Showing that $f^{-1}$ is not continuous.
Let $X$ be the half open interval $[0,2pi)$. Let $Y$ denote the unit circle in the plane.
Let $f$ be the map defined by $f(t)=(cos(t),sin(t))$. I checked that $f$ is continuous and bijective.
Since it is bijective, inverse exist. I want to show that $f^{-1}$ is not continuous at $(1,0)$ using multiple ways(just to check whether I know concepts well or not)
Using compactness. If $f^{-1}$ was continuous then it contradicts the fact that continuous image of a compact set is compact.
$epsilon-delta$ proof. (Idea: If we approach $(1,0)$ from below and above there is a jump from $0$ and $2pi$.)
How to use the most general definition of continuity(that inverse image of open set is open) to show that $f^{-1}$ is not continuous.
Thanks in advance.
real-analysis continuity
asked 7 mins ago
StammeringMathematician
2,2181322
add a comment |
Let $X$ be the half open interval $[0,2pi)$. Let $Y$ denote the unit circle in the plane.
Let $f$ be the map defined by $f(t)=(cos(t),sin(t))$. I checked that $f$ is continuous and bijective.
Since it is bijective, inverse exist. I want to show that $f^{-1}$ is not continuous at $(1,0)$ using multiple ways(just to check whether I know concepts well or not)
Using compactness. If $f^{-1}$ was continuous then it contradicts the fact that continuous image of a compact set is compact.
$epsilon-delta$ proof. (Idea: If we approach $(1,0)$ from below and above there is a jump from $0$ and $2pi$.)
How to use the most general definition of continuity(that inverse image of open set is open) to show that $f^{-1}$ is not continuous.
Thanks in advance.
real-analysis continuity
asked 7 mins ago
StammeringMathematician
2,2181322
add a comment |
Let $X$ be the half open interval $[0,2pi)$. Let $Y$ denote the unit circle in the plane.
Let $f$ be the map defined by $f(t)=(cos(t),sin(t))$. I checked that $f$ is continuous and bijective.
Since it is bijective, inverse exist. I want to show that $f^{-1}$ is not continuous at $(1,0)$ using multiple ways(just to check whether I know concepts well or not)
Using compactness. If $f^{-1}$ was continuous then it contradicts the fact that continuous image of a compact set is compact.
$epsilon-delta$ proof. (Idea: If we approach $(1,0)$ from below and above there is a jump from $0$ and $2pi$.)
How to use the most general definition of continuity(that inverse image of open set is open) to show that $f^{-1}$ is not continuous.
Thanks in advance.
real-analysis continuity
asked 7 mins ago
StammeringMathematician
2,2181322
Let $X$ be the half open interval $[0,2pi)$. Let $Y$ denote the unit circle in the plane.
Let $f$ be the map defined by $f(t)=(cos(t),sin(t))$. I checked that $f$ is continuous and bijective.
Since it is bijective, inverse exist. I want to show that $f^{-1}$ is not continuous at $(1,0)$ using multiple ways(just to check whether I know concepts well or not)
Using compactness. If $f^{-1}$ was continuous then it contradicts the fact that continuous image of a compact set is compact.
$epsilon-delta$ proof. (Idea: If we approach $(1,0)$ from below and above there is a jump from $0$ and $2pi$.)
How to use the most general definition of continuity(that inverse image of open set is open) to show that $f^{-1}$ is not continuous.
Thanks in advance.
real-analysis continuity
real-analysis continuity
asked 7 mins ago
StammeringMathematician
2,2181322
asked 7 mins ago
StammeringMathematician
2,2181322
asked 7 mins ago
StammeringMathematician
2,2181322
asked 7 mins ago
StammeringMathematician
2,2181322
asked 7 mins ago
StammeringMathematician
2,2181322
2,2181322
add a comment |
add a comment |
2 Answers
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For $0<epsilon <2pi$ not that $(f^{-1}) ^{-1} [0,epsilon)=f([0,epsilon)$ is not open even though $[0,epsilon)$ is open on $[0,2pi)$
answered 4 mins ago
Kavi Rama Murthy
50.7k31854
add a comment |
Look at a neighbourhood of $0$ in $X=[0,2pi)$, say $U=[0,1)$. Then $f(U)$ is not
open in $Y$. As $(1,0)in f(U)$ and every neighbourhood of $(1,0)$ in $Y$ contains
points $(x,y)$ with $y<0$, but $f(U)$ has no such points
then $U$ is not open.
