Is this conjecture about the boundary of a surface correct?












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I came up with an intuitive conjecture about boundaries of surfaces based on the idea that at a boundary point we can wrap a string across the edge, and the two halves of the string (on opposite sides of the surface) can be brought as close as we want:




Let $S subseteq mathbb{R}^n$ be a surface. A point $P in S$ is a boundary point iff there is a point $Q in S setminus {P}$ such that for all $epsilon>0$, there are functions $mathbf{f}, mathbf{g} : [0,1] to mathbb{R}^n$ such that:





  1. $mathbf{f}(0) = mathbf{g}(0)$ and $||mathbf{f}(0) - P|| < epsilon$ (string halves touch at a point close to $P$)


  2. $forall t in [0,1]: ||mathbf{f}(t)-mathbf{g}(t)|| < epsilon$ (string halves are close to each other)


  3. $forall t in [0,1]: ||mathbf{f}(t) - P|| > epsilon implies frac{mathbf{f}(t) + mathbf{g}(t)}{2} in S$ (string halves are separated by surface, except near $P$)


  4. $||mathbf{f}(1) - Q|| < epsilon$ and $||mathbf{g}(1) - Q|| < epsilon$ (string ends are close to $Q$)


  5. $(mathbf{f}([0,1]) cup mathbf{g}([0,1])) cap S = emptyset$ (string doesn't pass through the surface)



Is this conjecture correct? Does it have any useful applications? Is this a known theorem?






real-analysis general-topology manifolds-with-boundary







share|cite|improve this question











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    3












    $begingroup$


    I came up with an intuitive conjecture about boundaries of surfaces based on the idea that at a boundary point we can wrap a string across the edge, and the two halves of the string (on opposite sides of the surface) can be brought as close as we want:




    Let $S subseteq mathbb{R}^n$ be a surface. A point $P in S$ is a boundary point iff there is a point $Q in S setminus {P}$ such that for all $epsilon>0$, there are functions $mathbf{f}, mathbf{g} : [0,1] to mathbb{R}^n$ such that:





    1. $mathbf{f}(0) = mathbf{g}(0)$ and $||mathbf{f}(0) - P|| < epsilon$ (string halves touch at a point close to $P$)


    2. $forall t in [0,1]: ||mathbf{f}(t)-mathbf{g}(t)|| < epsilon$ (string halves are close to each other)


    3. $forall t in [0,1]: ||mathbf{f}(t) - P|| > epsilon implies frac{mathbf{f}(t) + mathbf{g}(t)}{2} in S$ (string halves are separated by surface, except near $P$)


    4. $||mathbf{f}(1) - Q|| < epsilon$ and $||mathbf{g}(1) - Q|| < epsilon$ (string ends are close to $Q$)


    5. $(mathbf{f}([0,1]) cup mathbf{g}([0,1])) cap S = emptyset$ (string doesn't pass through the surface)



    Is this conjecture correct? Does it have any useful applications? Is this a known theorem?






    real-analysis general-topology manifolds-with-boundary







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I came up with an intuitive conjecture about boundaries of surfaces based on the idea that at a boundary point we can wrap a string across the edge, and the two halves of the string (on opposite sides of the surface) can be brought as close as we want:




      Let $S subseteq mathbb{R}^n$ be a surface. A point $P in S$ is a boundary point iff there is a point $Q in S setminus {P}$ such that for all $epsilon>0$, there are functions $mathbf{f}, mathbf{g} : [0,1] to mathbb{R}^n$ such that:





      1. $mathbf{f}(0) = mathbf{g}(0)$ and $||mathbf{f}(0) - P|| < epsilon$ (string halves touch at a point close to $P$)


      2. $forall t in [0,1]: ||mathbf{f}(t)-mathbf{g}(t)|| < epsilon$ (string halves are close to each other)


      3. $forall t in [0,1]: ||mathbf{f}(t) - P|| > epsilon implies frac{mathbf{f}(t) + mathbf{g}(t)}{2} in S$ (string halves are separated by surface, except near $P$)


      4. $||mathbf{f}(1) - Q|| < epsilon$ and $||mathbf{g}(1) - Q|| < epsilon$ (string ends are close to $Q$)


      5. $(mathbf{f}([0,1]) cup mathbf{g}([0,1])) cap S = emptyset$ (string doesn't pass through the surface)



      Is this conjecture correct? Does it have any useful applications? Is this a known theorem?






      real-analysis general-topology manifolds-with-boundary







      share|cite|improve this question











      $endgroup$




      I came up with an intuitive conjecture about boundaries of surfaces based on the idea that at a boundary point we can wrap a string across the edge, and the two halves of the string (on opposite sides of the surface) can be brought as close as we want:




      Let $S subseteq mathbb{R}^n$ be a surface. A point $P in S$ is a boundary point iff there is a point $Q in S setminus {P}$ such that for all $epsilon>0$, there are functions $mathbf{f}, mathbf{g} : [0,1] to mathbb{R}^n$ such that:





      1. $mathbf{f}(0) = mathbf{g}(0)$ and $||mathbf{f}(0) - P|| < epsilon$ (string halves touch at a point close to $P$)


