Continuously chosen points on lines in $mathbbR^n$

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On each line $l subset mathbbR^n, n geqslant 2$ passing through $0$ we choose a point $a(l)$ such that $a(l)$ depends continuously on $l$. I have to prove that there exists a line with $a(l)=0$.

I show it using Borsuk-Ulam theorem. We can consider the set of lines passing through the origin in $mathbbR^n$ as sphere $mathbbS^n-1$ (I fix this ambiguity later). We assign to each point on a sphere a real number $lambda$ (point on a line passing through the origin and $x$ is $lambda x$ for some $lambda$). So we have a continuous function
$$varphi:mathbbS^n-1 to mathbbR$$ such that $varphi(-x) = -varphi (x)$, so it is odd. I take the embedding of $mathbbR$ into $mathbbR^n-1$ given by $x longmapsto (x,0,dots,0)$ which is continuous and this gives me a continuous odd function from $mathbbS^n-1 to mathbbR^n-1$. Finally, Borsuk-Ulam claims that there should be a point on a sphere with $varphi(x)=varphi(-x)$ and clearly, by oddity $varphi(x)=0$ for some $x$.
Am I correct? Thanks in advance.

geometry continuity fixed-point-theorems

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edited Jul 17 at 16:17

asked Jul 17 at 16:05

Nicholas S

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On each line $l subset mathbbR^n, n geqslant 2$ passing through $0$ we choose a point $a(l)$ such that $a(l)$ depends continuously on $l$. I have to prove that there exists a line with $a(l)=0$.

I show it using Borsuk-Ulam theorem. We can consider the set of lines passing through the origin in $mathbbR^n$ as sphere $mathbbS^n-1$ (I fix this ambiguity later). We assign to each point on a sphere a real number $lambda$ (point on a line passing through the origin and $x$ is $lambda x$ for some $lambda$). So we have a continuous function
$$varphi:mathbbS^n-1 to mathbbR$$ such that $varphi(-x) = -varphi (x)$, so it is odd. I take the embedding of $mathbbR$ into $mathbbR^n-1$ given by $x longmapsto (x,0,dots,0)$ which is continuous and this gives me a continuous odd function from $mathbbS^n-1 to mathbbR^n-1$. Finally, Borsuk-Ulam claims that there should be a point on a sphere with $varphi(x)=varphi(-x)$ and clearly, by oddity $varphi(x)=0$ for some $x$.
Am I correct? Thanks in advance.

geometry continuity fixed-point-theorems

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edited Jul 17 at 16:17

asked Jul 17 at 16:05

Nicholas S

17010

add a comment | 

up vote
0
down vote

favorite

up vote
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down vote

favorite

On each line $l subset mathbbR^n, n geqslant 2$ passing through $0$ we choose a point $a(l)$ such that $a(l)$ depends continuously on $l$. I have to prove that there exists a line with $a(l)=0$.

I show it using Borsuk-Ulam theorem. We can consider the set of lines passing through the origin in $mathbbR^n$ as sphere $mathbbS^n-1$ (I fix this ambiguity later). We assign to each point on a sphere a real number $lambda$ (point on a line passing through the origin and $x$ is $lambda x$ for some $lambda$). So we have a continuous function
$$varphi:mathbbS^n-1 to mathbbR$$ such that $varphi(-x) = -varphi (x)$, so it is odd. I take the embedding of $mathbbR$ into $mathbbR^n-1$ given by $x longmapsto (x,0,dots,0)$ which is continuous and this gives me a continuous odd function from $mathbbS^n-1 to mathbbR^n-1$. Finally, Borsuk-Ulam claims that there should be a point on a sphere with $varphi(x)=varphi(-x)$ and clearly, by oddity $varphi(x)=0$ for some $x$.
Am I correct? Thanks in advance.

geometry continuity fixed-point-theorems

share|cite|improve this question

edited Jul 17 at 16:17

asked Jul 17 at 16:05

Nicholas S

17010

On each line $l subset mathbbR^n, n geqslant 2$ passing through $0$ we choose a point $a(l)$ such that $a(l)$ depends continuously on $l$. I have to prove that there exists a line with $a(l)=0$.

