamical-systems ergodic-theory
asked Dec 31 '18 at 16:53
ChinaChina
1,4271029
1
How can that be that $Vert f^k(x) Vert to infty$ as $f$ takes its values in the compact $A$?
– mathcounterexamples.net
Dec 31 '18 at 17:03
@mathcounterexamples.net: The question is edited.
– China
Dec 31 '18 at 17:08
That still doesn't make sense...
– mathcounterexamples.net
Dec 31 '18 at 17:09
@mathcounterexamples.net: I mean the case of maps like $g$ chain transitive or not.
– China
Dec 31 '18 at 17:14
add a comment |
Let $f:mathbb R^2→mathbb R^2$ be a continuous self-map and let $δ$ be a positive real number. A (finite or infinite) sequence $(x_{n})_{n≥0}$ is a $δ$-chain if $$d(f(x_{n}),x_{n+1})<δ$$ for all $n$.
The map $f$ is called chain-transitive if for every $x,y∈mathbb R^2$ and every $δ>0$ there is a finite $δ$-chain $x₀,...,x_{n}$ such that $x₀=x$ and $x_{n}=y$.
Assuming a map g verify: $g^{k}(x)$ tends to infinity as $k→∞$ for all $x in mathbb R^2$, i.e., the successive iterations of $x$ by $g$ goes to infinity, or the orbit of the points $x in mathbb R^2$ diverge.
My question is: is this map $g$ chain-transitive?
geometry analysis differential-geometry dynamical-systems ergodic-theory
asked Dec 31 '18 at 16:53
ChinaChina
1,4271029
Let $f:mathbb R^2→mathbb R^2$ be a continuous self-map and let $δ$ be a positive real number. A (finite or infinite) sequence $(x_{n})_{n≥0}$ is a $δ$-chain if $$d(f(x_{n}),x_{n+1})<δ$$ for all $n$.
The map $f$ is called chain-transitive if for every $x,y∈mathbb R^2$ and every $δ>0$ there is a finite $δ$-chain $x₀,...,x_{n}$ such that $x₀=x$ and $x_{n}=y$.
Assuming a map g verify: $g^{k}(x)$ tends to infinity as $k→∞$ for all $x in mathbb R^2$, i.e., the successive iterations of $x$ by $g$ goes to infinity, or the orbit of the points $x in mathbb R^2$ diverge.
My question is: is this map $g$ chain-transitive?
geometry analysis differential-geometry dynamical-systems ergodic-theory
geometry analysis differential-geometry dynamical-systems ergodic-theory
asked Dec 31 '18 at 16:53
ChinaChina
1,4271029
asked Dec 31 '18 at 16:53
ChinaChina
1,4271029
edited Dec 31 '18 at 17:18
China
asked Dec 31 '18 at 16:53
ChinaChina
1,4271029
asked Dec 31 '18 at 16:53
ChinaChina
1,4271029
asked Dec 31 '18 at 16:53
ChinaChina
1,4271029
1,4271029
1
How can that be that $Vert f^k(x) Vert to infty$ as $f$ takes its values in the compact $A$?
– mathcounterexamples.net
Dec 31 '18 at 17:03
@mathcounterexamples.net: The question is edited.
– China
Dec 31 '18 at 17:08
That still doesn't make sense...
– mathcounterexamples.net
Dec 31 '18 at 17:09
@mathcounterexamples.net: I mean the case of maps like $g$ chain transitive or not.
– China
Dec 31 '18 at 17:14
add a comment |
1
How can that be that $Vert f^k(x) Vert to infty$ as $f$ takes its values in the compact $A$?
– mathcounterexamples.net
Dec 31 '18 at 17:03
@mathcounterexamples.net: The question is edited.
– China
Dec 31 '18 at 17:08
That still doesn't make sense...
– mathcounterexamples.net
Dec 31 '18 at 17:09
@mathcounterexamples.net: I mean the case of maps like $g$ chain transitive or not.