      2. $forall t in [0,1]: ||mathbf{f}(t)-mathbf{g}(t)|| < epsilon$ (string halves are close to each other)


      3. $forall t in [0,1]: ||mathbf{f}(t) - P|| > epsilon implies frac{mathbf{f}(t) + mathbf{g}(t)}{2} in S$ (string halves are separated by surface, except near $P$)


      4. $||mathbf{f}(1) - Q|| < epsilon$ and $||mathbf{g}(1) - Q|| < epsilon$ (string ends are close to $Q$)


      5. $(mathbf{f}([0,1]) cup mathbf{g}([0,1])) cap S = emptyset$ (string doesn't pass through the surface)



      Is this conjecture correct? Does it have any useful applications? Is this a known theorem?






      real-analysis general-topology manifolds-with-boundary




      real-analysis general-topology manifolds-with-boundary






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 7 '18 at 10:42







      Stefan

















      asked Dec 7 '18 at 7:09









      StefanStefan

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          $begingroup$

          I presume that by "surface" you mean a 2-dimensional submanifold, in which case this conjecture is incorrect when $n ge 4$: Your property is true for all $P in S$, not just for boundary points.



          On the other hand, I think the conjecture is true when $n=3$. I also think the conjecture would be true if, rather than assuming that $S$ is 2-dimensional, you assume instead that $S$ is of dimension $n-1$.



          A version of this property can be used to prove the theorem that if $S subset mathbb R^3$ is a 2-dimensional subsurface-with-boundary, if $S$ is connected, and if $partial S ne emptyset$, then $mathbb R^3 - S$ is path connected. But for this application you must prove the property for any $Q$ in $S$, instead of just for some $Q$. If you look up proofs of this theorem, you'll see objects similar to your two paths $bf f$, $bf g$.






          share|cite|improve this answer









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            $begingroup$

            I presume that by "surface" you mean a 2-dimensional submanifold, in which case this conjecture is incorrect when $n ge 4$: Your property is true for all $P in S$, not just for boundary points.



            On the other hand, I think the conjecture is true when $n=3$. I also think the conjecture would be true if, rather than assuming that $S$ is 2-dimensional, you assume instead that $S$ is of dimension $n-1$.



            A version of this property can be used to prove the theorem that if $S subset mathbb R^3$ is a 2-dimensional subsurface-with-boundary, if $S$ is connected, and if $partial S ne emptyset$, then $mathbb R^3 - S$ is path connected. But for this application you must prove the property for any $Q$ in $S$, instead of just for some $Q$. If you look up proofs of this theorem, you'll see objects similar to your two paths $bf f$, $bf g$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              I presume that by "surface" you mean a 2-dimensional submanifold, in which case this conjecture is incorrect when $n ge 4$: Your property is true for all $P in S$, not just for boundary points.



              On the other hand, I think the conjecture is true when $n=3$. I also think the conjecture would be true if, rather than assuming that $S$ is 2-dimensional, you assume instead that $S$ is of dimension $n-1$.



              A version of this property can be used to prove the theorem that if $S subset mathbb R^3$ is a 2-dimensional subsurface-with-boundary, if $S$ is connected, and if $partial S ne emptyset$, then $mathbb R^3 - S$ is path connected. But for this application you must prove the property for any $Q$ in $S$, instead of just for some $Q$. If you look up proofs of this theorem, you'll see objects similar to your two paths $bf f$, $bf g$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                I presume that by "surface" you mean a 2-dimensional submanifold, in which case this conjecture is incorrect when $n ge 4$: Your property is true for all $P in S$, not just for boundary points.



                On the other hand, I think the conjecture is true when $n=3$. I also think the conjecture would be true if, rather than assuming that $S$ is 2-dimensional, you assume instead that $S$ is of dimension $n-1$.



                A version of this property can be used to prove the theorem that if $S subset mathbb R^3$ is a 2-dimensional subsurface-with-boundary, if $S$ is connected, and if $partial S ne emptyset$, then $mathbb R^3 - S$ is path connected. But for this application you must prove the property for any $Q$ in $S$, instead of just for some $Q$. If you look up proofs of this theorem, you'll see objects similar to your two paths $bf f$, $bf g$.






                share|cite|improve this answer









                $endgroup$



                I presume that by "surface" you mean a 2-dimensional submanifold, in which case this conjecture is incorrect when $n ge 4$: Your property is true for all $P in S$, not just for boundary points.



                On the other hand, I think the conjecture is true when $n=3$. I also think the conjecture would be true if, rather than assuming that $S$ is 2-dimensional, you assume instead that $S$ is of dimension $n-1$.



                A version of this property can be used to prove the theorem that if $S subset mathbb R^3$ is a 2-dimensional subsurface-with-boundary, if $S$ is connected, and if $partial S ne emptyset$, then $mathbb R^3 - S$ is path connected. But for this application you must prove the property for any $Q$ in $S$, instead of just for some $Q$. If you look up proofs of this theorem, you'll see objects similar to your two paths $bf f$, $bf g$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 17:44









                Lee MosherLee Mosher

                48.8k33683




                48.8k33683






























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