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Nonwandering set (equivalent definitions)
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Katok, Hasselblatt: "Modern theory of dynamical systems", p. 129:
Definition 3.3.3. A point $xin X$ is nonwandering with respect to the map $fcolon Xto X$ if for any open set $Uni x$ there is an $N>0$ such that $f^N(U)cap Uneqemptyset$. The set of all nonwandering points of $f$ is denoted by $NW(f)$.
Equivalently, one may assume that $N$ is arbitrary big. For, if for every $U$, $f^N(U)cap U=emptyset$ for $Ngeq N_0$, then $x$ is not periodic. Hence one can find a neighborhood $Vni x$ such that $f^i(V)cap V=emptyset$ for $i=0,1,ldots,N_0$, and $x$ cannot be nonwandering.
I do not understand this proof of the equivalence.
Let $N_0>0$ be arbitrary.
We suppose that $x$ is nonwandering, i.e. for any open set $Uni x$ there exists some $N>0$ such that $f^N(U)cap Uneqemptyset$.
I guess, we then suppose there exists some open set $Vni x$ such that for all $Ngeq N_0$, we have $f^N(V)cap V=emptyset$. But I dont see how to get the contradiction to $x$ is nonwandering.
dynamical-systems
asked Jul 16 at 9:32
Rhjg
260214
add a comment |Â
up vote
0
down vote
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Katok, Hasselblatt: "Modern theory of dynamical systems", p. 129:
Definition 3.3.3. A point $xin X$ is nonwandering with respect to the map $fcolon Xto X$ if for any open set $Uni x$ there is an $N>0$ such that $f^N(U)cap Uneqemptyset$. The set of all nonwandering points of $f$ is denoted by $NW(f)$.
Equivalently, one may assume that $N$ is arbitrary big. For, if for every $U$, $f^N(U)cap U=emptyset$ for $Ngeq N_0$, then $x$ is not periodic. Hence one can find a neighborhood $Vni x$ such that $f^i(V)cap V=emptyset$ for $i=0,1,ldots,N_0$, and $x$ cannot be nonwandering.
I do not understand this proof of the equivalence.
Let $N_0>0$ be arbitrary.
We suppose that $x$ is nonwandering, i.e. for any open set $Uni x$ there exists some $N>0$ such that $f^N(U)cap Uneqemptyset$.
I guess, we then suppose there exists some open set $Vni x$ such that for all $Ngeq N_0$, we have $f^N(V)cap V=emptyset$. But I dont see how to get the contradiction to $x$ is nonwandering.
dynamical-systems
asked Jul 16 at 9:32
Rhjg
260214
1
$f^N(U) cap U = emptyset$ for all open sets $U$ and all $N = N_0, N_0+1, N_0+2, dots$. So it is enough to find a nbhd $V$ such that $f^i(V) cap V = emptyset$ for finitely many $i=1,2,dots,N_0-1$. And such a $V$ can be found because $f$ is continuous and $f^i(x)$ are distinct. Can you continue now?
– user539887
Jul 16 at 11:23
For what do we need the continuity? To make sure that preimages of open sets are open?
– Rhjg
Jul 16 at 16:53
I would say that the whole "spirit" of the concept of nonwandering point is topological in nature: a nonwandering point need not return to its original position, but (some) nearby points must.
– user539887
Jul 16 at 19:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Katok, Hasselblatt: "Modern theory of dynamical systems", p. 129:
Definition 3.3.3. A point $xin X$ is nonwandering with respect to the map $fcolon Xto X$ if for any open set $Uni x$ there is an $N>0$ such that $f^N(U)cap Uneqemptyset$. The set of all nonwandering points of $f$ is denoted by $NW(f)$.
Equivalently, one may assume that $N$ is arbitrary big. For, if for every $U$, $f^N(U)cap U=emptyset$ for $Ngeq N_0$, then $x$ is not periodic. Hence one can find a neighborhood $Vni x$ such that $f^i(V)cap V=emptyset$ for $i=0,1,ldots,N_0$, and $x$ cannot be nonwandering.
