Mixing
local homeomorphism of the circle
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Let $f:S^1to S^1$ be a local homeomorphism, in the following we try to show that there is $delta>0$ such that for $x,yin S^1$, if $operatornamediam operatornamearc(x,y)<delta$, then $f$ is increasing (or decreasing) on $operatornamearc(x, y)$.
Since $f$ is local homeomorphism, for every $xin S^1$, there is arc $I_x$ such that $f$ is increasing (or decreasing) on $I_x$. Let $delta>0$ be Lebesgue number for cover $I_x: xin S^1$. Hence we can say that for $x,yin S^1$, if $operatornamediam operatornamearc(x,y)<delta$, then $f$ is increasing (or decreasing) on $operatornamearc(x, y)$.
Q. The proof is true?
Thanks for your help
calculus general-topology analysis
edited Jul 20 at 17:30
Michael Hardy
204k23186462
asked Jul 20 at 16:23
user479859
626
add a comment |Â
up vote
0
down vote
favorite
Let $f:S^1to S^1$ be a local homeomorphism, in the following we try to show that there is $delta>0$ such that for $x,yin S^1$, if $operatornamediam operatornamearc(x,y)<delta$, then $f$ is increasing (or decreasing) on $operatornamearc(x, y)$.
Since $f$ is local homeomorphism, for every $xin S^1$, there is arc $I_x$ such that $f$ is increasing (or decreasing) on $I_x$. Let $delta>0$ be Lebesgue number for cover $I_x: xin S^1$. Hence we can say that for $x,yin S^1$, if $operatornamediam operatornamearc(x,y)<delta$, then $f$ is increasing (or decreasing) on $operatornamearc(x, y)$.
Q. The proof is true?
Thanks for your help
calculus general-topology analysis
edited Jul 20 at 17:30
Michael Hardy
204k23186462
asked Jul 20 at 16:23
user479859
626
What is $diamarc(x,y)$?
– leibnewtz
Jul 20 at 16:35
@leibnewtz $diam arc(x, y)= min, 1-$, that $S^1=[0,1]/$
– user479859
Jul 20 at 16:38
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:S^1to S^1$ be a local homeomorphism, in the following we try to show that there is $delta>0$ such that for $x,yin S^1$, if $operatornamediam operatornamearc(x,y)<delta$, then $f$ is increasing (or decreasing) on $operatornamearc(x, y)$.
Since $f$ is local homeomorphism, for every $xin S^1$, there is arc $I_x$ such that $f$ is increasing (or decreasing) on $I_x$. Let $delta>0$ be Lebesgue number for cover $I_x: xin S^1$. Hence we can say that for $x,yin S^1$, if $operatornamediam operatornamearc(x,y)<delta$, then $f$ is increasing (or decreasing) on $operatornamearc(x, y)$.
Q. The proof is true?
Thanks for your help
calculus general-topology analysis
edited Jul 20 at 17:30
Michael Hardy
204k23186462
asked Jul 20 at 16:23
user479859
626
Let $f:S^1to S^1$ be a local homeomorphism, in the following we try to show that there is $delta>0$ such that for $x,yin S^1$, if $operatornamediam operatornamearc(x,y)<delta$, then $f$ is increasing (or decreasing) on $operatornamearc(x, y)$.
Since $f$ is local homeomorphism, for every $xin S^1$, there is arc $I_x$ such that $f$ is increasing (or decreasing) on $I_x$. Let $delta>0$ be Lebesgue number for cover $I_x: xin S^1$. Hence we can say that for $x,yin S^1$, if $operatornamediam operatornamearc(x,y)<delta$, then $f$ is increasing (or decreasing) on $operatornamearc(x, y)$.
Q. The proof is true?
Thanks for your help
calculus general-topology analysis
edited Jul 20 at 17:30
Michael Hardy
204k23186462
asked Jul 20 at 16:23
user479859
626
edited Jul 20 at 17:30
Michael Hardy
204k23186462
edited Jul 20 at 17:30
Michael Hardy
204k23186462
edited Jul 20 at 17:30
Michael Hardy
204k23186462
204k23186462
asked Jul 20 at 16:23
user479859
626
asked Jul 20 at 16:23
user479859
626
asked Jul 20 at 16:23
user479859
626
626
What is $diamarc(x,y)$?
– leibnewtz
Jul 20 at 16:35
@leibnewtz $diam arc(x, y)= min, 1-$, that $S^1=[0,1]/$
– user479859
Jul 20 at 16:38
add a comment |Â
What is $diamarc(x,y)$?
– leibnewtz
Jul 20 at 16:35
@leibnewtz $diam arc(x, y)= min, 1-$, that $S^1=[0,1]/$
– user479859
Jul 20 at 16:38
What is $diamarc(x,y)$?
