Mixing
How to prove that if $f:mathbb{R} to mathbb{R}^2$ is of class $C^1$, then $f$ is not onto.
$begingroup$
In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that
If $f:mathbb{R} to mathbb{R}^2$ is of class $C^1$, show that $f$
does not carry $mathbb{R}$ onto $mathbb{R}^2$. In fact show that
$f(mathbb{R})$ contains no open subset of $R^2$.
I have started with assuming that $f(mathbb{R})$ contains an open set $U$ of $mathbb{R}^2$, and by the continuity of $f$, I have argued that $f^{-1}(U)$ is open in $mathbb{R}$; however, after this point, I am stuck, so I would appreciate any hint or help.
real-analysis general-topology derivatives metric-spaces
edited Jan 1 at 21:13
José Carlos Santos
153k22123225
asked Jul 16 '18 at 13:24
onurcanbektasonurcanbektas
3,3541936
$endgroup$
add a comment |
$begingroup$
In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that
If $f:mathbb{R} to mathbb{R}^2$ is of class $C^1$, show that $f$
does not carry $mathbb{R}$ onto $mathbb{R}^2$. In fact show that
$f(mathbb{R})$ contains no open subset of $R^2$.
I have started with assuming that $f(mathbb{R})$ contains an open set $U$ of $mathbb{R}^2$, and by the continuity of $f$, I have argued that $f^{-1}(U)$ is open in $mathbb{R}$; however, after this point, I am stuck, so I would appreciate any hint or help.
real-analysis general-topology derivatives metric-spaces
edited Jan 1 at 21:13
José Carlos Santos
153k22123225
asked Jul 16 '18 at 13:24
onurcanbektasonurcanbektas
3,3541936
$endgroup$
add a comment |
$begingroup$
In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that
If $f:mathbb{R} to mathbb{R}^2$ is of class $C^1$, show that $f$
does not carry $mathbb{R}$ onto $mathbb{R}^2$. In fact show that
$f(mathbb{R})$ contains no open subset of $R^2$.
I have started with assuming that $f(mathbb{R})$ contains an open set $U$ of $mathbb{R}^2$, and by the continuity of $f$, I have argued that $f^{-1}(U)$ is open in $mathbb{R}$; however, after this point, I am stuck, so I would appreciate any hint or help.
real-analysis general-topology derivatives metric-spaces
edited Jan 1 at 21:13
José Carlos Santos
153k22123225
asked Jul 16 '18 at 13:24
onurcanbektasonurcanbektas
3,3541936
$endgroup$
In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that
If $f:mathbb{R} to mathbb{R}^2$ is of class $C^1$, show that $f$
does not carry $mathbb{R}$ onto $mathbb{R}^2$. In fact show that
$f(mathbb{R})$ contains no open subset of $R^2$.
I have started with assuming that $f(mathbb{R})$ contains an open set $U$ of $mathbb{R}^2$, and by the continuity of $f$, I have argued that $f^{-1}(U)$ is open in $mathbb{R}$; however, after this point, I am stuck, so I would appreciate any hint or help.
real-analysis general-topology derivatives metric-spaces
real-analysis general-topology derivatives metric-spaces
edited Jan 1 at 21:13
José Carlos Santos
153k22123225
asked Jul 16 '18 at 13:24
onurcanbektasonurcanbektas
3,3541936
edited Jan 1 at 21:13
José Carlos Santos
153k22123225
asked Jul 16 '18 at 13:24
onurcanbektasonurcanbektas
3,3541936
edited Jan 1 at 21:13
José Carlos Santos
153k22123225
edited Jan 1 at 21:13
José Carlos Santos
153k22123225
edited Jan 1 at 21:13
José Carlos Santos
153k22123225
153k22123225
asked Jul 16 '18 at 13:24
onurcanbektasonurcanbektas
3,3541936
asked Jul 16 '18 at 13:24
onurcanbektasonurcanbektas
3,3541936
asked Jul 16 '18 at 13:24
onurcanbektasonurcanbektas
3,3541936
3,3541936
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the map$$begin{array}{rccc}Fcolon&mathbb{R}^2&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&f(x).end{array}$$Then $F$ is of class $C^1$. Therefore, since $mathbb{R}times{0}$ has measure $0$, $Fbigl(mathbb{R}times{0}bigr)$ has measure $0$ too (here, I am using the fact that functions of class $C^1$ map sets of measure $0$ into sets of measure $0$, which is lemma 18.1 in Munkres' textbook). But $Fbigl(mathbb{R}times{0}bigr)=f(mathbb{R})$ and a subset of $mathbb{R}^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 '18 at 13:43
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Well, that was a clever trick :) Thanks a lot for the answer sir.
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:44
add a comment |
$begingroup$
Here's another approach.