I do not understand this proof of the equivalence.
Let $N_0>0$ be arbitrary.
We suppose that $x$ is nonwandering, i.e. for any open set $Uni x$ there exists some $N>0$ such that $f^N(U)cap Uneqemptyset$.
I guess, we then suppose there exists some open set $Vni x$ such that for all $Ngeq N_0$, we have $f^N(V)cap V=emptyset$. But I dont see how to get the contradiction to $x$ is nonwandering.
dynamical-systems
asked Jul 16 at 9:32
Rhjg
260214
Katok, Hasselblatt: "Modern theory of dynamical systems", p. 129:
Definition 3.3.3. A point $xin X$ is nonwandering with respect to the map $fcolon Xto X$ if for any open set $Uni x$ there is an $N>0$ such that $f^N(U)cap Uneqemptyset$. The set of all nonwandering points of $f$ is denoted by $NW(f)$.
Equivalently, one may assume that $N$ is arbitrary big. For, if for every $U$, $f^N(U)cap U=emptyset$ for $Ngeq N_0$, then $x$ is not periodic. Hence one can find a neighborhood $Vni x$ such that $f^i(V)cap V=emptyset$ for $i=0,1,ldots,N_0$, and $x$ cannot be nonwandering.
I do not understand this proof of the equivalence.
Let $N_0>0$ be arbitrary.
We suppose that $x$ is nonwandering, i.e. for any open set $Uni x$ there exists some $N>0$ such that $f^N(U)cap Uneqemptyset$.
I guess, we then suppose there exists some open set $Vni x$ such that for all $Ngeq N_0$, we have $f^N(V)cap V=emptyset$. But I dont see how to get the contradiction to $x$ is nonwandering.
dynamical-systems
asked Jul 16 at 9:32
Rhjg
260214
edited Jul 16 at 10:47
asked Jul 16 at 9:32
Rhjg
260214
asked Jul 16 at 9:32
Rhjg
260214
asked Jul 16 at 9:32
Rhjg
260214
260214
1
$f^N(U) cap U = emptyset$ for all open sets $U$ and all $N = N_0, N_0+1, N_0+2, dots$. So it is enough to find a nbhd $V$ such that $f^i(V) cap V = emptyset$ for finitely many $i=1,2,dots,N_0-1$. And such a $V$ can be found because $f$ is continuous and $f^i(x)$ are distinct. Can you continue now?
– user539887
Jul 16 at 11:23
For what do we need the continuity? To make sure that preimages of open sets are open?
– Rhjg
Jul 16 at 16:53
I would say that the whole "spirit" of the concept of nonwandering point is topological in nature: a nonwandering point need not return to its original position, but (some) nearby points must.
– user539887
Jul 16 at 19:12
add a comment |Â
1
$f^N(U) cap U = emptyset$ for all open sets $U$ and all $N = N_0, N_0+1, N_0+2, dots$. So it is enough to find a nbhd $V$ such that $f^i(V) cap V = emptyset$ for finitely many $i=1,2,dots,N_0-1$. And such a $V$ can be found because $f$ is continuous and $f^i(x)$ are distinct. Can you continue now?
– user539887
Jul 16 at 11:23
For what do we need the continuity? To make sure that preimages of open sets are open?
– Rhjg
Jul 16 at 16:53
I would say that the whole "spirit" of the concept of nonwandering point is topological in nature: a nonwandering point need not return to its original position, but (some) nearby points must.
– user539887
Jul 16 at 19:12
1
1
$f^N(U) cap U = emptyset$ for all open sets $U$ and all $N = N_0, N_0+1, N_0+2, dots$. So it is enough to find a nbhd $V$ such that $f^i(V) cap V = emptyset$ for finitely many $i=1,2,dots,N_0-1$. And such a $V$ can be found because $f$ is continuous and $f^i(x)$ are distinct. Can you continue now?
