$Usubseteqmathbb{R}^n$ is open and connected $f:Uto mathbb{R}^m$ differentiable.$Df(x)=0$ $forall x in U$....
Attempt:
Since U is open and connected, U is path connected. then there exists a path $gamma(t):[0,1]to U$ between any points $a,bin U$ such that $gamma(1)=b$ and $gamma(0)=a$
Define $h(t)=(fcircgamma)(t):mathbb{R}to mathbb{R}^m$
What I want to do:
$h(1)-h(0)=int_0^1 frac{d}{dx}f(gamma(t))dt=int_0^1Df(gamma(t))gamma^prime(t)dt=0$
then $f(b)=f(a)$ $forall a,bin U$ so $f$ is constant. But I don't know that this path is differentiable, so this approach doesn't work.
multivariable-calculus connectedness
asked Nov 24 at 2:07
AColoredReptile
1788
|
show 4 more comments
Attempt:
Since U is open and connected, U is path connected. then there exists a path $gamma(t):[0,1]to U$ between any points $a,bin U$ such that $gamma(1)=b$ and $gamma(0)=a$
Define $h(t)=(fcircgamma)(t):mathbb{R}to mathbb{R}^m$
What I want to do:
$h(1)-h(0)=int_0^1 frac{d}{dx}f(gamma(t))dt=int_0^1Df(gamma(t))gamma^prime(t)dt=0$
then $f(b)=f(a)$ $forall a,bin U$ so $f$ is constant. But I don't know that this path is differentiable, so this approach doesn't work.
multivariable-calculus connectedness
asked Nov 24 at 2:07
AColoredReptile
1788
What is to stop you from simply choosing a differentiable path?
– Robert Lewis
Nov 24 at 2:15
2
The definition I have of path connected simply says there is a path for each a,b not that there is a differentiable one.
– AColoredReptile
Nov 24 at 2:19
That is true. I think though that you can show there is a differentiable path by starting with a continuous one, constructing an open cover, using compactness to find a finite open cover, an then in each element of that open cover explicitly presenting a differentiable path, and then making sure they "hook up" correctly.
– Robert Lewis
Nov 24 at 2:32
2
@AColoredReptile $U$ is open, the path can be chosen to be differentiable.
– Jacky Chong
Nov 24 at 2:39
@Jacky Chong Is there a name for this theorem or some proof of this result?
– AColoredReptile
Nov 24 at 2:42
|
show 4 more comments
Attempt:
Since U is open and connected, U is path connected. then there exists a path $gamma(t):[0,1]to U$ between any points $a,bin U$ such that $gamma(1)=b$ and $gamma(0)=a$
Define $h(t)=(fcircgamma)(t):mathbb{R}to mathbb{R}^m$
What I want to do:
$h(1)-h(0)=int_0^1 frac{d}{dx}f(gamma(t))dt=int_0^1Df(gamma(t))gamma^prime(t)dt=0$
then $f(b)=f(a)$ $forall a,bin U$ so $f$ is constant. But I don't know that this path is differentiable, so this approach doesn't work.
multivariable-calculus connectedness
asked Nov 24 at 2:07
AColoredReptile
1788
Attempt:
Since U is open and connected, U is path connected. then there exists a path $gamma(t):[0,1]to U$ between any points $a,bin U$ such that $gamma(1)=b$ and $gamma(0)=a$
Define $h(t)=(fcircgamma)(t):mathbb{R}to mathbb{R}^m$
What I want to do:
$h(1)-h(0)=int_0^1 frac{d}{dx}f(gamma(t))dt=int_0^1Df(gamma(t))gamma^prime(t)dt=0$
then $f(b)=f(a)$ $forall a,bin U$ so $f$ is constant. But I don't know that this path is differentiable, so this approach doesn't work.
multivariable-calculus connectedness
multivariable-calculus connectedness
asked Nov 24 at 2:07
AColoredReptile
1788
asked Nov 24 at 2:07
AColoredReptile
1788
asked Nov 24 at 2:07
AColoredReptile
1788
asked Nov 24 at 2:07
AColoredReptile
1788
asked Nov 24 at 2:07
AColoredReptile
1788
1788
What is to stop you from simply choosing a differentiable path?
