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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enter image description here



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.






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    What’s the question?
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    – Randall
    Dec 9 '18 at 18:31
















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$begingroup$


enter image description here



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







share|cite|improve this question









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  • 1




    $begingroup$
    What’s the question?
    $endgroup$
    – Randall
    Dec 9 '18 at 18:31














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1



$begingroup$


enter image description here



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







share|cite|improve this question









$endgroup$




enter image description here



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






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asked Dec 9 '18 at 18:28









david Ddavid D

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$begingroup$
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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.






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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 $T(x_0)=T(-x_0).$






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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.






      share|cite|improve this answer









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        1












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Dec 9 '18 at 18:31









          TravisTravis

          60k767146




          60k767146























              0












              $begingroup$

              $D(x)=T(x)-T(-x)$ 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 $T(x_0)=T(-x_0).$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $D(x)=T(x)-T(-x)$ 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 $T(x_0)=T(-x_0).$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $D(x)=T(x)-T(-x)$ 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 $T(x_0)=T(-x_0).$






                  share|cite|improve this answer









                  $endgroup$



                  $D(x)=T(x)-T(-x)$ 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 $T(x_0)=T(-x_0).$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 18:56









                  John_WickJohn_Wick

                  1,486111




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