For each $Ninmathbb N$, the restriction $f_N$ of $f$ to $[-N,N]$ is Lipschitz. Use this to show that $f_N([-N,N])$ does not contain a rectangle, hence has empty interior. Then since $f(mathbb R)=cup_Nf([-N,N])$, the Baire category theorem yields the result.
answered Jul 16 '18 at 13:48
AweyganAweygan
13.6k21441
$endgroup$
1
$begingroup$
Well, in that approach, there is two problems that I have no idea how to overcome.First is that how to show a set does not contain a rectangle, and second is that what is Baire category theorem :)
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:52
$begingroup$
If it the range contains a square, divide it into smaller squares, and sum up distances from centers of squares to obtain a contradiction. The Baire category theorem says that a complete metric space is not the union of countably many nowhere dense sets.
$endgroup$
– Aweygan
Jul 16 '18 at 14:02
$begingroup$
By Lipschitz condition, the width of each square in the image bounded by the Lipschitz constant times the distance between the pre-images, and this will be true for any opposite points in the boundary, each of which is in the image of a $C^1$ map ... .I'm lost sir, do you have any light ?
$endgroup$
– onurcanbektas
Jul 16 '18 at 15:50
1
$begingroup$
Since the distance between centers is at least $frac{A}{n}$, we have $$frac{A}{n}(n^2-1)leqsum_{i=2}^{n^2-1}|f(x_i)-f(x_{i-1})|leq Msum_{i=2}^{n^2-1}x_i-x_{i-1}leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
$endgroup$
– Aweygan
Jul 16 '18 at 15:59
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jul 16 '18 at 17:08
|
show 4 more comments
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
votes
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oldest
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$begingroup$
Consider the map$$begin{array}{rccc}Fcolon&mathbb{R}^2&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&f(x).end{array}$$Then $F$ is of class $C^1$. Therefore, since $mathbb{R}times{0}$ has measure $0$, $Fbigl(mathbb{R}times{0}bigr)$ has measure $0$ too (here, I am using the fact that functions of class $C^1$ map sets of measure $0$ into sets of measure $0$, which is lemma 18.1 in Munkres' textbook). But $Fbigl(mathbb{R}times{0}bigr)=f(mathbb{R})$ and a subset of $mathbb{R}^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 '18 at 13:43
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Well, that was a clever trick :) Thanks a lot for the answer sir.
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:44
add a comment |
$begingroup$
Consider the map$$begin{array}{rccc}Fcolon&mathbb{R}^2&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&f(x).end{array}$$Then $F$ is of class $C^1$. Therefore, since $mathbb{R}times{0}$ has measure $0$, $Fbigl(mathbb{R}times{0}bigr)$ has measure $0$ too (here, I am using the fact that functions of class $C^1$ map sets of measure $0$ into sets of measure $0$, which is lemma 18.1 in Munkres' textbook). But $Fbigl(mathbb{R}times{0}bigr)=f(mathbb{R})$ and a subset of $mathbb{R}^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 '18 at 13:43
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Well, that was a clever trick :) Thanks a lot for the answer sir.
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:44
add a comment |
$begingroup$
Consider the map$$begin{array}{rccc}Fcolon&mathbb{R}^2&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&f(x).end{array}$$Then $F$ is of class $C^1$. Therefore, since $mathbb{R}times{0}$ has measure $0$, $Fbigl(mathbb{R}times{0}bigr)$ has measure $0$ too (here, I am using the fact that functions of class $C^1$ map sets of measure $0$ into sets of measure $0$, which is lemma 18.1 in Munkres' textbook). But $Fbigl(mathbb{R}times{0}bigr)=f(mathbb{R})$ and a subset of $mathbb{R}^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 '18 at 13:43
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
Consider the map$$begin{array}{rccc}Fcolon&mathbb{R}^2&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&f(x).end{array}$$Then $F$ is of class $C^1$. Therefore, since $mathbb{R}times{0}$ has measure $0$, $Fbigl(mathbb{R}times{0}bigr)$ has measure $0$ too (here, I am using the fact that functions of class $C^1$ map sets of measure $0$ into sets of measure $0$, which is lemma 18.1 in Munkres' textbook). But $Fbigl(mathbb{R}times{0}bigr)=f(mathbb{R})$ and a subset of $mathbb{R}^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 '18 at 13:43
José Carlos SantosJosé Carlos Santos
153k22123225
edited Dec 29 '18 at 11:36
answered Jul 16 '18 at 13:43
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jul 16 '18 at 13:43
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jul 16 '18 at 13:43
José Carlos SantosJosé Carlos Santos
153k22123225
153k22123225
$begingroup$
Well, that was a clever trick :) Thanks a lot for the answer sir.
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:44
add a comment |
$begingroup$
Well, that was a clever trick :) Thanks a lot for the answer sir.
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:44
$begingroup$
Well, that was a clever trick :) Thanks a lot for the answer sir.