– Robert Lewis
Nov 24 at 2:15
2
The definition I have of path connected simply says there is a path for each a,b not that there is a differentiable one.
– AColoredReptile
Nov 24 at 2:19
That is true. I think though that you can show there is a differentiable path by starting with a continuous one, constructing an open cover, using compactness to find a finite open cover, an then in each element of that open cover explicitly presenting a differentiable path, and then making sure they "hook up" correctly.
– Robert Lewis
Nov 24 at 2:32
2
@AColoredReptile $U$ is open, the path can be chosen to be differentiable.
– Jacky Chong
Nov 24 at 2:39
@Jacky Chong Is there a name for this theorem or some proof of this result?
– AColoredReptile
Nov 24 at 2:42
|
show 4 more comments
What is to stop you from simply choosing a differentiable path?
– Robert Lewis
Nov 24 at 2:15
2
The definition I have of path connected simply says there is a path for each a,b not that there is a differentiable one.
– AColoredReptile
Nov 24 at 2:19
That is true. I think though that you can show there is a differentiable path by starting with a continuous one, constructing an open cover, using compactness to find a finite open cover, an then in each element of that open cover explicitly presenting a differentiable path, and then making sure they "hook up" correctly.
– Robert Lewis
Nov 24 at 2:32
2
@AColoredReptile $U$ is open, the path can be chosen to be differentiable.
– Jacky Chong
Nov 24 at 2:39
@Jacky Chong Is there a name for this theorem or some proof of this result?
– AColoredReptile
Nov 24 at 2:42
What is to stop you from simply choosing a differentiable path?
– Robert Lewis
Nov 24 at 2:15
What is to stop you from simply choosing a differentiable path?
– Robert Lewis
Nov 24 at 2:15
2
2
The definition I have of path connected simply says there is a path for each a,b not that there is a differentiable one.
– AColoredReptile
Nov 24 at 2:19
The definition I have of path connected simply says there is a path for each a,b not that there is a differentiable one.
– AColoredReptile
Nov 24 at 2:19
That is true. I think though that you can show there is a differentiable path by starting with a continuous one, constructing an open cover, using compactness to find a finite open cover, an then in each element of that open cover explicitly presenting a differentiable path, and then making sure they "hook up" correctly.
– Robert Lewis
Nov 24 at 2:32
That is true. I think though that you can show there is a differentiable path by starting with a continuous one, constructing an open cover, using compactness to find a finite open cover, an then in each element of that open cover explicitly presenting a differentiable path, and then making sure they "hook up" correctly.
– Robert Lewis
Nov 24 at 2:32
2
2
@AColoredReptile $U$ is open, the path can be chosen to be differentiable.
– Jacky Chong
Nov 24 at 2:39
@AColoredReptile $U$ is open, the path can be chosen to be differentiable.
– Jacky Chong
Nov 24 at 2:39
@Jacky Chong Is there a name for this theorem or some proof of this result?
– AColoredReptile
Nov 24 at 2:42
@Jacky Chong Is there a name for this theorem or some proof of this result?
– AColoredReptile
Nov 24 at 2:42
|
show 4 more comments
2 Answers
2
active
oldest
votes
It reduces to the case $m=1$. Let $a$ in $U$ and let $b=f(a)$. We will prove that $A={xin U:f(x)=b}$ is non-empty, open and closed in $U$. By definition, $a$ is in $A$ so $A$ is non-empty. Also $A=f^{-1}({b})$ is closed because it is the preimage of a closed set under a continuous function. Finally, if $p$ is in $A$, there exists a positive real number $epsilon$ such that the open ball $B_epsilon(p)$ of center $p$ and radius $epsilon$ is included in $U$. Let $q$ in $Usetminus{p}$. Consider the continuous function $g:tin [0,1]mapsto f(tq+(1-t)p)in B_epsilon(p)$. Then $g$ is differentiable in $(0,1)$ and its derivative is $g'(t)=D_{g(t)}f.(q-p)=0$. So $g$ is constant and in particular $g(0)=f(p)=f(q)=g(1)$. Then $q$ is in $A$ and thus $B_epsilon(p)subseteq A$. So $A$ is open. We conclude that $A=U$ because $U$ is connected.
answered Nov 24 at 2:47
Dante Grevino
90819
Thanks, can you explain why you can reduce this to m=1?