You need $f$ to be an open map in order for $f^{-1}$ to be continuous.
answered 1 min ago
Lord Shark the Unknown
101k958132
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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For $0<epsilon <2pi$ not that $(f^{-1}) ^{-1} [0,epsilon)=f([0,epsilon)$ is not open even though $[0,epsilon)$ is open on $[0,2pi)$
answered 4 mins ago
Kavi Rama Murthy
50.7k31854
add a comment |
For $0<epsilon <2pi$ not that $(f^{-1}) ^{-1} [0,epsilon)=f([0,epsilon)$ is not open even though $[0,epsilon)$ is open on $[0,2pi)$
answered 4 mins ago
Kavi Rama Murthy
50.7k31854
add a comment |
For $0<epsilon <2pi$ not that $(f^{-1}) ^{-1} [0,epsilon)=f([0,epsilon)$ is not open even though $[0,epsilon)$ is open on $[0,2pi)$
answered 4 mins ago
Kavi Rama Murthy
50.7k31854
For $0<epsilon <2pi$ not that $(f^{-1}) ^{-1} [0,epsilon)=f([0,epsilon)$ is not open even though $[0,epsilon)$ is open on $[0,2pi)$
answered 4 mins ago
Kavi Rama Murthy
50.7k31854
answered 4 mins ago
Kavi Rama Murthy
50.7k31854
answered 4 mins ago
Kavi Rama Murthy
50.7k31854
answered 4 mins ago
Kavi Rama Murthy
50.7k31854
50.7k31854
add a comment |
add a comment |
Look at a neighbourhood of $0$ in $X=[0,2pi)$, say $U=[0,1)$. Then $f(U)$ is not
open in $Y$. As $(1,0)in f(U)$ and every neighbourhood of $(1,0)$ in $Y$ contains
points $(x,y)$ with $y<0$, but $f(U)$ has no such points
then $U$ is not open.
You need $f$ to be an open map in order for $f^{-1}$ to be continuous.
answered 1 min ago
Lord Shark the Unknown
101k958132
add a comment |
Look at a neighbourhood of $0$ in $X=[0,2pi)$, say $U=[0,1)$. Then $f(U)$ is not
open in $Y$. As $(1,0)in f(U)$ and every neighbourhood of $(1,0)$ in $Y$ contains
points $(x,y)$ with $y<0$, but $f(U)$ has no such points
then $U$ is not open.
You need $f$ to be an open map in order for $f^{-1}$ to be continuous.
answered 1 min ago
Lord Shark the Unknown
101k958132
add a comment |
Look at a neighbourhood of $0$ in $X=[0,2pi)$, say $U=[0,1)$. Then $f(U)$ is not
open in $Y$. As $(1,0)in f(U)$ and every neighbourhood of $(1,0)$ in $Y$ contains
points $(x,y)$ with $y<0$, but $f(U)$ has no such points
then $U$ is not open.
You need $f$ to be an open map in order for $f^{-1}$ to be continuous.
answered 1 min ago
Lord Shark the Unknown
101k958132
Look at a neighbourhood of $0$ in $X=[0,2pi)$, say $U=[0,1)$. Then $f(U)$ is not
open in $Y$. As $(1,0)in f(U)$ and every neighbourhood of $(1,0)$ in $Y$ contains
points $(x,y)$ with $y<0$, but $f(U)$ has no such points
then $U$ is not open.
You need $f$ to be an open map in order for $f^{-1}$ to be continuous.
answered 1 min ago
Lord Shark the Unknown
101k958132
answered 1 min ago
Lord Shark the Unknown
101k958132
answered 1 min ago
Lord Shark the Unknown
101k958132
answered 1 min ago
Lord Shark the Unknown
101k958132
101k958132
add a comment |
add a comment |
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