– China
Dec 31 '18 at 17:14
1
1
How can that be that $Vert f^k(x) Vert to infty$ as $f$ takes its values in the compact $A$?
– mathcounterexamples.net
Dec 31 '18 at 17:03
How can that be that $Vert f^k(x) Vert to infty$ as $f$ takes its values in the compact $A$?
– mathcounterexamples.net
Dec 31 '18 at 17:03
@mathcounterexamples.net: The question is edited.
– China
Dec 31 '18 at 17:08
@mathcounterexamples.net: The question is edited.
– China
Dec 31 '18 at 17:08
That still doesn't make sense...
– mathcounterexamples.net
Dec 31 '18 at 17:09
That still doesn't make sense...
– mathcounterexamples.net
Dec 31 '18 at 17:09
@mathcounterexamples.net: I mean the case of maps like $g$ chain transitive or not.
– China
Dec 31 '18 at 17:14
@mathcounterexamples.net: I mean the case of maps like $g$ chain transitive or not.
– China
Dec 31 '18 at 17:14
add a comment |
1 Answer
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Such a $g$ might not be Chain transitive.
Counter-example: $g(a, b) = (a+1, b)$. Then there does not exists a $delta$ transitive chain between $x = (0,0)$ and $y = (1,0)$ whenever $delta <1/2$: No matter what one choose $x_n = (a_n, b_n)$, one has $a_n ge n/2$.
answered Jan 1 at 3:25
Arctic CharArctic Char
15317
add a comment |
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Such a $g$ might not be Chain transitive.
Counter-example: $g(a, b) = (a+1, b)$. Then there does not exists a $delta$ transitive chain between $x = (0,0)$ and $y = (1,0)$ whenever $delta <1/2$: No matter what one choose $x_n = (a_n, b_n)$, one has $a_n ge n/2$.
answered Jan 1 at 3:25
Arctic CharArctic Char
15317
add a comment |
Such a $g$ might not be Chain transitive.
Counter-example: $g(a, b) = (a+1, b)$. Then there does not exists a $delta$ transitive chain between $x = (0,0)$ and $y = (1,0)$ whenever $delta <1/2$: No matter what one choose $x_n = (a_n, b_n)$, one has $a_n ge n/2$.
answered Jan 1 at 3:25
Arctic CharArctic Char
15317
add a comment |
Such a $g$ might not be Chain transitive.
Counter-example: $g(a, b) = (a+1, b)$. Then there does not exists a $delta$ transitive chain between $x = (0,0)$ and $y = (1,0)$ whenever $delta <1/2$: No matter what one choose $x_n = (a_n, b_n)$, one has $a_n ge n/2$.
answered Jan 1 at 3:25
Arctic CharArctic Char
15317
Such a $g$ might not be Chain transitive.
Counter-example: $g(a, b) = (a+1, b)$. Then there does not exists a $delta$ transitive chain between $x = (0,0)$ and $y = (1,0)$ whenever $delta <1/2$: No matter what one choose $x_n = (a_n, b_n)$, one has $a_n ge n/2$.
answered Jan 1 at 3:25
Arctic CharArctic Char
15317
answered Jan 1 at 3:25
Arctic CharArctic Char
15317
answered Jan 1 at 3:25
Arctic CharArctic Char
15317
answered Jan 1 at 3:25
Arctic CharArctic Char
15317
15317
add a comment |
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1
How can that be that $Vert f^k(x) Vert to infty$ as $f$ takes its values in the compact $A$?
– mathcounterexamples.net
Dec 31 '18 at 17:03
@mathcounterexamples.net: The question is edited.
– China
Dec 31 '18 at 17:08
That still doesn't make sense...
– mathcounterexamples.net
Dec 31 '18 at 17:09
@mathcounterexamples.net: I mean the case of maps like $g$ chain transitive or not.
– China
Dec 31 '18 at 17:14