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:44
$begingroup$
Well, that was a clever trick :) Thanks a lot for the answer sir.
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:44
add a comment |
$begingroup$
Here's another approach.
For each $Ninmathbb N$, the restriction $f_N$ of $f$ to $[-N,N]$ is Lipschitz. Use this to show that $f_N([-N,N])$ does not contain a rectangle, hence has empty interior. Then since $f(mathbb R)=cup_Nf([-N,N])$, the Baire category theorem yields the result.
answered Jul 16 '18 at 13:48
AweyganAweygan
13.6k21441
$endgroup$
1
$begingroup$
Well, in that approach, there is two problems that I have no idea how to overcome.First is that how to show a set does not contain a rectangle, and second is that what is Baire category theorem :)
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:52
$begingroup$
If it the range contains a square, divide it into smaller squares, and sum up distances from centers of squares to obtain a contradiction. The Baire category theorem says that a complete metric space is not the union of countably many nowhere dense sets.
$endgroup$
– Aweygan
Jul 16 '18 at 14:02
$begingroup$
By Lipschitz condition, the width of each square in the image bounded by the Lipschitz constant times the distance between the pre-images, and this will be true for any opposite points in the boundary, each of which is in the image of a $C^1$ map ... .I'm lost sir, do you have any light ?
$endgroup$
– onurcanbektas
Jul 16 '18 at 15:50
1
$begingroup$
Since the distance between centers is at least $frac{A}{n}$, we have $$frac{A}{n}(n^2-1)leqsum_{i=2}^{n^2-1}|f(x_i)-f(x_{i-1})|leq Msum_{i=2}^{n^2-1}x_i-x_{i-1}leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
$endgroup$
– Aweygan
Jul 16 '18 at 15:59
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jul 16 '18 at 17:08
|
show 4 more comments
$begingroup$
Here's another approach.
For each $Ninmathbb N$, the restriction $f_N$ of $f$ to $[-N,N]$ is Lipschitz. Use this to show that $f_N([-N,N])$ does not contain a rectangle, hence has empty interior. Then since $f(mathbb R)=cup_Nf([-N,N])$, the Baire category theorem yields the result.
answered Jul 16 '18 at 13:48
AweyganAweygan
13.6k21441
$endgroup$
1
$begingroup$
Well, in that approach, there is two problems that I have no idea how to overcome.First is that how to show a set does not contain a rectangle, and second is that what is Baire category theorem :)
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:52
$begingroup$
If it the range contains a square, divide it into smaller squares, and sum up distances from centers of squares to obtain a contradiction. The Baire category theorem says that a complete metric space is not the union of countably many nowhere dense sets.
$endgroup$
– Aweygan
Jul 16 '18 at 14:02
$begingroup$
By Lipschitz condition, the width of each square in the image bounded by the Lipschitz constant times the distance between the pre-images, and this will be true for any opposite points in the boundary, each of which is in the image of a $C^1$ map ... .I'm lost sir, do you have any light ?
$endgroup$
– onurcanbektas
Jul 16 '18 at 15:50
1
$begingroup$
Since the distance between centers is at least $frac{A}{n}$, we have $$frac{A}{n}(n^2-1)leqsum_{i=2}^{n^2-1}|f(x_i)-f(x_{i-1})|leq Msum_{i=2}^{n^2-1}x_i-x_{i-1}leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
$endgroup$
– Aweygan
Jul 16 '18 at 15:59
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jul 16 '18 at 17:08
|
show 4 more comments
$begingroup$
Here's another approach.
For each $Ninmathbb N$, the restriction $f_N$ of $f$ to $[-N,N]$ is Lipschitz. Use this to show that $f_N([-N,N])$ does not contain a rectangle, hence has empty interior. Then since $f(mathbb R)=cup_Nf([-N,N])$, the Baire category theorem yields the result.
answered Jul 16 '18 at 13:48
AweyganAweygan
13.6k21441
$endgroup$
Here's another approach.
For each $Ninmathbb N$, the restriction $f_N$ of $f$ to $[-N,N]$ is Lipschitz. Use this to show that $f_N([-N,N])$ does not contain a rectangle, hence has empty interior. Then since $f(mathbb R)=cup_Nf([-N,N])$, the Baire category theorem yields the result.
answered Jul 16 '18 at 13:48
AweyganAweygan
13.6k21441
answered Jul 16 '18 at 13:48
AweyganAweygan
13.6k21441
answered Jul 16 '18 at 13:48
AweyganAweygan
13.6k21441
answered Jul 16 '18 at 13:48
AweyganAweygan
13.6k21441
13.6k21441
1
$begingroup$
Well, in that approach, there is two problems that I have no idea how to overcome.First is that how to show a set does not contain a rectangle, and second is that what is Baire category theorem :)
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:52
$begingroup$
If it the range contains a square, divide it into smaller squares, and sum up distances from centers of squares to obtain a contradiction. The Baire category theorem says that a complete metric space is not the union of countably many nowhere dense sets.