– leibnewtz
Jul 20 at 16:35
What is $diamarc(x,y)$?
– leibnewtz
Jul 20 at 16:35
@leibnewtz $diam arc(x, y)= min, 1-$, that $S^1=[0,1]/$
– user479859
Jul 20 at 16:38
@leibnewtz $diam arc(x, y)= min, 1-$, that $S^1=[0,1]/$
– user479859
Jul 20 at 16:38
add a comment |Â
1 Answer
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You should first make precise what you mean by a (closed/open?) arc and by increasing (decreasing).
Obviously $operatornamearc(x,y)$ is defined for any two points $x = e^is, y= e^it$ such that $s < t$ and $t -s < 2 pi$; it is the image of $[s,t] subset mathbbR$ under the map $e : mathbbR to S^1, e(x) = e^ix$ (if you mean the closed arc). But $e$ is a covering so that $f circ e : mathbbR to S^1$ lifts to a map $tildef : mathbbR to mathbbR$ such that $e circ tildef = f circ e$. This lift is not unique, but if $tildef_i$ are any two lifts, we have $tildef_2(x) = tildef_1(x) + 2 k pi$ for some $k in mathbbZ$.
What you mean by "$f$ is increasing on $operatornamearc(x,y)$" is that $tildef$ is increasing on $[s,t]$.
With these clarifications your proof is correct, but we can say more. The map $f$ is in fact a covering, see When is a local homeomorphism a covering map?. This implies that $tildef$ is increasing or decreasing on all of $mathbbR$.
You can see this also directly. W.l.o.g. let $tildef$ be increasing on some interval $[a_0,b_0]$.
Let $A = inf a le a_0 mid tildef text increasing on [a,b_0] $, $B = sup b ge b_0 mid tildef text increasing on [a_0,b] $. Assume that $B < infty$. Then $tildef$ is increasing on $[a_0,B]$. But for some $epsilon > 0$ the map $tildef$ must be increasing or decreasing on $[B - epsilon, B + epsilon]$. Hence it must be increasing. This contradicts the definition of $B$ and we conclude $B = infty$. Similarly $A = -infty$.
answered Jul 23 at 14:55
Paul Frost
3,658420
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
You should first make precise what you mean by a (closed/open?) arc and by increasing (decreasing).
Obviously $operatornamearc(x,y)$ is defined for any two points $x = e^is, y= e^it$ such that $s < t$ and $t -s < 2 pi$; it is the image of $[s,t] subset mathbbR$ under the map $e : mathbbR to S^1, e(x) = e^ix$ (if you mean the closed arc). But $e$ is a covering so that $f circ e : mathbbR to S^1$ lifts to a map $tildef : mathbbR to mathbbR$ such that $e circ tildef = f circ e$. This lift is not unique, but if $tildef_i$ are any two lifts, we have $tildef_2(x) = tildef_1(x) + 2 k pi$ for some $k in mathbbZ$.
What you mean by "$f$ is increasing on $operatornamearc(x,y)$" is that $tildef$ is increasing on $[s,t]$.
With these clarifications your proof is correct, but we can say more. The map $f$ is in fact a covering, see When is a local homeomorphism a covering map?. This implies that $tildef$ is increasing or decreasing on all of $mathbbR$.
You can see this also directly. W.l.o.g. let $tildef$ be increasing on some interval $[a_0,b_0]$.
Let $A = inf a le a_0 mid tildef text increasing on [a,b_0] $, $B = sup b ge b_0 mid tildef text increasing on [a_0,b] $. Assume that $B < infty$. Then $tildef$ is increasing on $[a_0,B]$. But for some $epsilon > 0$ the map $tildef$ must be increasing or decreasing on $[B - epsilon, B + epsilon]$. Hence it must be increasing. This contradicts the definition of $B$ and we conclude $B = infty$. Similarly $A = -infty$.
answered Jul 23 at 14:55
Paul Frost
3,658420
add a comment |Â
up vote
1
down vote
You should first make precise what you mean by a (closed/open?) arc and by increasing (decreasing).
Obviously $operatornamearc(x,y)$ is defined for any two points $x = e^is, y= e^it$ such that $s < t$ and $t -s < 2 pi$; it is the image of $[s,t] subset mathbbR$ under the map $e : mathbbR to S^1, e(x) = e^ix$ (if you mean the closed arc). But $e$ is a covering so that $f circ e : mathbbR to S^1$ lifts to a map $tildef : mathbbR to mathbbR$ such that $e circ tildef = f circ e$. This lift is not unique, but if $tildef_i$ are any two lifts, we have $tildef_2(x) = tildef_1(x) + 2 k pi$ for some $k in mathbbZ$.
What you mean by "$f$ is increasing on $operatornamearc(x,y)$" is that $tildef$ is increasing on $[s,t]$.