– AColoredReptile
Nov 24 at 3:01
Yes, sure. You can prove the result for each coordinate $f_i$, with $1leq i leq m$, of $f$. Do you see it? Recall that the rows of the differential matrix of $f$ are the gradients of the coordinates $f_i$.
– Dante Grevino
Nov 24 at 3:45
add a comment |
We can prove the result by contradiction as follows:
If possible let $ f$ is not a constant.
Then the cardinality of $text{Image (f) or Im(f)}$ must be greater than $1$.
Let $ v in Im(f)$ and consider the following two sets:
$ A=f^{-1}{v }$ and $B=f^{-1} (Im(f) setminus {v })$.
Now it is your task to show that (use continuity of $f$):
$(i) A,B $ are non-empty sets,
$(ii) A, B $ are both open sets,
$(iii) A cup B=U$,
$(iv) A cap B=phi$.
By separation axiom $ U$ is disconnected because given that $U$ is connected.
Hence a contradiction.
This prove that $ f$ must be constant.
answered Nov 24 at 3:30
M. A. SARKAR
2,1391619
add a comment |
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2 Answers
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2 Answers
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It reduces to the case $m=1$. Let $a$ in $U$ and let $b=f(a)$. We will prove that $A={xin U:f(x)=b}$ is non-empty, open and closed in $U$. By definition, $a$ is in $A$ so $A$ is non-empty. Also $A=f^{-1}({b})$ is closed because it is the preimage of a closed set under a continuous function. Finally, if $p$ is in $A$, there exists a positive real number $epsilon$ such that the open ball $B_epsilon(p)$ of center $p$ and radius $epsilon$ is included in $U$. Let $q$ in $Usetminus{p}$. Consider the continuous function $g:tin [0,1]mapsto f(tq+(1-t)p)in B_epsilon(p)$. Then $g$ is differentiable in $(0,1)$ and its derivative is $g'(t)=D_{g(t)}f.(q-p)=0$. So $g$ is constant and in particular $g(0)=f(p)=f(q)=g(1)$. Then $q$ is in $A$ and thus $B_epsilon(p)subseteq A$. So $A$ is open. We conclude that $A=U$ because $U$ is connected.
answered Nov 24 at 2:47
Dante Grevino
90819
Thanks, can you explain why you can reduce this to m=1?
– AColoredReptile
Nov 24 at 3:01
Yes, sure. You can prove the result for each coordinate $f_i$, with $1leq i leq m$, of $f$. Do you see it? Recall that the rows of the differential matrix of $f$ are the gradients of the coordinates $f_i$.
– Dante Grevino
Nov 24 at 3:45
add a comment |
It reduces to the case $m=1$. Let $a$ in $U$ and let $b=f(a)$. We will prove that $A={xin U:f(x)=b}$ is non-empty, open and closed in $U$. By definition, $a$ is in $A$ so $A$ is non-empty. Also $A=f^{-1}({b})$ is closed because it is the preimage of a closed set under a continuous function. Finally, if $p$ is in $A$, there exists a positive real number $epsilon$ such that the open ball $B_epsilon(p)$ of center $p$ and radius $epsilon$ is included in $U$. Let $q$ in $Usetminus{p}$. Consider the continuous function $g:tin [0,1]mapsto f(tq+(1-t)p)in B_epsilon(p)$. Then $g$ is differentiable in $(0,1)$ and its derivative is $g'(t)=D_{g(t)}f.(q-p)=0$. So $g$ is constant and in particular $g(0)=f(p)=f(q)=g(1)$. Then $q$ is in $A$ and thus $B_epsilon(p)subseteq A$. So $A$ is open. We conclude that $A=U$ because $U$ is connected.
answered Nov 24 at 2:47
Dante Grevino
90819
Thanks, can you explain why you can reduce this to m=1?