$endgroup$
– Aweygan
Jul 16 '18 at 14:02
$begingroup$
By Lipschitz condition, the width of each square in the image bounded by the Lipschitz constant times the distance between the pre-images, and this will be true for any opposite points in the boundary, each of which is in the image of a $C^1$ map ... .I'm lost sir, do you have any light ?
$endgroup$
– onurcanbektas
Jul 16 '18 at 15:50
1
$begingroup$
Since the distance between centers is at least $frac{A}{n}$, we have $$frac{A}{n}(n^2-1)leqsum_{i=2}^{n^2-1}|f(x_i)-f(x_{i-1})|leq Msum_{i=2}^{n^2-1}x_i-x_{i-1}leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
$endgroup$
– Aweygan
Jul 16 '18 at 15:59
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jul 16 '18 at 17:08
|
show 4 more comments
1
$begingroup$
Well, in that approach, there is two problems that I have no idea how to overcome.First is that how to show a set does not contain a rectangle, and second is that what is Baire category theorem :)
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:52
$begingroup$
If it the range contains a square, divide it into smaller squares, and sum up distances from centers of squares to obtain a contradiction. The Baire category theorem says that a complete metric space is not the union of countably many nowhere dense sets.
$endgroup$
– Aweygan
Jul 16 '18 at 14:02
$begingroup$
By Lipschitz condition, the width of each square in the image bounded by the Lipschitz constant times the distance between the pre-images, and this will be true for any opposite points in the boundary, each of which is in the image of a $C^1$ map ... .I'm lost sir, do you have any light ?
$endgroup$
– onurcanbektas
Jul 16 '18 at 15:50
1
$begingroup$
Since the distance between centers is at least $frac{A}{n}$, we have $$frac{A}{n}(n^2-1)leqsum_{i=2}^{n^2-1}|f(x_i)-f(x_{i-1})|leq Msum_{i=2}^{n^2-1}x_i-x_{i-1}leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
$endgroup$
– Aweygan
Jul 16 '18 at 15:59
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jul 16 '18 at 17:08
1
1
$begingroup$
Well, in that approach, there is two problems that I have no idea how to overcome.First is that how to show a set does not contain a rectangle, and second is that what is Baire category theorem :)
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:52
$begingroup$
Well, in that approach, there is two problems that I have no idea how to overcome.First is that how to show a set does not contain a rectangle, and second is that what is Baire category theorem :)
$endgroup$
– onurcanbektas
Jul 16 '18 at 13:52
$begingroup$
If it the range contains a square, divide it into smaller squares, and sum up distances from centers of squares to obtain a contradiction. The Baire category theorem says that a complete metric space is not the union of countably many nowhere dense sets.
$endgroup$
– Aweygan
Jul 16 '18 at 14:02
$begingroup$
If it the range contains a square, divide it into smaller squares, and sum up distances from centers of squares to obtain a contradiction. The Baire category theorem says that a complete metric space is not the union of countably many nowhere dense sets.
$endgroup$
– Aweygan
Jul 16 '18 at 14:02
$begingroup$
By Lipschitz condition, the width of each square in the image bounded by the Lipschitz constant times the distance between the pre-images, and this will be true for any opposite points in the boundary, each of which is in the image of a $C^1$ map ... .I'm lost sir, do you have any light ?
$endgroup$
– onurcanbektas
Jul 16 '18 at 15:50
$begingroup$
By Lipschitz condition, the width of each square in the image bounded by the Lipschitz constant times the distance between the pre-images, and this will be true for any opposite points in the boundary, each of which is in the image of a $C^1$ map ... .I'm lost sir, do you have any light ?
$endgroup$
– onurcanbektas
Jul 16 '18 at 15:50
1
1
$begingroup$
Since the distance between centers is at least $frac{A}{n}$, we have $$frac{A}{n}(n^2-1)leqsum_{i=2}^{n^2-1}|f(x_i)-f(x_{i-1})|leq Msum_{i=2}^{n^2-1}x_i-x_{i-1}leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
$endgroup$
– Aweygan
Jul 16 '18 at 15:59
$begingroup$
Since the distance between centers is at least $frac{A}{n}$, we have $$frac{A}{n}(n^2-1)leqsum_{i=2}^{n^2-1}|f(x_i)-f(x_{i-1})|leq Msum_{i=2}^{n^2-1}x_i-x_{i-1}leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
$endgroup$
– Aweygan
Jul 16 '18 at 15:59
1
1
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jul 16 '18 at 17:08
$begingroup$
You're welcome, glad to help!
$endgroup$
– Aweygan
Jul 16 '18 at 17:08
|
show 4 more comments
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