– user539887
Jul 16 at 11:23
$f^N(U) cap U = emptyset$ for all open sets $U$ and all $N = N_0, N_0+1, N_0+2, dots$. So it is enough to find a nbhd $V$ such that $f^i(V) cap V = emptyset$ for finitely many $i=1,2,dots,N_0-1$. And such a $V$ can be found because $f$ is continuous and $f^i(x)$ are distinct. Can you continue now?
– user539887
Jul 16 at 11:23
For what do we need the continuity? To make sure that preimages of open sets are open?
– Rhjg
Jul 16 at 16:53
For what do we need the continuity? To make sure that preimages of open sets are open?
– Rhjg
Jul 16 at 16:53
I would say that the whole "spirit" of the concept of nonwandering point is topological in nature: a nonwandering point need not return to its original position, but (some) nearby points must.
– user539887
Jul 16 at 19:12
I would say that the whole "spirit" of the concept of nonwandering point is topological in nature: a nonwandering point need not return to its original position, but (some) nearby points must.
– user539887
Jul 16 at 19:12
add a comment |Â
1 Answer
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1
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Laying out the proof in a little more detail, suppose $x$ is nonwandering and that there exists some $N_0$ so that $N > N_0$ implies $f^N(U) cap U = emptyset$. This implies $x$ is not periodic. Therefore, we can pick $V subseteq U$ about $x$ which is small enough so that all of the first $N_0$ iterates $f^i(V)$ do not intersect $V$ (this seems worth verifying instead of taking for granted, but it is necessary for the proof). Now we know that $f^N(V) subseteq f^N(U)$, so $f^N(U) cap U = emptyset$ implies $f^N(V) cap V = emptyset$. We already know that for $0 < i leq N_0$ we have $f^i(V) cap V = emptyset$, so putting this together demonstrates that for all $n > 0$ we have $f^n(V) cap V = emptyset$, contradicting our claim that $x$ is nonwandering.
answered Jul 16 at 11:25
Cory Griffith
729411
By the way: The fact that, in the definition 3.3.3 we can choose $N$ arbitrary large, i.e. reformulate the definition as "[...] for any $Uni x$ there exists some $Ngeq N_0$ such that ..." means that for any neighborhood of $x$, the function "returns infintely often" to this neighborhood. Right?
– Rhjg
Jul 16 at 11:49
It might be more correct to say "nearly returns infinitely often" or something along those lines, but that's the right spirit. I'm quibbling because the point $x$ itself need never return to $U$. For example, consider the plane and a continuous open function which sends $0$ to some point on the unit circle, and fixes the unit circle. Take $U$ to be the unit ball about $0$. Then $0$ is nonwandering, and $f^i(0) not in U$ for every $i > 0$.
– Cory Griffith
Jul 16 at 21:59
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Laying out the proof in a little more detail, suppose $x$ is nonwandering and that there exists some $N_0$ so that $N > N_0$ implies $f^N(U) cap U = emptyset$. This implies $x$ is not periodic. Therefore, we can pick $V subseteq U$ about $x$ which is small enough so that all of the first $N_0$ iterates $f^i(V)$ do not intersect $V$ (this seems worth verifying instead of taking for granted, but it is necessary for the proof). Now we know that $f^N(V) subseteq f^N(U)$, so $f^N(U) cap U = emptyset$ implies $f^N(V) cap V = emptyset$. We already know that for $0 < i leq N_0$ we have $f^i(V) cap V = emptyset$, so putting this together demonstrates that for all $n > 0$ we have $f^n(V) cap V = emptyset$, contradicting our claim that $x$ is nonwandering.
answered Jul 16 at 11:25
Cory Griffith
729411
By the way: The fact that, in the definition 3.3.3 we can choose $N$ arbitrary large, i.e. reformulate the definition as "[...] for any $Uni x$ there exists some $Ngeq N_0$ such that ..." means that for any neighborhood of $x$, the function "returns infintely often" to this neighborhood. Right?