With these clarifications your proof is correct, but we can say more. The map $f$ is in fact a covering, see When is a local homeomorphism a covering map?. This implies that $tildef$ is increasing or decreasing on all of $mathbbR$.
You can see this also directly. W.l.o.g. let $tildef$ be increasing on some interval $[a_0,b_0]$.
Let $A = inf a le a_0 mid tildef text increasing on [a,b_0] $, $B = sup b ge b_0 mid tildef text increasing on [a_0,b] $. Assume that $B < infty$. Then $tildef$ is increasing on $[a_0,B]$. But for some $epsilon > 0$ the map $tildef$ must be increasing or decreasing on $[B - epsilon, B + epsilon]$. Hence it must be increasing. This contradicts the definition of $B$ and we conclude $B = infty$. Similarly $A = -infty$.
answered Jul 23 at 14:55
Paul Frost
3,658420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You should first make precise what you mean by a (closed/open?) arc and by increasing (decreasing).
Obviously $operatornamearc(x,y)$ is defined for any two points $x = e^is, y= e^it$ such that $s < t$ and $t -s < 2 pi$; it is the image of $[s,t] subset mathbbR$ under the map $e : mathbbR to S^1, e(x) = e^ix$ (if you mean the closed arc). But $e$ is a covering so that $f circ e : mathbbR to S^1$ lifts to a map $tildef : mathbbR to mathbbR$ such that $e circ tildef = f circ e$. This lift is not unique, but if $tildef_i$ are any two lifts, we have $tildef_2(x) = tildef_1(x) + 2 k pi$ for some $k in mathbbZ$.
What you mean by "$f$ is increasing on $operatornamearc(x,y)$" is that $tildef$ is increasing on $[s,t]$.
With these clarifications your proof is correct, but we can say more. The map $f$ is in fact a covering, see When is a local homeomorphism a covering map?. This implies that $tildef$ is increasing or decreasing on all of $mathbbR$.
You can see this also directly. W.l.o.g. let $tildef$ be increasing on some interval $[a_0,b_0]$.
Let $A = inf a le a_0 mid tildef text increasing on [a,b_0] $, $B = sup b ge b_0 mid tildef text increasing on [a_0,b] $. Assume that $B < infty$. Then $tildef$ is increasing on $[a_0,B]$. But for some $epsilon > 0$ the map $tildef$ must be increasing or decreasing on $[B - epsilon, B + epsilon]$. Hence it must be increasing. This contradicts the definition of $B$ and we conclude $B = infty$. Similarly $A = -infty$.
answered Jul 23 at 14:55
Paul Frost
3,658420
You should first make precise what you mean by a (closed/open?) arc and by increasing (decreasing).
Obviously $operatornamearc(x,y)$ is defined for any two points $x = e^is, y= e^it$ such that $s < t$ and $t -s < 2 pi$; it is the image of $[s,t] subset mathbbR$ under the map $e : mathbbR to S^1, e(x) = e^ix$ (if you mean the closed arc). But $e$ is a covering so that $f circ e : mathbbR to S^1$ lifts to a map $tildef : mathbbR to mathbbR$ such that $e circ tildef = f circ e$. This lift is not unique, but if $tildef_i$ are any two lifts, we have $tildef_2(x) = tildef_1(x) + 2 k pi$ for some $k in mathbbZ$.
What you mean by "$f$ is increasing on $operatornamearc(x,y)$" is that $tildef$ is increasing on $[s,t]$.
With these clarifications your proof is correct, but we can say more. The map $f$ is in fact a covering, see When is a local homeomorphism a covering map?. This implies that $tildef$ is increasing or decreasing on all of $mathbbR$.
You can see this also directly. W.l.o.g. let $tildef$ be increasing on some interval $[a_0,b_0]$.
Let $A = inf a le a_0 mid tildef text increasing on [a,b_0] $, $B = sup b ge b_0 mid tildef text increasing on [a_0,b] $. Assume that $B < infty$. Then $tildef$ is increasing on $[a_0,B]$. But for some $epsilon > 0$ the map $tildef$ must be increasing or decreasing on $[B - epsilon, B + epsilon]$. Hence it must be increasing. This contradicts the definition of $B$ and we conclude $B = infty$. Similarly $A = -infty$.
answered Jul 23 at 14:55
Paul Frost
3,658420
edited Jul 23 at 16:01
answered Jul 23 at 14:55
Paul Frost
3,658420
answered Jul 23 at 14:55
Paul Frost
3,658420
answered Jul 23 at 14:55
Paul Frost
3,658420
3,658420
add a comment |Â
add a comment |Â
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What is $diamarc(x,y)$?
– leibnewtz
Jul 20 at 16:35
@leibnewtz $diam arc(x, y)= min, 1-$, that $S^1=[0,1]/$
– user479859
Jul 20 at 16:38