I show it using Borsuk-Ulam theorem. We can consider the set of lines passing through the origin in $mathbbR^n$ as sphere $mathbbS^n-1$ (I fix this ambiguity later). We assign to each point on a sphere a real number $lambda$ (point on a line passing through the origin and $x$ is $lambda x$ for some $lambda$). So we have a continuous function
$$varphi:mathbbS^n-1 to mathbbR$$ such that $varphi(-x) = -varphi (x)$, so it is odd. I take the embedding of $mathbbR$ into $mathbbR^n-1$ given by $x longmapsto (x,0,dots,0)$ which is continuous and this gives me a continuous odd function from $mathbbS^n-1 to mathbbR^n-1$. Finally, Borsuk-Ulam claims that there should be a point on a sphere with $varphi(x)=varphi(-x)$ and clearly, by oddity $varphi(x)=0$ for some $x$.
Am I correct? Thanks in advance.

geometry continuity fixed-point-theorems

share|cite|improve this question

edited Jul 17 at 16:17

asked Jul 17 at 16:05

Nicholas S

17010

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 16:17

edited Jul 17 at 16:17

edited Jul 17 at 16:17

asked Jul 17 at 16:05

Nicholas S

17010

asked Jul 17 at 16:05

Nicholas S

17010

asked Jul 17 at 16:05

Nicholas S

17010

17010

add a comment | 

add a comment | 

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You have to make precise the meaning of "$a(l)$ depends continuously on $l$". The only reasonable explanation is that you take the set $mathbbRP^n-1$ of all lines through $0$ (i.e. of all one-dimensional subspaces) and give it the identification topology induced by the surjection $p : mathbbR^n backslash 0 to mathbbRP^n-1, p(x) = $ line through $0$ and $x$. This space is the real projective space of dimension $n-1$.

Then your claim is that for each continuous $a : mathbbRP^n-1 to mathbbR^n$ such that $a(l) in l$ for all $l$ there exists $l$ such that $a(l) = 0$.

Assume there exists a map $a$ such that $0 ne a(l) in l$ for all $l$. Define $a' : mathbbRP^n-1 to S^n-1, a'(x) = a(x)/lVert a(x) rVert$. Then $p mid_S^n-1 circ phantom . a' = id$. This is impossible, but I do not see that it is a consequence of the Borsuk-Ulam theorem. One can easily show that $p' : p mid_S^n-1 : S^n-1 to mathbbRP^n-1$ is a two-sheeted covering (between connected spaces). But $a'$ is section of $p'$ which implies that $p'$ is one-sheeted (see If a covering map has a section, is it a $1$-fold cover?). This contradiction shows that our assumption was wrong.

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answered Jul 17 at 17:17

Paul Frost

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You have to make precise the meaning of "$a(l)$ depends continuously on $l$". The only reasonable explanation is that you take the set $mathbbRP^n-1$ of all lines through $0$ (i.e. of all one-dimensional subspaces) and give it the identification topology induced by the surjection $p : mathbbR^n backslash 0 to mathbbRP^n-1, p(x) = $ line through $0$ and $x$. This space is the real projective space of dimension $n-1$.

Then your claim is that for each continuous $a : mathbbRP^n-1 to mathbbR^n$ such that $a(l) in l$ for all $l$ there exists $l$ such that $a(l) = 0$.

Assume there exists a map $a$ such that $0 ne a(l) in l$ for all $l$. Define $a' : mathbbRP^n-1 to S^n-1, a'(x) = a(x)/lVert a(x) rVert$. Then $p mid_S^n-1 circ phantom . a' = id$. This is impossible, but I do not see that it is a consequence of the Borsuk-Ulam theorem. One can easily show that $p' : p mid_S^n-1 : S^n-1 to mathbbRP^n-1$ is a two-sheeted covering (between connected spaces). But $a'$ is section of $p'$ which implies that $p'$ is one-sheeted (see If a covering map has a section, is it a $1$-fold cover?). This contradiction shows that our assumption was wrong.

share|cite|improve this answer

answered Jul 17 at 17:17

Paul Frost

3,703420

add a comment | 

up vote
0
down vote

You have to make precise the meaning of "$a(l)$ depends continuously on $l$". The only reasonable explanation is that you take the set $mathbbRP^n-1$ of all lines through $0$ (i.e. of all one-dimensional subspaces) and give it the identification topology induced by the surjection $p : mathbbR^n backslash 0 to mathbbRP^n-1, p(x) = $ line through $0$ and $x$. This space is the real projective space of dimension $n-1$.