– Rhjg
Jul 16 at 11:49
It might be more correct to say "nearly returns infinitely often" or something along those lines, but that's the right spirit. I'm quibbling because the point $x$ itself need never return to $U$. For example, consider the plane and a continuous open function which sends $0$ to some point on the unit circle, and fixes the unit circle. Take $U$ to be the unit ball about $0$. Then $0$ is nonwandering, and $f^i(0) not in U$ for every $i > 0$.
– Cory Griffith
Jul 16 at 21:59
add a comment |Â
up vote
1
down vote
accepted
Laying out the proof in a little more detail, suppose $x$ is nonwandering and that there exists some $N_0$ so that $N > N_0$ implies $f^N(U) cap U = emptyset$. This implies $x$ is not periodic. Therefore, we can pick $V subseteq U$ about $x$ which is small enough so that all of the first $N_0$ iterates $f^i(V)$ do not intersect $V$ (this seems worth verifying instead of taking for granted, but it is necessary for the proof). Now we know that $f^N(V) subseteq f^N(U)$, so $f^N(U) cap U = emptyset$ implies $f^N(V) cap V = emptyset$. We already know that for $0 < i leq N_0$ we have $f^i(V) cap V = emptyset$, so putting this together demonstrates that for all $n > 0$ we have $f^n(V) cap V = emptyset$, contradicting our claim that $x$ is nonwandering.
answered Jul 16 at 11:25
Cory Griffith
729411
By the way: The fact that, in the definition 3.3.3 we can choose $N$ arbitrary large, i.e. reformulate the definition as "[...] for any $Uni x$ there exists some $Ngeq N_0$ such that ..." means that for any neighborhood of $x$, the function "returns infintely often" to this neighborhood. Right?
– Rhjg
Jul 16 at 11:49
It might be more correct to say "nearly returns infinitely often" or something along those lines, but that's the right spirit. I'm quibbling because the point $x$ itself need never return to $U$. For example, consider the plane and a continuous open function which sends $0$ to some point on the unit circle, and fixes the unit circle. Take $U$ to be the unit ball about $0$. Then $0$ is nonwandering, and $f^i(0) not in U$ for every $i > 0$.
– Cory Griffith
Jul 16 at 21:59
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Laying out the proof in a little more detail, suppose $x$ is nonwandering and that there exists some $N_0$ so that $N > N_0$ implies $f^N(U) cap U = emptyset$. This implies $x$ is not periodic. Therefore, we can pick $V subseteq U$ about $x$ which is small enough so that all of the first $N_0$ iterates $f^i(V)$ do not intersect $V$ (this seems worth verifying instead of taking for granted, but it is necessary for the proof). Now we know that $f^N(V) subseteq f^N(U)$, so $f^N(U) cap U = emptyset$ implies $f^N(V) cap V = emptyset$. We already know that for $0 < i leq N_0$ we have $f^i(V) cap V = emptyset$, so putting this together demonstrates that for all $n > 0$ we have $f^n(V) cap V = emptyset$, contradicting our claim that $x$ is nonwandering.
answered Jul 16 at 11:25
Cory Griffith
729411
Laying out the proof in a little more detail, suppose $x$ is nonwandering and that there exists some $N_0$ so that $N > N_0$ implies $f^N(U) cap U = emptyset$. This implies $x$ is not periodic. Therefore, we can pick $V subseteq U$ about $x$ which is small enough so that all of the first $N_0$ iterates $f^i(V)$ do not intersect $V$ (this seems worth verifying instead of taking for granted, but it is necessary for the proof). Now we know that $f^N(V) subseteq f^N(U)$, so $f^N(U) cap U = emptyset$ implies $f^N(V) cap V = emptyset$. We already know that for $0 < i leq N_0$ we have $f^i(V) cap V = emptyset$, so putting this together demonstrates that for all $n > 0$ we have $f^n(V) cap V = emptyset$, contradicting our claim that $x$ is nonwandering.
answered Jul 16 at 11:25
Cory Griffith
729411
answered Jul 16 at 11:25
Cory Griffith
729411
answered Jul 16 at 11:25
Cory Griffith
729411
answered Jul 16 at 11:25
Cory Griffith
729411
729411
By the way: The fact that, in the definition 3.3.3 we can choose $N$ arbitrary large, i.e. reformulate the definition as "[...] for any $Uni x$ there exists some $Ngeq N_0$ such that ..." means that for any neighborhood of $x$, the function "returns infintely often" to this neighborhood. Right?
