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How to prove that if $f:mathbbR to mathbbR^2$ is of class $C^1$, then $f$ is not onto.
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In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that
If $f:mathbbR to mathbbR^2$ is of class $C^1$, show that $f$
does not carry $mathbbR$ onto $mathbbR^2$. In fact show that
$f(mathbbR)$ contains no open subset of $R^2$.
I have started with assuming that $f(mathbbR)$ contains an open set $U$ of $mathbbR^2$, and by the continuity of $f$, I have argued that $f^-1(U)$ is open in $mathbbR$; however, after this point, I have stuck, so I would appreciate any hint or help.
real-analysis general-topology derivatives metric-spaces
asked Jul 16 at 13:24
onurcanbektas
3,1061834
add a comment |Â
up vote
2
down vote
favorite
In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that
If $f:mathbbR to mathbbR^2$ is of class $C^1$, show that $f$
does not carry $mathbbR$ onto $mathbbR^2$. In fact show that
$f(mathbbR)$ contains no open subset of $R^2$.
I have started with assuming that $f(mathbbR)$ contains an open set $U$ of $mathbbR^2$, and by the continuity of $f$, I have argued that $f^-1(U)$ is open in $mathbbR$; however, after this point, I have stuck, so I would appreciate any hint or help.
real-analysis general-topology derivatives metric-spaces
asked Jul 16 at 13:24
onurcanbektas
3,1061834
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that
If $f:mathbbR to mathbbR^2$ is of class $C^1$, show that $f$
does not carry $mathbbR$ onto $mathbbR^2$. In fact show that
$f(mathbbR)$ contains no open subset of $R^2$.
I have started with assuming that $f(mathbbR)$ contains an open set $U$ of $mathbbR^2$, and by the continuity of $f$, I have argued that $f^-1(U)$ is open in $mathbbR$; however, after this point, I have stuck, so I would appreciate any hint or help.
real-analysis general-topology derivatives metric-spaces
asked Jul 16 at 13:24
onurcanbektas
3,1061834
In the book of Analysis on Manifolds by Munkres, at page 160, question 1.b, it is asked that
If $f:mathbbR to mathbbR^2$ is of class $C^1$, show that $f$
does not carry $mathbbR$ onto $mathbbR^2$. In fact show that
$f(mathbbR)$ contains no open subset of $R^2$.
I have started with assuming that $f(mathbbR)$ contains an open set $U$ of $mathbbR^2$, and by the continuity of $f$, I have argued that $f^-1(U)$ is open in $mathbbR$; however, after this point, I have stuck, so I would appreciate any hint or help.
real-analysis general-topology derivatives metric-spaces
asked Jul 16 at 13:24
onurcanbektas
3,1061834
asked Jul 16 at 13:24
onurcanbektas
3,1061834
asked Jul 16 at 13:24
onurcanbektas
3,1061834
asked Jul 16 at 13:24
onurcanbektas
3,1061834
3,1061834
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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up vote
10
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accepted
Consider the map$$beginarrayrcccFcolon&mathbbR^2&longrightarrow&mathbbR^2\&(x,y)&mapsto&f(x).endarray$$Then $F$ is of class $C^1$. Therefore, since $mathbbRtimes0$ has measure $0$, $F(mathbbRtimes0)$ 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 $F(mathbbRtimes0)=f(mathbbR)$ and a subset of $mathbbR^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 at 13:43
José Carlos Santos
114k1698177
Well, that was a clever trick :) Thanks a lot for the answer sir.
– onurcanbektas
Jul 16 at 13:44
add a comment |Â
up vote
3
down vote
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 at 13:48
Aweygan
11.9k21437
1
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 :)
– onurcanbektas
Jul 16 at 13:52
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.
– Aweygan
Jul 16 at 14:02
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 ?
– onurcanbektas
Jul 16 at 15:50
1
Since the distance between centers is at least $fracAn$, we have $$fracAn(n^2-1)leqsum_i=2^n^2-1|f(x_i)-f(x_i-1)|leq Msum_i=2^n^2-1x_i-x_i-1leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
– Aweygan
Jul 16 at 15:59
1
You're welcome, glad to help!
– Aweygan
Jul 16 at 17:08
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
Consider the map$$beginarrayrcccFcolon&mathbbR^2&longrightarrow&mathbbR^2\&(x,y)&mapsto&f(x).endarray$$Then $F$ is of class $C^1$. Therefore, since $mathbbRtimes0$ has measure $0$, $F(mathbbRtimes0)$ 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 $F(mathbbRtimes0)=f(mathbbR)$ and a subset of $mathbbR^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 at 13:43
José Carlos Santos
114k1698177
Well, that was a clever trick :) Thanks a lot for the answer sir.
– onurcanbektas
Jul 16 at 13:44
add a comment |Â
up vote
10
down vote
accepted
Consider the map$$beginarrayrcccFcolon&mathbbR^2&longrightarrow&mathbbR^2\&(x,y)&mapsto&f(x).endarray$$Then $F$ is of class $C^1$. Therefore, since $mathbbRtimes0$ has measure $0$, $F(mathbbRtimes0)$ 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 $F(mathbbRtimes0)=f(mathbbR)$ and a subset of $mathbbR^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 at 13:43
José Carlos Santos
114k1698177
Well, that was a clever trick :) Thanks a lot for the answer sir.