Then your claim is that for each continuous $a : mathbbRP^n-1 to mathbbR^n$ such that $a(l) in l$ for all $l$ there exists $l$ such that $a(l) = 0$.

Assume there exists a map $a$ such that $0 ne a(l) in l$ for all $l$. Define $a' : mathbbRP^n-1 to S^n-1, a'(x) = a(x)/lVert a(x) rVert$. Then $p mid_S^n-1 circ phantom . a' = id$. This is impossible, but I do not see that it is a consequence of the Borsuk-Ulam theorem. One can easily show that $p' : p mid_S^n-1 : S^n-1 to mathbbRP^n-1$ is a two-sheeted covering (between connected spaces). But $a'$ is section of $p'$ which implies that $p'$ is one-sheeted (see If a covering map has a section, is it a $1$-fold cover?). This contradiction shows that our assumption was wrong.

share|cite|improve this answer

answered Jul 17 at 17:17

Paul Frost

3,703420

add a comment | 

up vote
0
down vote

up vote
0
down vote

You have to make precise the meaning of "$a(l)$ depends continuously on $l$". The only reasonable explanation is that you take the set $mathbbRP^n-1$ of all lines through $0$ (i.e. of all one-dimensional subspaces) and give it the identification topology induced by the surjection $p : mathbbR^n backslash 0 to mathbbRP^n-1, p(x) = $ line through $0$ and $x$. This space is the real projective space of dimension $n-1$.

Then your claim is that for each continuous $a : mathbbRP^n-1 to mathbbR^n$ such that $a(l) in l$ for all $l$ there exists $l$ such that $a(l) = 0$.

Assume there exists a map $a$ such that $0 ne a(l) in l$ for all $l$. Define $a' : mathbbRP^n-1 to S^n-1, a'(x) = a(x)/lVert a(x) rVert$. Then $p mid_S^n-1 circ phantom . a' = id$. This is impossible, but I do not see that it is a consequence of the Borsuk-Ulam theorem. One can easily show that $p' : p mid_S^n-1 : S^n-1 to mathbbRP^n-1$ is a two-sheeted covering (between connected spaces). But $a'$ is section of $p'$ which implies that $p'$ is one-sheeted (see If a covering map has a section, is it a $1$-fold cover?). This contradiction shows that our assumption was wrong.

share|cite|improve this answer

answered Jul 17 at 17:17

Paul Frost

3,703420

You have to make precise the meaning of "$a(l)$ depends continuously on $l$". The only reasonable explanation is that you take the set $mathbbRP^n-1$ of all lines through $0$ (i.e. of all one-dimensional subspaces) and give it the identification topology induced by the surjection $p : mathbbR^n backslash 0 to mathbbRP^n-1, p(x) = $ line through $0$ and $x$. This space is the real projective space of dimension $n-1$.

Then your claim is that for each continuous $a : mathbbRP^n-1 to mathbbR^n$ such that $a(l) in l$ for all $l$ there exists $l$ such that $a(l) = 0$.

Assume there exists a map $a$ such that $0 ne a(l) in l$ for all $l$. Define $a' : mathbbRP^n-1 to S^n-1, a'(x) = a(x)/lVert a(x) rVert$. Then $p mid_S^n-1 circ phantom . a' = id$. This is impossible, but I do not see that it is a consequence of the Borsuk-Ulam theorem. One can easily show that $p' : p mid_S^n-1 : S^n-1 to mathbbRP^n-1$ is a two-sheeted covering (between connected spaces). But $a'$ is section of $p'$ which implies that $p'$ is one-sheeted (see If a covering map has a section, is it a $1$-fold cover?). This contradiction shows that our assumption was wrong.

share|cite|improve this answer

answered Jul 17 at 17:17

Paul Frost

3,703420

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 17:17

Paul Frost

3,703420

answered Jul 17 at 17:17

Paul Frost

3,703420

answered Jul 17 at 17:17

Paul Frost

3,703420

3,703420

add a comment | 

add a comment | 

 

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