– Rhjg
Jul 16 at 11:49
It might be more correct to say "nearly returns infinitely often" or something along those lines, but that's the right spirit. I'm quibbling because the point $x$ itself need never return to $U$. For example, consider the plane and a continuous open function which sends $0$ to some point on the unit circle, and fixes the unit circle. Take $U$ to be the unit ball about $0$. Then $0$ is nonwandering, and $f^i(0) not in U$ for every $i > 0$.
– Cory Griffith
Jul 16 at 21:59
add a comment |Â
By the way: The fact that, in the definition 3.3.3 we can choose $N$ arbitrary large, i.e. reformulate the definition as "[...] for any $Uni x$ there exists some $Ngeq N_0$ such that ..." means that for any neighborhood of $x$, the function "returns infintely often" to this neighborhood. Right?
– Rhjg
Jul 16 at 11:49
It might be more correct to say "nearly returns infinitely often" or something along those lines, but that's the right spirit. I'm quibbling because the point $x$ itself need never return to $U$. For example, consider the plane and a continuous open function which sends $0$ to some point on the unit circle, and fixes the unit circle. Take $U$ to be the unit ball about $0$. Then $0$ is nonwandering, and $f^i(0) not in U$ for every $i > 0$.
– Cory Griffith
Jul 16 at 21:59
By the way: The fact that, in the definition 3.3.3 we can choose $N$ arbitrary large, i.e. reformulate the definition as "[...] for any $Uni x$ there exists some $Ngeq N_0$ such that ..." means that for any neighborhood of $x$, the function "returns infintely often" to this neighborhood. Right?
– Rhjg
Jul 16 at 11:49
By the way: The fact that, in the definition 3.3.3 we can choose $N$ arbitrary large, i.e. reformulate the definition as "[...] for any $Uni x$ there exists some $Ngeq N_0$ such that ..." means that for any neighborhood of $x$, the function "returns infintely often" to this neighborhood. Right?
– Rhjg
Jul 16 at 11:49
It might be more correct to say "nearly returns infinitely often" or something along those lines, but that's the right spirit. I'm quibbling because the point $x$ itself need never return to $U$. For example, consider the plane and a continuous open function which sends $0$ to some point on the unit circle, and fixes the unit circle. Take $U$ to be the unit ball about $0$. Then $0$ is nonwandering, and $f^i(0) not in U$ for every $i > 0$.
– Cory Griffith
Jul 16 at 21:59
It might be more correct to say "nearly returns infinitely often" or something along those lines, but that's the right spirit. I'm quibbling because the point $x$ itself need never return to $U$. For example, consider the plane and a continuous open function which sends $0$ to some point on the unit circle, and fixes the unit circle. Take $U$ to be the unit ball about $0$. Then $0$ is nonwandering, and $f^i(0) not in U$ for every $i > 0$.
– Cory Griffith
Jul 16 at 21:59
add a comment |Â
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1
$f^N(U) cap U = emptyset$ for all open sets $U$ and all $N = N_0, N_0+1, N_0+2, dots$. So it is enough to find a nbhd $V$ such that $f^i(V) cap V = emptyset$ for finitely many $i=1,2,dots,N_0-1$. And such a $V$ can be found because $f$ is continuous and $f^i(x)$ are distinct. Can you continue now?
– user539887
Jul 16 at 11:23
For what do we need the continuity? To make sure that preimages of open sets are open?
– Rhjg
Jul 16 at 16:53
I would say that the whole "spirit" of the concept of nonwandering point is topological in nature: a nonwandering point need not return to its original position, but (some) nearby points must.
– user539887
Jul 16 at 19:12