For any point x on the Earth (or any sphere really) the antipode, often written as −x, is the point exactly...
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I know this is like an easier version proof of Borsuk–Ulam theorem. However, the proof to Borsuk–Ulam theorem is a little bit difficult for me to follow.
real-analysis
asked Dec 9 '18 at 18:28
david Ddavid D
875
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add a comment |
$begingroup$
I know this is like an easier version proof of Borsuk–Ulam theorem. However, the proof to Borsuk–Ulam theorem is a little bit difficult for me to follow.
real-analysis
asked Dec 9 '18 at 18:28
david Ddavid D
875
$endgroup$
1
$begingroup$
What’s the question?
$endgroup$
– Randall
Dec 9 '18 at 18:31
add a comment |
$begingroup$
I know this is like an easier version proof of Borsuk–Ulam theorem. However, the proof to Borsuk–Ulam theorem is a little bit difficult for me to follow.
real-analysis
asked Dec 9 '18 at 18:28
david Ddavid D
875
$endgroup$
I know this is like an easier version proof of Borsuk–Ulam theorem. However, the proof to Borsuk–Ulam theorem is a little bit difficult for me to follow.
real-analysis
real-analysis
asked Dec 9 '18 at 18:28
david Ddavid D
875
asked Dec 9 '18 at 18:28
david Ddavid D
875
asked Dec 9 '18 at 18:28
david Ddavid D
875
asked Dec 9 '18 at 18:28
david Ddavid D
875
asked Dec 9 '18 at 18:28
david Ddavid D
875
875
1
$begingroup$
What’s the question?
$endgroup$
– Randall
Dec 9 '18 at 18:31
add a comment |
1
$begingroup$
What’s the question?
$endgroup$
– Randall
Dec 9 '18 at 18:31
1
1
$begingroup$
What’s the question?
$endgroup$
– Randall
Dec 9 '18 at 18:31
$begingroup$
What’s the question?
$endgroup$
– Randall
Dec 9 '18 at 18:31
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2 Answers
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Hint If $e$ is the equator, consider the function $$D : e to Bbb R, qquad D(x) := T(x) - T(-x) .$$ By definition, it suffices to show that $D$ has a zero.
answered Dec 9 '18 at 18:31
TravisTravis
60k767146
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$D(x)=T(x)-T(-x)$ is a continuous function on $xin E text{ (equator)}. $ If $exists xin E$ such that $D(x)=0$ we are done. If not then $D(x)neq 0$. Wlog, $D(x)>0.$ Then $D(-x)<0$ gives that there is a point $x_0$ on $E$ such that $D(x_0)=0$ and hence $T(x_0)=T(-x_0).$
answered Dec 9 '18 at 18:56
John_WickJohn_Wick
1,486111
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2 Answers
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2 Answers
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$begingroup$
Hint If $e$ is the equator, consider the function $$D : e to Bbb R, qquad D(x) := T(x) - T(-x) .$$ By definition, it suffices to show that $D$ has a zero.
answered Dec 9 '18 at 18:31
TravisTravis
60k767146
$endgroup$
add a comment |
$begingroup$
Hint If $e$ is the equator, consider the function $$D : e to Bbb R, qquad D(x) := T(x) - T(-x) .$$ By definition, it suffices to show that $D$ has a zero.
answered Dec 9 '18 at 18:31
TravisTravis
60k767146
$endgroup$
add a comment |
$begingroup$
Hint If $e$ is the equator, consider the function $$D : e to Bbb R, qquad D(x) := T(x) - T(-x) .$$ By definition, it suffices to show that $D$ has a zero.
answered Dec 9 '18 at 18:31
TravisTravis
60k767146
$endgroup$
Hint If $e$ is the equator, consider the function $$D : e to Bbb R, qquad D(x) := T(x) - T(-x) .$$ By definition, it suffices to show that $D$ has a zero.
answered Dec 9 '18 at 18:31
TravisTravis
60k767146
answered Dec 9 '18 at 18:31
TravisTravis
60k767146
answered Dec 9 '18 at 18:31
TravisTravis
60k767146
answered Dec 9 '18 at 18:31
TravisTravis
60k767146
60k767146
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$begingroup$
$D(x)=T(x)-T(-x)$ is a continuous function on $xin E text{ (equator)}. $ If $exists xin E$ such that $D(x)=0$ we are done. If not then $D(x)neq 0$. Wlog, $D(x)>0.$ Then $D(-x)<0$ gives that there is a point $x_0$ on $E$ such that $D(x_0)=0$ and hence $T(x_0)=T(-x_0).$
answered Dec 9 '18 at 18:56
John_WickJohn_Wick
1,486111
$endgroup$
add a comment |
$begingroup$
$D(x)=T(x)-T(-x)$ is a continuous function on $xin E text{ (equator)}. $ If $exists xin E$ such that $D(x)=0$ we are done. If not then $D(x)neq 0$. Wlog, $D(x)>0.$ Then $D(-x)<0$ gives that there is a point $x_0$ on $E$ such that $D(x_0)=0$ and hence $T(x_0)=T(-x_0).$
answered Dec 9 '18 at 18:56
John_WickJohn_Wick
1,486111
$endgroup$
add a comment |
$begingroup$
$D(x)=T(x)-T(-x)$ is a continuous function on $xin E text{ (equator)}. $ If $exists xin E$ such that $D(x)=0$ we are done. If not then $D(x)neq 0$. Wlog, $D(x)>0.$ Then $D(-x)<0$ gives that there is a point $x_0$ on $E$ such that $D(x_0)=0$ and hence $T(x_0)=T(-x_0).$
answered Dec 9 '18 at 18:56
John_WickJohn_Wick
1,486111
$endgroup$
$D(x)=T(x)-T(-x)$ is a continuous function on $xin E text{ (equator)}. $ If $exists xin E$ such that $D(x)=0$ we are done. If not then $D(x)neq 0$. Wlog, $D(x)>0.$ Then $D(-x)<0$ gives that there is a point $x_0$ on $E$ such that $D(x_0)=0$ and hence $T(x_0)=T(-x_0).$
answered Dec 9 '18 at 18:56
John_WickJohn_Wick
1,486111
answered Dec 9 '18 at 18:56
John_WickJohn_Wick
1,486111
answered Dec 9 '18 at 18:56
John_WickJohn_Wick
1,486111
answered Dec 9 '18 at 18:56
John_WickJohn_Wick
1,486111
1,486111
add a comment |
add a comment |
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What’s the question?
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– Randall
Dec 9 '18 at 18:31