– onurcanbektas
Jul 16 at 13:44
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
Consider the map$$beginarrayrcccFcolon&mathbbR^2&longrightarrow&mathbbR^2\&(x,y)&mapsto&f(x).endarray$$Then $F$ is of class $C^1$. Therefore, since $mathbbRtimes0$ has measure $0$, $F(mathbbRtimes0)$ 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 $F(mathbbRtimes0)=f(mathbbR)$ and a subset of $mathbbR^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 at 13:43
José Carlos Santos
114k1698177
Consider the map$$beginarrayrcccFcolon&mathbbR^2&longrightarrow&mathbbR^2\&(x,y)&mapsto&f(x).endarray$$Then $F$ is of class $C^1$. Therefore, since $mathbbRtimes0$ has measure $0$, $F(mathbbRtimes0)$ 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 $F(mathbbRtimes0)=f(mathbbR)$ and a subset of $mathbbR^2$ which contains a non-empty open subset cannot have measure $0$.
answered Jul 16 at 13:43
José Carlos Santos
114k1698177
edited Jul 17 at 6:35
answered Jul 16 at 13:43
José Carlos Santos
114k1698177
answered Jul 16 at 13:43
José Carlos Santos
114k1698177
answered Jul 16 at 13:43
José Carlos Santos
114k1698177
114k1698177
Well, that was a clever trick :) Thanks a lot for the answer sir.
– onurcanbektas
Jul 16 at 13:44
add a comment |Â
Well, that was a clever trick :) Thanks a lot for the answer sir.
– onurcanbektas
Jul 16 at 13:44
Well, that was a clever trick :) Thanks a lot for the answer sir.
– onurcanbektas
Jul 16 at 13:44
Well, that was a clever trick :) Thanks a lot for the answer sir.
– onurcanbektas
Jul 16 at 13:44
add a comment |Â
up vote
3
down vote
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 at 13:48
Aweygan
11.9k21437
1
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 :)
– onurcanbektas
Jul 16 at 13:52
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.
– Aweygan
Jul 16 at 14:02
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 ?
– onurcanbektas
Jul 16 at 15:50
1
Since the distance between centers is at least $fracAn$, we have $$fracAn(n^2-1)leqsum_i=2^n^2-1|f(x_i)-f(x_i-1)|leq Msum_i=2^n^2-1x_i-x_i-1leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
– Aweygan
Jul 16 at 15:59
1
You're welcome, glad to help!
– Aweygan
Jul 16 at 17:08
 |Â
show 4 more comments
up vote
3
down vote
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 at 13:48
Aweygan
11.9k21437
1
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 :)
– onurcanbektas
Jul 16 at 13:52
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.
– Aweygan
Jul 16 at 14:02
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 ?
– onurcanbektas
Jul 16 at 15:50
1
Since the distance between centers is at least $fracAn$, we have $$fracAn(n^2-1)leqsum_i=2^n^2-1|f(x_i)-f(x_i-1)|leq Msum_i=2^n^2-1x_i-x_i-1leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
– Aweygan
Jul 16 at 15:59
1
You're welcome, glad to help!
– Aweygan
Jul 16 at 17:08
 |Â
show 4 more comments
up vote
3
down vote
up vote
3
down vote
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 at 13:48
Aweygan
11.9k21437
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 at 13:48
Aweygan
11.9k21437
answered Jul 16 at 13:48
Aweygan
11.9k21437
answered Jul 16 at 13:48
Aweygan
11.9k21437
answered Jul 16 at 13:48
Aweygan
11.9k21437
11.9k21437
1
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 :)
– onurcanbektas
Jul 16 at 13:52
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.
– Aweygan
Jul 16 at 14:02
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 ?
– onurcanbektas
Jul 16 at 15:50
1
Since the distance between centers is at least $fracAn$, we have $$fracAn(n^2-1)leqsum_i=2^n^2-1|f(x_i)-f(x_i-1)|leq Msum_i=2^n^2-1x_i-x_i-1leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
– Aweygan
Jul 16 at 15:59
1
You're welcome, glad to help!
– Aweygan
Jul 16 at 17:08
 |Â
show 4 more comments
1
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 :)
– onurcanbektas
Jul 16 at 13:52
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.
– Aweygan
Jul 16 at 14:02
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 ?
– onurcanbektas
Jul 16 at 15:50
1
Since the distance between centers is at least $fracAn$, we have $$fracAn(n^2-1)leqsum_i=2^n^2-1|f(x_i)-f(x_i-1)|leq Msum_i=2^n^2-1x_i-x_i-1leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
– Aweygan
Jul 16 at 15:59
1
You're welcome, glad to help!
– Aweygan
Jul 16 at 17:08
1
1
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 :)
– onurcanbektas
Jul 16 at 13:52
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 :)
– onurcanbektas
Jul 16 at 13:52
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.
– Aweygan
Jul 16 at 14:02
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.
– Aweygan
Jul 16 at 14:02
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 ?
– onurcanbektas
Jul 16 at 15:50
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 ?
– onurcanbektas
Jul 16 at 15:50
1
1
Since the distance between centers is at least $fracAn$, we have $$fracAn(n^2-1)leqsum_i=2^n^2-1|f(x_i)-f(x_i-1)|leq Msum_i=2^n^2-1x_i-x_i-1leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
– Aweygan
Jul 16 at 15:59
Since the distance between centers is at least $fracAn$, we have $$fracAn(n^2-1)leqsum_i=2^n^2-1|f(x_i)-f(x_i-1)|leq Msum_i=2^n^2-1x_i-x_i-1leq2MN.$$ But choosing $n$ sufficiently large, we obtain a contradiction.
– Aweygan
Jul 16 at 15:59
1
1
You're welcome, glad to help!
– Aweygan
Jul 16 at 17:08
You're welcome, glad to help!
– Aweygan
Jul 16 at 17:08
 |Â
show 4 more comments
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