– AColoredReptile
Nov 24 at 3:01
Yes, sure. You can prove the result for each coordinate $f_i$, with $1leq i leq m$, of $f$. Do you see it? Recall that the rows of the differential matrix of $f$ are the gradients of the coordinates $f_i$.
– Dante Grevino
Nov 24 at 3:45
add a comment |
It reduces to the case $m=1$. Let $a$ in $U$ and let $b=f(a)$. We will prove that $A={xin U:f(x)=b}$ is non-empty, open and closed in $U$. By definition, $a$ is in $A$ so $A$ is non-empty. Also $A=f^{-1}({b})$ is closed because it is the preimage of a closed set under a continuous function. Finally, if $p$ is in $A$, there exists a positive real number $epsilon$ such that the open ball $B_epsilon(p)$ of center $p$ and radius $epsilon$ is included in $U$. Let $q$ in $Usetminus{p}$. Consider the continuous function $g:tin [0,1]mapsto f(tq+(1-t)p)in B_epsilon(p)$. Then $g$ is differentiable in $(0,1)$ and its derivative is $g'(t)=D_{g(t)}f.(q-p)=0$. So $g$ is constant and in particular $g(0)=f(p)=f(q)=g(1)$. Then $q$ is in $A$ and thus $B_epsilon(p)subseteq A$. So $A$ is open. We conclude that $A=U$ because $U$ is connected.
answered Nov 24 at 2:47
Dante Grevino
90819
It reduces to the case $m=1$. Let $a$ in $U$ and let $b=f(a)$. We will prove that $A={xin U:f(x)=b}$ is non-empty, open and closed in $U$. By definition, $a$ is in $A$ so $A$ is non-empty. Also $A=f^{-1}({b})$ is closed because it is the preimage of a closed set under a continuous function. Finally, if $p$ is in $A$, there exists a positive real number $epsilon$ such that the open ball $B_epsilon(p)$ of center $p$ and radius $epsilon$ is included in $U$. Let $q$ in $Usetminus{p}$. Consider the continuous function $g:tin [0,1]mapsto f(tq+(1-t)p)in B_epsilon(p)$. Then $g$ is differentiable in $(0,1)$ and its derivative is $g'(t)=D_{g(t)}f.(q-p)=0$. So $g$ is constant and in particular $g(0)=f(p)=f(q)=g(1)$. Then $q$ is in $A$ and thus $B_epsilon(p)subseteq A$. So $A$ is open. We conclude that $A=U$ because $U$ is connected.
answered Nov 24 at 2:47
Dante Grevino
90819
answered Nov 24 at 2:47
Dante Grevino
90819
answered Nov 24 at 2:47
Dante Grevino
90819
answered Nov 24 at 2:47
Dante Grevino
90819
90819
Thanks, can you explain why you can reduce this to m=1?
– AColoredReptile
Nov 24 at 3:01
Yes, sure. You can prove the result for each coordinate $f_i$, with $1leq i leq m$, of $f$. Do you see it? Recall that the rows of the differential matrix of $f$ are the gradients of the coordinates $f_i$.
– Dante Grevino
Nov 24 at 3:45
add a comment |
Thanks, can you explain why you can reduce this to m=1?
– AColoredReptile
Nov 24 at 3:01
Yes, sure. You can prove the result for each coordinate $f_i$, with $1leq i leq m$, of $f$. Do you see it? Recall that the rows of the differential matrix of $f$ are the gradients of the coordinates $f_i$.
– Dante Grevino
Nov 24 at 3:45
Thanks, can you explain why you can reduce this to m=1?
– AColoredReptile
Nov 24 at 3:01
Thanks, can you explain why you can reduce this to m=1?
– AColoredReptile
Nov 24 at 3:01
Yes, sure. You can prove the result for each coordinate $f_i$, with $1leq i leq m$, of $f$. Do you see it? Recall that the rows of the differential matrix of $f$ are the gradients of the coordinates $f_i$.
– Dante Grevino
Nov 24 at 3:45
Yes, sure. You can prove the result for each coordinate $f_i$, with $1leq i leq m$, of $f$. Do you see it? Recall that the rows of the differential matrix of $f$ are the gradients of the coordinates $f_i$.
