Problem with simply connected 3D domains

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is a path-connected domain where one can continuously shrink any simple closed curve into a point while remaining in the domain "



I thought I understood it and my understanding went well with the bellow 2D domains:



2d domain



I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)



But now when I see the bellow 3D domains, I get confused.



enter image description here



I don't understand why the $2^nd$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?



Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.







share|cite|improve this question
























    up vote
    1
    down vote

    favorite












    I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is a path-connected domain where one can continuously shrink any simple closed curve into a point while remaining in the domain "



    I thought I understood it and my understanding went well with the bellow 2D domains:



    2d domain



    I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)



    But now when I see the bellow 3D domains, I get confused.



    enter image description here



    I don't understand why the $2^nd$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?



    Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is a path-connected domain where one can continuously shrink any simple closed curve into a point while remaining in the domain "



      I thought I understood it and my understanding went well with the bellow 2D domains:



      2d domain



      I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)



      But now when I see the bellow 3D domains, I get confused.



      enter image description here



      I don't understand why the $2^nd$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?



      Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.







      share|cite|improve this question












      I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is a path-connected domain where one can continuously shrink any simple closed curve into a point while remaining in the domain "



      I thought I understood it and my understanding went well with the bellow 2D domains:



      2d domain



      I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)



      But now when I see the bellow 3D domains, I get confused.



      enter image description here



      I don't understand why the $2^nd$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?



      Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 21 at 14:08









      paulplusx

      15812




      15812




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.






          share|cite|improve this answer




















          • Wow! I didn't know that. This makes perfect sense. Thanks.
            – paulplusx
            Aug 21 at 15:15










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );








           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2889914%2fproblem-with-simply-connected-3d-domains%23new-answer', 'question_page');

          );

          Post as a guest






























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.






          share|cite|improve this answer




















          • Wow! I didn't know that. This makes perfect sense. Thanks.
            – paulplusx
            Aug 21 at 15:15














          up vote
          1
          down vote



          accepted










          I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.






          share|cite|improve this answer




















          • Wow! I didn't know that. This makes perfect sense. Thanks.
            – paulplusx
            Aug 21 at 15:15












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.






          share|cite|improve this answer












          I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 21 at 14:16









          Kusma

          2,373213




          2,373213











          • Wow! I didn't know that. This makes perfect sense. Thanks.
            – paulplusx
            Aug 21 at 15:15
















          • Wow! I didn't know that. This makes perfect sense. Thanks.
            – paulplusx
            Aug 21 at 15:15















          Wow! I didn't know that. This makes perfect sense. Thanks.
          – paulplusx
          Aug 21 at 15:15




          Wow! I didn't know that. This makes perfect sense. Thanks.
          – paulplusx
          Aug 21 at 15:15












           

          draft saved


          draft discarded


























           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2889914%2fproblem-with-simply-connected-3d-domains%23new-answer', 'question_page');

          );

          Post as a guest













































































          這個網誌中的熱門文章

          How to combine Bézier curves to a surface?

          Mutual Information Always Non-negative

          Why am i infinitely getting the same tweet with the Twitter Search API?