– Dante Grevino
Nov 24 at 3:45
add a comment |
We can prove the result by contradiction as follows:
If possible let $ f$ is not a constant.
Then the cardinality of $text{Image (f) or Im(f)}$ must be greater than $1$.
Let $ v in Im(f)$ and consider the following two sets:
$ A=f^{-1}{v }$ and $B=f^{-1} (Im(f) setminus {v })$.
Now it is your task to show that (use continuity of $f$):
$(i) A,B $ are non-empty sets,
$(ii) A, B $ are both open sets,
$(iii) A cup B=U$,
$(iv) A cap B=phi$.
By separation axiom $ U$ is disconnected because given that $U$ is connected.
Hence a contradiction.
This prove that $ f$ must be constant.
answered Nov 24 at 3:30
M. A. SARKAR
2,1391619
add a comment |
We can prove the result by contradiction as follows:
If possible let $ f$ is not a constant.
Then the cardinality of $text{Image (f) or Im(f)}$ must be greater than $1$.
Let $ v in Im(f)$ and consider the following two sets:
$ A=f^{-1}{v }$ and $B=f^{-1} (Im(f) setminus {v })$.
Now it is your task to show that (use continuity of $f$):
$(i) A,B $ are non-empty sets,
$(ii) A, B $ are both open sets,
$(iii) A cup B=U$,
$(iv) A cap B=phi$.
By separation axiom $ U$ is disconnected because given that $U$ is connected.
Hence a contradiction.
This prove that $ f$ must be constant.
answered Nov 24 at 3:30
M. A. SARKAR
2,1391619
add a comment |
We can prove the result by contradiction as follows:
If possible let $ f$ is not a constant.
Then the cardinality of $text{Image (f) or Im(f)}$ must be greater than $1$.
Let $ v in Im(f)$ and consider the following two sets:
$ A=f^{-1}{v }$ and $B=f^{-1} (Im(f) setminus {v })$.
Now it is your task to show that (use continuity of $f$):
$(i) A,B $ are non-empty sets,
$(ii) A, B $ are both open sets,
$(iii) A cup B=U$,
$(iv) A cap B=phi$.
By separation axiom $ U$ is disconnected because given that $U$ is connected.
Hence a contradiction.
This prove that $ f$ must be constant.
answered Nov 24 at 3:30
M. A. SARKAR
2,1391619
We can prove the result by contradiction as follows:
If possible let $ f$ is not a constant.
Then the cardinality of $text{Image (f) or Im(f)}$ must be greater than $1$.
Let $ v in Im(f)$ and consider the following two sets:
$ A=f^{-1}{v }$ and $B=f^{-1} (Im(f) setminus {v })$.
Now it is your task to show that (use continuity of $f$):
$(i) A,B $ are non-empty sets,
$(ii) A, B $ are both open sets,
$(iii) A cup B=U$,
$(iv) A cap B=phi$.
By separation axiom $ U$ is disconnected because given that $U$ is connected.
Hence a contradiction.
This prove that $ f$ must be constant.
answered Nov 24 at 3:30
M. A. SARKAR
2,1391619
answered Nov 24 at 3:30
M. A. SARKAR
2,1391619
answered Nov 24 at 3:30
M. A. SARKAR
2,1391619
answered Nov 24 at 3:30
M. A. SARKAR
2,1391619
2,1391619
add a comment |
add a comment |
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What is to stop you from simply choosing a differentiable path?
– Robert Lewis
Nov 24 at 2:15
2
The definition I have of path connected simply says there is a path for each a,b not that there is a differentiable one.
– AColoredReptile
Nov 24 at 2:19
That is true. I think though that you can show there is a differentiable path by starting with a continuous one, constructing an open cover, using compactness to find a finite open cover, an then in each element of that open cover explicitly presenting a differentiable path, and then making sure they "hook up" correctly.
– Robert Lewis
Nov 24 at 2:32
2
@AColoredReptile $U$ is open, the path can be chosen to be differentiable.
– Jacky Chong
Nov 24 at 2:39
@Jacky Chong Is there a name for this theorem or some proof of this result?
– AColoredReptile
Nov 24 at 2:42